5
Integration
The Quabbin Reservoir is a huge man-made lake in Massachusetts. Four complete towns were relocated when the reservoir was
constructed in the 1930s and when full it holds approximately 412 billion gallons of water. Calculus can be used to estimate the
amount of water that flows into, and out of, the reservoir during a certain time period. Note that there are lots of AP Calculus
problems that involve water flowing into and out of pipes, tanks, pails, urns, etc.
Contents
5.1 Antiderivatives
5.2 Riemann Sums
5.3 The Definite Integral
5.4 The Fundamental Theorem of
Calculus
5.5 The Indefinite Integral
5.6 The Method of Substitution
THE TANGENT AND VELOCITY PROBLEMS were used to motivate and introduce
the concept of the derivative, which is thes central idea in differential calculus. In a
similar manner, the area and distance problems can be used to develop the idea of a
definite integral, which is the basic concept of integral calculus.
And, there is a very important connection between integral calculus and differential calculus. The most important idea in the course, the Fundamental Theorem of Calculus,
connects the integral to the derivative. This theorem will also help simplify the solution
to many problems.
1
2 CHAPTER 5 Integration
5.1 Antiderivatives
Learning
Objectives
Essential
Knowledge
LO 3.1A
EK 3.1A1
EK 3.1A2
EK 3.5A1
EK 2.2A1
EK 2.2A2
LO 3.5A
LO 2.2A
Review
• Differentiation formulas.
Preview
• Find a function whose derivative is f .
In the previous chapters, we were given a function f and asked to find the derivative
f ′ . However, often we need to work backward. That is, given the derivative, or rate of
change, find the original function. For example, a physicist may use the velocity of a
particle to find its position at a certain time. An engineer who can measure the variable
rate at which water is leaking from a tank can find the total amount leaked over a certain
time period. A biologist who knows the rate at which a bacteria population is increasing
can determine the size of the population at some future time. In each case, the problem is
to find a function F whose derivative is a known function f . If such a function F exists,
it is called an antidervative of f .
Definition 1
A function F is called an antiderivative of f on an interval I if F ′ (x) = f (x) for all x
in I.
For example, let f (x) = x2 . It’s pretty easy to find an antiderivative of f if we think about
the Power Rule.
1
If F(x) = x3 , then F ′ (x) = 13 · 3 · x2 = x2 = f (x).
3
1
The function G(x) = x3 + 28 is also an antiderivative of f : G′ (x) = x2 .
3
1
In fact, any function of the form H(x) = x3 + C, where C is a constant, is an antideriva3
tive of f . This raises an important question. Are there any other antiderivatives?
In Chapter 4 we used the Mean Value Theorem to prove that if two functions have the
same derivative on an interval, then they must differ by a constant. Therefore, if F and
G are any two antiderivatives of f , then
F ′ (x) = f (x) = G′ (x)
and G(x) − F(x) = C, where C is a constant. We can rewrite this expression as G(x) =
F(x) + C, which leads to the following result.
Theorem 1
If F is an antiderivative of f on an interval I, then the most general antiderivative of
f on I is
F(x) + C
where C is an arbitrary constant.
1
The most general antiderivative of f (x) = x2 is the function F(x) = x3 + C. By as3
signing specific values to the constant C, we obtain a family of functions whose graphs
are vertical translations of one another (see Figure 5.1). This makes sense because each
graph must have the same slope (derivative) at any given value of x.
SECTION 5.1 Antiderivatives
3
3
2
1
-3
-2
-1
1
2
-1
-2
F IGURE 5.1
Graphs of several members of
the family of antiderivatives of
f (x) = x2 .
-3
-4
Example 1 Find Some Antiderivatives
Find the most general antiderivative of each of the following functions.
(a) f (x) = sin x
(b) f (x) =
1
x
(c) f (x) = xn , n 6= −1
SOLUTION
(a) If F(x) = − cos x, then F ′ (x) = sin x.
F(x) = cos x is an antiderivative of sin x.
The most general antiderivative is G(x) = − cosx + C.
(b) In Chapter 3 we learned
1
d
(ln x) =
dx
x
Therefore, on the interval (0, ∞), the general antiderivative of
1
is ln x + C.
x
d
1
(ln |x|) = for all x 6= 0.
dx
x
1
The most general antiderivative of f (x) = is ln |x| + C on any interval that does
x
not contain 0.
We also learned that
In particular, this is true on each of the intervals (−∞, 0) and (0, ∞).
So, the general antiderivative of f can be written as
(
ln x + C1
if x > 0
F(x) =
ln(−x) + C2 if x < 0
(c) Think about the Power Rule to discover an antiderivative of xn .
d xn+1
xn
If n 6= −1, then
= xn
= (n + 1) ·
dx n + 1
n+1
xn+1
+C
n+1
This is valid for n ≥ 0 because f (x) = xn is defined on an interval. If n is negative
(but n 6= −1), it is valid on any interval that does not contain 0.
Therefore, the most general antiderivative of f (x) = xn is F(x) =
As Example 1 suggests, every differentiation formula leads to an antidifferentiation formula. Table 5.1 lists some specific antiderivatives. Each formula is true because the
derivative of the function in the right column appears in the left column. In particular,
4 CHAPTER 5 Integration
the first formula says that the antiderivative of a constant times a function is the constant
times the antiderivative of the function. The second formula says that the antiderivative
of a sum is the sum or the antiderivatives. The following notation is used: F ′ = f , G′ = g.
TABLE 5.1
Table of Antidifferentiation
Formulas. To obtain the most
general antiderivative from a
particular one, add a constant (or
constants) as in Example 1.
Function
c f (x)
Particular antiderivative
cF(x)
Function
sin x
f (x) + g(x)
F(x) + G(x)
cos x
xn+1
xn (n 6= −1)
n+1
1
x
ln |x|
ex
ex
bx
bx
ln b
Particular antiderivative
− cos x
sin x
sec2 x
tan x
sec x tan x
sec x
1
sin−1 x
√
1 − x2
1
1 + x2
tan−1 x
A Closer Look
(1) There are no formulas for the antiderivative of products or quotients.
(2) The process of calculating an antiderivative F(x) from a given f (x) is called antidifferentiation.
Example 2 Use The Table of Antidifferentiation Formulas
√
2x5 − x
′
Find all functions g such that g (x) = 4 sin x +
x
SOLUTION
Rewrite the derivative.
g′ (x) = 4 sin x +
√
2x5
x
1
−
= 4 sin x + 2x4 − √ = 4 sin x + 2x4 − x−1/2
x
x
x
Use the formulas in Table 5.1 and Theorem 1.
g(x) = 4(− cosx) + 2
x5 x1/2
− 1 +C
5
2
√
2
= −4 cosx + x5 − 2 x + C
5
In applications of calculus, it is very common to have a situation as in Example 2, where
we need to find a function given knowledge about its derivative. An equation involving the derivatives of a function is called a differential equation (DE). We will study
these in greater detail in Chapter 8. For now, we can solve some elementary differential
equations. The general solution of a differential equation involves an arbitrary constant
(or constants). However, there may be some extra conditions given that will allow us to
determine the constants and therefore uniquely specify the solution.
Example 3 Solve a Differential Equation
Find f if f ′ (x) = ex +
20
and f (0) = −2.
1 + x2
SOLUTION
Find the general antiderivative.
f (x) = ex + 20 tan−1 x + C
To determine the value of C, use the given information, f (0) = −2.
Table 5.1 formulas.
SECTION 5.1 Antiderivatives
f (0) = e0 + 20 tan−1 0 + C = −2
5
Let x = 0.
C = −2 − 1 = −3
Solve for C.
−1
x
The particular solution is f (x) = e + 20 tan
Example 4 A DE Involving f ′′
x−3
Find f if f ′′ (x) = 12x2 + 6x − 4, f (0) = 4, and f (1) = 1.
SOLUTION
Find the general antiderivative of f ′′ (x).
f ′ (x) = 12
x2
x3
+ 6 − 4x + C
3
2
Table 5.1 formulas.
= 4x3 + 3x2 − 4x + C
Simplify.
Use the antidifferentiation rules again.
f (x) = 4
x4
x3
x2
+ 3 − 4 + Cx + D
4
3
2
= x4 + x3 − 2x2 + Cx + D
Use the initial conditions to determine C and D: f (0) = 4, f (1) = 1.
f (0) = 0 + D = 4 =⇒ D = 4
f (1) = 1 + 1 − 2 + C+ 4 = 1 =⇒ C = −3
Therefore, f (x) = x4 + x3 − 2x2 − 3x + 4
If we are given the graph of a function f , it seems reasonable that we should be able to
sketch the graph of an antiderivative F. Suppose, for example, that we are given that
F(0) = 1. Then we have a place to start the graph of F, at the point (0, 1). The direction
in which we draw the graph over each interval is given by the derivative F ′ (x) = f (x).
In the next example we use the principles of this chapter to draw the graph of F even
when we do not have a formula for f . This occurs, for example, when f (x) is determined
by experiential data.
Example 5 Sketch an Antiderivative
The graph of a function f is given in Figure 5.2. Make a rough sketch of an antiderivative
F, given that F(0) = 2.
SOLUTION
We are guided by the central fact that the slope of y = F(x) is f (x).
3
3
2
2
1
1
2
3
4
5
1
-1
-2
1
-3
2
3
-1
F IGURE 5.2
Graph of the derivative, f .
F IGURE 5.3
Graph of the antiderivative, F.
Start at the point (0, 2).
Draw F initially decreasing since f (x) is negative when 0 < x < 1.
4
5
6 CHAPTER 5 Integration
Notice that f (1) = f (3) = 0. F has horizontal tangents when x = 1 and x = 3.
For 1 < x < 3, f (x) is positive, so F is increasing.
F has a local minimum when x = 1 and a local maximum when x = 3.
For x > 3, f (x) is negative, so F is decreasing on (3, ∞).
Since f (x) → 0 as x → ∞, the graph of F becomes flatter as x → ∞.
Notice that F ′′ (x) = f ′ (x) changes from positive to negative at x = 2 and from negative
to positive at x = 4, so F has inflection points when x = 2 and x = 4.
The graph of the antiderivative is given in Figure 5.3.
Rectilinear Motion
Antidifferentiation is very useful in analyzing the motion of an object moving in a straight
line. Recall that if the position of the object at time t is given by s(t), then the velocity
function is v(t) = s′ (t). This means that the position function is an antiderivative of the
velocity function. Similarly, the acceleration function is a(t) = v′ (t), so the velocity function is an antiderivative of the acceleration. If the acceleration and the initial values s(0)
and v(0) are known then the position function can be determined by antidifferentiation,
twice.
Example 6 Particle Motion
A particle moves in a straight line so that its acceleration at time t is given by a(t) =
6t + 4. Its initial velocity is v(0) = −6 cm/s and its initial position is s(0) = 9 cm. Find
its position function s(t).
SOLUTION
Find the general antiderivative of a.
t2
+ 4t + C = 3t 2 + 4t + C
2
Use v(0) = −6
v(t) = 6
v(0) = 0 + 0 + C + −6 =⇒ C = −6 =⇒ v(t) = 3t 2 + 4t − 6
Find the general antiderivative of v.
t3
t2
+ 4 − 6t + D = t 3 + 2t 2 − 6t + D
3
2
Use s(0) = 9.
s(t) = 3
s(0) = 0 + 0 − 0 + D = 9 =⇒ D = 9
The position function is s(t) = t 3 + 2t 2 − 6t + 9.
Acceleration Due To Gravity
An object near the surface of the earth is subject to a gravitational force that produces
a downward acceleration denoted by g. For motion close to the ground we may assume
that g is constant and its value is approximately 9.8 m/s2 (or 32 ft/s2 ).
Example 7 What Goes Up, Must Come Down
A ball is thrown upward with a velocity of 48 ft/s from the edge of a cliff 432 feet above
the ground. Find its height above the ground t seconds later. When does it reach its
maximum height? When does it hit the ground?
SOLUTION
The motion of the ball is vertical and we generally choose the positive direction to be
upward.
SECTION 5.1 Antiderivatives
7
Let s(t) be the distance above the ground.
The only force acting on the ball is gravity, pulling it downward.
dv
= −32
dt
Find the general antiderivative of a. v(t) = −32t + C
Therefore, a(t) =
Use the given information, v(0) = 48, to determine C.
v(0) = 0 + C = 48 =⇒ C = 48
Therefore, the velocity function is v(t) = −32t + 48
The maximum height is reached when v(t) = 0.
v(t) = 0 =⇒ −32t + 48 = 0 =⇒ t =
48
= 1.5 seconds
32
Find the general antiderivative of v.
t2
+ 48t + D = −16t 2 + 48t + D
2
Use the given information, s(0) = 432, to determine D.
s(t) = −32
s(0) = 0 + 0 + D = 432 =⇒ D = 432
Therefore, the position function is s(t) = −16t 2 + 48t + 432
The position function is valid until the ball hits the ground.
This happens when s(t) = 0.
s(t) = 0 =⇒ −16t 2 + 48t + 432 = 0 =⇒ t 2 − 3t − 27 = 0
Use technology to solve this equation.
As shown in Figure 5.4, the ball hits the ground at t = 6.908 seconds.
Figure 5.5 shows a graph of the position function for the ball. The graph confirms the
solutions in this problem. The ball reaches its maximum height after 1.5 seconds and hits
the ground at approximately 6.908 seconds.
400
F IGURE 5.4
Use the numeric solver to find the
value of t when the ball hits the
ground.
300
200
100
F IGURE 5.5
Graph of the position function.
0
0
1
2
3
4
5
6
7
8 CHAPTER 5 Integration
5.1 EXERCISES
38. f ′ (t) = t +
Concepts and Vocabulary
1. True or False If F and G are two antiderivatives of a funciton
f , then F − G = C, where C is a constant.
2. True or False If f ′′ (x) = kx4 and the coefficient of x6 in f (x) is
1, then k = 30.
3. True or False If
f ′′ (x) = g′′ (x),
1
,
t3
t > 0,
f (1) = 6
39. f ′ (x) = 5x2/3 ,
f (8) = 21
x+1
40. f (x) = √ ,
x
f (1) = 5
41. f ′ (t) = sect(sect + tant),
then f (x) = g(x) +C.
4. If F is a particular antiderivative of f , then an antiderivative of
c · f (x) is c · F(x).
3
42. f ′ (t) = 3t − ,
t
−
f (1) = 2,
Practice
43. f ′′ (x) = −2 + 12x − 12x2 ,
Find the most general antiderivative of each function. Check your
answer by differentiation.
45. f ′′ (θ ) = sin θ + cos θ ,
44. f ′′ (x) = 8x3 + 5,
1
,
t2
6. f (x) = x2 − 3x + 2
46. f ′′ (t) = t 2 +
2
7. 2x3 − x2 + 5x
3
8. f (x) = 6x5 − 8x4 − 9x2
47. f ′′ (x) = 4 + 6x + 24x2 ,
10. f (x) = (x − 5)2
9. f (x) = x(12x + 8)
11. f (x) = 7x2/5 + 8x−4/5
√
13. f (x) = 2
√
√
15. f (x) = 3 x − 2 3 x
17. f (x) =
19. g(t) =
1 2
−
5 x
18. f (t) =
1 + t + t2
22. g(v) = 2 cos v − √
2x4 + 4x3 − x
x3
,
3
1 − v2
3
24. f (x) = 1 + 2 sin x + √
x
x>0
2x2 + 5
x2 + 1
Find the antiderivative F of f that satisfies the given condition.
Check your answer by comparing the graphs of f and F.
27. f (x) = 5x4 − 2x5 ,
28. f (x) = 4 −
F(0) = 4
3
,
1 + x2
F(1) = 0
= −1
f ′ (1) = 8
f ′ (0) = 4
f (2) = 3,
f ′ (1) = 2
x > 0,
f (1) = 0,
f (0) = 1,
AP 52. If f ′ (x) = cos x and f
(A) − sin x + 1
f (0) = 3,
f (2) = 0
f ′ (0) = 2,
f ′′ (0) = 3
π = 0, then f (x) =
2
(B) sin x − 1
(C) sin x − x
(D) sin x + x
1
AP 53. If f ′ (x) = 2x − and f (−1) = 1, then f (x) =
(A) x2 + 1
x
(B) x2 + ln x
(C) x2 − ln |x| + 1
(D) x2 − ln |x|
AP 54. If f ′′ (x) = 6x, then which of the following cannot be f (x)?
x
(D) x3 + 6x
2
1
AP 55. If F is a particular antiderivative of f (x) = 2 such that
x
F(1) = 2, then F(x) =
1
1
(A) ln x2 + 2
(B) − + 3
(C) + 1
(D) ln |x| + 2
x
x
(A) x3 + 1
(B) x3 − x
(C) 2x3 +
56. Suppose the graph of f passes through the point (2, 5) and that
the slope of the tangent line to the graph of f at the point
(x, f (x)) is 3 − 4x. Find f (1).
57. Find a function f such that f ′ (x) = x3 and the graph of the line
defined by x + y = 0 is tangent to the graph of f .
Find f .
29. f ′′ (x) = 20x3 − 12x2 + 6x
The graph of a function f is shown in each figure. Which graph is an
antiderivative of f ? Explain your reasoning.
30. f ′′ (x) = x6 − 4x4 + x + 1
33. f ′′′ (t) = 12 + sin t
√
35. f ′ (x) = 1 + 3 x, f (4) = 25
36. f ′ (x) = 5x4 − 3x2 + 4,
4
,
1 + t2
1
x2
√
34. f ′′′ (t) = t − 2 cos t
32. f ′′ (x) =
31. f ′′ (x) = 2x + 3ex
37. f ′ (t) =
1
,
x2
51. f ′′′ (x) = cos x,
3t 4 − t 3 + 6t 2
t4
4
f (1) = 10
π 48. f ′′ (x) = ex − 2 sin x, f (0) = 3, f
=0
2
√
49. f ′′ (t) = 3 t − cost, f (0) = 2, f (1) = 2
50. f ′′ (x) =
20. f (θ ) = sec θ tan tan θ − 2eθ
√
t
23. f (x) = 2x + x2
26. f (x) =
2−1
14. f (x) = e2
√
√
3
16. f (x) = x2 + x x
21. h(θ ) = 2 sin θ − sec2 θ
25. f (x) =
√
12. f (x) = x3.4 − 2x
π f ′ (0) = 12
f (0) = 4,
f (0) = 3,
t > 0,
f
f (−1) = 1
f (1) = 0,
5. f (x) = 4x + 7
π
π
<t < ,
2
2
f (−1) = 2
f (1) = 0
58.
SECTION 5.1 Antiderivatives
9
AP 63. The graph of a function f is shown in the figure below.
59.
2
1
1
2
3
4
-1
-2
-3
60. The graph of the function f is shown in the figure below.
Which of the following could be the graph of it antiderivative,
F, given that F(0) = 1?
(A)
2
1
1
2
3
4
1
2
3
4
1
2
3
4
1
2
3
4
1
2
3
4
-1
-2
Make a rough sketch of an antiderivative F, given that
F(0) = 1.
-3
(B)
61. The graph of the velocity function of a particle is shown in the
figure below.
2
1
-1
-2
-3
(C)
2
1
Sketch a graph of a corresponding position function.
62. The graph of f ′ is shown in the figure below.
-1
-2
-3
2
(D)
2
1
1
1
2
3
-1
-1
-2
-3
-2
Sketch the graph of f if f is continuous and f (0) = −1.
√
64. Let f (x) = 2x − 3 x.
(a) Sketch the graph of f .
(b) Use the graph in part (a) to sketch a rough graph of the
antiderivative F such that F(0) = 1.
(c) Use the antidifferentiation formulas in this section to find
an expression for F(x).
(d) Graph F using the expression in part (c). Compare this
with your sketch in part (b).
10 CHAPTER 5 Integration
Draw a graph of f and use it to mae a rough sketch of the
antiderivative that passes through the origin.
sin x
, −2π ≤ x ≤ 2π
1 + x2
√
66. f (x) = x4 − 2x2 + 2 − 2, −3 ≤ x ≤ 3
65. f (x) =
A particle is moving along a straight line with the given information.
Find the position of the particle.
67. v(t) = sint − cos t, s(0) = 0
√
68. v(t) = t 2 − 3 t, s(4) = 8
69. a(t) = 2t + 1,
s(0) = 3,
70. a(t) = 3 cost − 2 sint,
71. a(t) = 10 sint + 3 cos t,
72.
a(t) = t 2 − 4t + 6,
78. If a diver of mass m stands at the end of a diving board with
length L and linear density ρ , then the board takes on the shape
of the graph of y = f (x), where
1
EIy′′ = mg(L − x) + ρ g(L − x)2
2
E and I are positive constants that depend on the material of the
board and g (< 0) is the acceleration due to gravity.
(a) Find an expression for the y.
(b) Use f (L) to estimate the distance below the horizontal at
the end of the board.
v(0) = −2
s(0) = 0,
s(0) = 0,
s(0) = 0,
v(0) = 4
2(2π ) = 12
s(1) = 20
73. A stone is dropped from the upper observation deck (the Space
Deck) of the CN Tower, 450 m above the ground.
(a) Find the distance of the stone above the ground at time t.
(b) How long does it take the stone to reach the ground?
(c) With what velocity does it strike the ground?
(d) If the stone is thrown downward with a velocity of 5 m/s,
how long does it take to reach the ground?
AP 74. Oil is leaking from a tank at a rate of r(t) barrels per hour. Let
A(t) be the number of barrels of oil that have leaked out after t
hours, and suppose r′ (t), the rate of change of r(t), is given by
1 t2
r′ (t) = 2 + . It is also know that r(1) = 3 and A(1) = 4.
4
t
(a) Write an equation for r(t), the rate oil is leaking from the
tank at time t. Using correct units, find r(2).
(b) Write an equation for A(t), the total amount of oil that has
leaked after t hours.
(c) At time t = 2, is the rate of leakage increasing or
decreasing? Justify your answer.
75. Show that for motion in a straight line with constant
acceleration a, initial velocity v0 , and initial position s0 , the
position at time t is
1
s = at 2 + v0 t + s0
2
76. An object is projected upward with initial velocity v0 meters
per second from a point s0 meters above the ground. Show that
[v(t)]2 = v20 − 19.6[s(t) − s0 ]
77. Two balls are thrown upward from the edge of a cliff as in
Example 7. The first is thrown with a velocity of 48 ft/s and the
other is thrown a second later with a speed of 24 ft/s. Do the
balls ever pass each other?
79. A company estimates that the marginal cost (in dollars per
item) of producing x items is 1.92 − 0.002x. If the cost of
producing one item is $562, find the cost of producing 100
items.
80. The linear density of a rod of length 1 m is given by
1
ρ (x) = √ , in grams per centimeter, where x is measured in
x
centimeters from one end of the rod. Find the mass of the rod.
81. Since raindrops grow as they fall, their surface area increases
and therefore the resistance to their falling increases. A
raindrop has an initial downward velocity of 10 m/s and its
downward acceleration is
(
9 − 0.9t if 0 ≤ t ≤ 10
a(t) =
0
if t > 10
If the raindrop is initially 500 m above the ground, how long
does it take to fall?
82. A car is traveling at 50 mi/h when the brakes are fully applied,
producing a constant deceleration of 22 ft/s2 . What is the
distance traveled before the care comes to a complete stop?
83. What constant acceleration is required to increase the speed of
a car from 30 mi/h to 50 mi/h in 5 seconds?
84. A car braked with a constant deceleration of 16 ft/s2 , producing
skid marks measuring 200 feet before coming to a stop. How
fast was the car traveling when the brakes were first applied?
85. A car is traveling at 100 km/h when the driver sees an accident
80 m ahead and slams on the brakes. What constant
deceleration is required to stop the car in time to avoid the
pileup?
SECTION 5.1 Antiderivatives
86. A model rocket is fired vertically upward from rest. Its
acceleration for the first three seconds is a(t) = 60t, at which
time the fule is exhausted and it become a freely falling body.
Fourteen seconds later, the rocket’s parachute opens, and the
(downward) velocity slows linearly to −18 ft/s in 5 seconds.
The rocket then floats to the ground at that rate.
(a) Determine the position function s and the velocity
function v (for all times t). Sketch the graphs of s and v.
(b) At what time does the rocket reach its maximum height,
and what is that height?
(c) At what time does the rocket land?
11
87. A high-speed bullet train accelerates and decelerates at the rate
of 4 ft/s2 . Its maximum cruising speed is 90 mi/h.
(a) What is the maximum distance the train can travel if it
accelerates from rest until it reaches its cruising speed and
then runs at that speed for 15 minutes?
(b) Suppose that the train starts from rest and must come to a
complete stop in 15 minutes. What is the maximum
distance it can travel under these conditions?
(c) Find the minimum time that the train takes to travel
between two consecutive stations that are 45 miles apart.
(d) The trip from one station to the next takes 37.5 minutes.
How far apart are the stations?
12 CHAPTER 5 Integration
5.2 Riemann Sums
Learning
Objectives
Essential
Knowledge
LO 3.2B
EK 3.2B1
EK 3.2B2
Review
• The area problem.
• Summation notation; common summation formulas.
Preview
• Approximate the area under a curve.
• Approximate distance traveled.
• Riemann sums.
One of the most important unifying problems in calculus is the area problem. Suppose
f is a continuous function and f (x) ≥ 0. Find the area of the region S bounded by the
graph of y = f (x), the x-axis, and the lines x = a and x = b. See Figure 5.6.
F IGURE 5.6
The region S is bounded by the
graph of y = f (x), the x-axis, and
the lines x = a and x = b.
Before we can begin to solve this problem, we need some assumptions about area, and a
review of summation notation. Here are some facts about area that we assume to be true.
(1) Area is positive, that is, the area of any plane region is a positive number. It doesn’t
really make sense to say the area of a region is negative.
(2) Area is additive. If we divide a region into non-overlapping pieces, then the area of
the region is the sum of the areas of the individual pieces.
(3) If region 1 is contained in region 2, then the area of region 1 is less than or equal to
the area of region 2.
(4) We know the area of a rectangle: area = length × width.
These assumptions suggest that we will need to add the areas of a lot of rectangles.
Therefore, we need a compact way to write long sums.
Sigma Notation
The sum of n terms a1 , a2 , a3 , . . . , an is written as
n
∑ ai = a1 + a2 + a3 + · · · + an
i=1
A Closer Look
(1) This is called sigma notation because Σ is the Greek capital letter sigma, or S, and S
stands for Sum.
(2) The letter i is called the index of summation or summation variable. It is an arbitrary, or dummy variable. The number 1 is the initial value, or lower bound. The
number n is the final value, or upper bound (of summation).
SECTION 5.2 Riemann Sums 13
(3) The notation ai is shorthand for a(i), a function of i.
There are also some important properties of summations. Suppose k is any constant.
n
n
(1)
∑ kai = k ∑ ai
i=1
i=1
An easy way to remember this property is that constants pass freely through summation symbols.
n
n
(2)
n
∑ (ai ± bi) = ∑ ai ± ∑ bi
i=1
i=1
i=1
This property allows us to split a sum or difference into two separate summations.
There are also several common formulas that allow us to simplify lengthy summations.
Summation Formulas
Let c be a constant and n a positive integer.
n
n
(a)
∑1 = n
(b)
n
n(n + 1)
2
i=1
n
n(n + 1) 2
3
(e) ∑ i =
2
i=1
(c)
∑i =
∑ c = cn
i=1
i=1
n
(d)
∑ i2 =
i=1
n
(f)
∑ i4 =
i=1
n(n + 1)(2n + 1)
6
n(n + 1)(2n + 1)(3n2 + 3n − 1)
30
In the following example, we will use the area assumptions and the properties of summations to estimate the area of a plane region.
Example 1 Estimate the Area
Estimate the area of the region S bounded by the graph of y = x2 , the x-axis, and the lines
x = 0 and x = 1. See Figure 5.7.
1
F IGURE 5.7
Estimate the area of the region S.
1
SOLUTION
Note that the area of S must be between 0 and 1 because the region S is contained in a
square with side of length 1. But we can certainly find a better estimate than this.
Divide the region S into four pieces (or subregions) S1 , S2 , S3 , and S4 by drawing vertical
1
3
1
lines at x = , x = , and x = , as shown in Figure 5.8.
4
2
4
14 CHAPTER 5 Integration
1
1
F IGURE 5.8
Divide the region S into four pieces.
F IGURE 5.9
Approximate the area of each piece by using a
rectangle.
We can approximate the area of each piece by using a rectangle that has the same base
as each subregion and whose height is the same as the right edge of the subregion. See
Figure 5.9. Therefore, the height of each rectangle is the value of the function f (x) = x2
at the right endpoint of the corresponding subinterval;
1
1 1
1 3
3
0, ,
, ,
, , or
,1 .
4
4 2
2 4
4
2 2 2
1
3
1
1
,
,
, and 12 .
Each rectangle has width and the heights are
4
4
2
4
Let R4 be the sum of the areas of these approximating rectangles. Then
2
2
2
1
1
1
1
1
3
1
+ ·
+ ·
+ · 12 = 0.46875
R4 = ·
4
4
4
2
4
4
4
Figure 5.9 shows that the area, A, of the region S is less than R4 , so A < 0.46875.
We can also use rectangles constructed in a different way in order to obtain another
approximation of the area of S. For example, suppose we use rectangles with heights
equal to the values of f at the left endpoint of each subinterval. See Figure 5.10.
1
F IGURE 5.10
An approximation to the area of
S using rectangles whose heights
are the values of f at the left
endpoints of each subinterval.
Note that the leftmost rectangle
has height 0.
The sum of the areas of these approximating rectangles is
2
2
2
1
1
1
1
3
1
1
+ ·
+ ·
= 0.21875
L4 = · 02 + ·
4
4
4
4
2
4
4
The area of S is certainly larger than L4 . Therefore, we now have lower and upper
SECTION 5.2 Riemann Sums 15
estimates for A:
0.21875 < A < 0.46875
It seems reasonable that in order to obtain a better estimate of the area of S, we should
divide the region into smaller pieces, and use smaller (thinner) rectangles. Figures 5.11
and 5.12 show better approximations using eight strips (rectangles) of equal width.
1
1
F IGURE 5.11
Approximating the area of S with eight rectangles
using left endpoints.
F IGURE 5.12
Approximating the area of S with eight rectangles
using right endpoints.
The sum of the areas of the smaller rectangles, L8 , and the sum of the areas of the larger
rectangles, R8 , provide better lower and upper estimates for A.
0.2734375 < A < 0.3984375
We can obtain even better estimates by increasing the number of subregions, or equivalently, by increasing the number of rectangles. Table 5.2 shows the results of similar
calculations using n rectangles whose heights are found using the left endpoints, Ln , or
right endpoints, Rn .
TABLE 5.2
Table of estimates using left
endpoints and right endpoints for
various values of n, the number
of rectangles.
These approximation techniques
using rectangles are reasonable, and
easy to visualize. However, we really
need a quick method to find the exact
area of regions like S.
n
10
20
30
50
100
1000
Ln
0.2850000
0.3087500
0.3168519
0.3234000
0.3283500
0.3328335
Rn
0.3850000
0.3587500
0.3501852
0.3434000
0.3383500
0.3338335
If we use 50 rectangles, then the area lies between 0.3234 and 0.3434. Using 1000
rectangles, A lies between 0.3328335 and 0.3338335. This process suggests a good
estimate for the area of S is the arithmetic mean of these two values, A ≈ 0.3333335.
1
The values in this table suggest that Rn and Ln are approaching .
3
There are certainly other ways to approximate the area of each subregion, then add the areas to approximate the area of S. Let’s consider some generalizations of the two methods
used in Example 1 and two additional approximating sums.
Consider a plane region S bounded above by the graph of a nonnegative, continuous
function y = f (x), below by the x-axis, on the left by x = a and on the right by x = b.
See Figure 5.13
Divide the interval [a, b] into n nonoverlapping subintervals as defined by the values
16 CHAPTER 5 Integration
F IGURE 5.13
We want to approximate the area of the region S
using various sums..
F IGURE 5.14
A visualization of a partition P; divide the interval
[a, b] into n subintervals.
a = x0 < x1 < x2 < . . . < xn = b. This is called a partition of the interval [a, b], denoted
by P = {x0 , x1 , x2 , . . . , xn }
Let ∆xi be the length of the ith subinterval: ∆xi = xi − xi−1 .
Let x∗i be an aribtrary point in the ith subinterval: x∗i ∈ [xi−1 , xi ].
A Riemann sum of the function f over the interval [a, b] with partition P is
n
n
i=1
i=1
∑ f (x∗i )(xi − xi−1) = ∑ f (x∗i ) ∆xi
For a given partition P, different Riemann sums are obtained depending upon how the
values x∗i are selected.
(1) If x∗i is the left endpoint of each subinterval, x∗i = xi−1 for all i, then the sum is called
a left Riemann sum, denoted Ln .
n
Ln = ∑ f (xi−1 ) ∆xi
i=1
(2) If x∗i is the right endpoint of each subinterval, x∗i = xi for all i, then the sum is called
a right Riemann sum, denoted Rn .
n
Rn = ∑ f (xi ) ∆xi
i=1
1
(3) If x∗i is the midpoint of each subinterval, x∗i = x̄i = (xi + xi−1 ) for all i, then the sum
2
is called a midpoint Riemann sum, denoted Mn .
n
Mn = ∑ f (x̄i ) ∆xi
i=1
Instead of using rectangles to estimate the area of each subregion, we could also use
trapezoids. See Figure 5.15.
F IGURE 5.15
We can approximate the area of
each subregion by using
trapezoids.
Recall that the area of a trapezoid is
one-half times the height times the
1
sum of the bases: A = h(b1 + b2 )
2
For a given partition the sum of the areas of the trapezoids is called a trapezoidal sum,
SECTION 5.2 Riemann Sums 17
denoted Tn .
n
1
Tn = ∑ ( f (xi−1 ) + f (xi ))∆xi
2
i=1
A trapezoidal sum is the arithmetic mean of the corresponding left and right Riemann
1
sum: Tn = (Ln + Rn ). You can visualize this result by recognizing how the top portion
2
of each trapezoid divides the area of the corresponding left and right rectangles.
If we divide the interval [a, b] into equal subintervals, then the partition is called regular.
For a regular partition P,
b−a
n
and the points in the partition are a, a + ∆x, a + 2∆x, . . ., a + (n − 1)∆x, b
∆xi = ∆x =
In this case the four sums given above can be written as
Left Riemann Sum
n
∑ f (xi−1 ) ∆x
Ln =
i=1
= ∆x[ f (a) + f (a + ∆x) + f (a + 2∆x) + · · ·+ f (a + (n − 1)∆x)]
Right Riemann Sum
n
∑ f (xi ) ∆x
Rn =
i=1
= ∆x[ f (a + ∆x) + f (a + 2∆x) + · · ·+ f (b)]
Midpoint Sum
n
Mn =
∑ f (x̄i ) ∆x
i=1
∆x
3∆x
(2n − 1)∆x
= ∆x f a +
+ f a+
+ ···+ f a +
2
2
2
Trapezoidal Sum
n
Tn =
1
∑ 2 [ f (xi−1 + f (xi )] ∆x
i=1
=
1
∆x [( f (a) + f (x1 )) + ( f (x1 ) + f (x2 )) + f (x2 ) + f (x3 )) + · · · + ( f (xn−1 + f (b))]
2
=
1
∆x [ f (a) + 2 f (x1 ) + 2 f (x2 ) + · · · + 2 f (xi−1 ) + f (b)]
2
=
1
∆x [ f (a) + 2 f (a + ∆x) + 2 f (a + 2∆x) + · · ·+ 2 f (a + (n − 1)∆x) + f (b)]
2
Example 2 More Sums
Unless otherwise stated, assume the
partition is regualr, that the
subintervals are of equal length.
For the region S in Example 1 and a regular partition, find the midpoint sum and the
trapezoidal sum for n = 4.
SOLUTION
Figure 5.16 shows the rectangles associated with the midpoint sum.
18 CHAPTER 5 Integration
1
1
F IGURE 5.16
An approximation to the area of S using a
midpoint sum.
F IGURE 5.17
An approximation to the area of S using a
trapezoidal sum.
2
2
2
2
1
1
1
1
1
3
5
7
+ ·
+ ·
+ ·
= 0.328125
M4 = ·
4
8
4
8
4
8
4
8
Figure 5.17 illustrates the trapezoidal sum.
#
"
2
2
2
1
1
3
1 1 2
2
0 +2·
+2·
+2·
+ 1 = 0.34375
T4 = ·
2 4
4
2
4
Table 5.3 extends Table 5.2 to include midpoint sums and trapezoidal sums.
TABLE 5.3
Extended table of estimates
including midpoint sums and
trapezoidal sums.
n
10
20
30
50
100
1000
Ln
0.2850000
0.3087500
0.3168519
0.3234000
0.3283500
0.3328335
Rn
0.3850000
0.3587500
0.3501852
0.3434000
0.3383500
0.3338335
Mn
0.3325000
0.3331250
0.3332407
0.3333333
0.3333250
0.3333333
Tn
0.3350000
0.3337500
0.3335185
0.3334000
0.3333500
0.3333335
There are two very important observations from Table 5.3.
1
(1) Each of the sums appears to be zeroing in on, or approaching, the same value, .
3
(2) The midpoint sum and the trapezoidal sum appear to be approaching this limiting
value faster.
1
Let’s try to confirm that the limiting value of Rn is .
3
Example 3 The Limiting Value of Rn
For the region S in Example 1, show that right Riemann sum corresponding to a regular
1
partition approaches , that is
3
1
lim Rn =
n→∞
3
SOLUTION
Divide the interval [0, 1] into n equal subintervals, each of width ∆x =
n−1
1 2 3
,1
The partition is 0, , , , . . . ,
n n n
n
1−0 1
=
n
n
SECTION 5.2 Riemann Sums 19
Use the definition of Rn and simplify as much as possible to find a closed form expression.
n
Rn =
∑ f (xi ) ∆x
i=1
2
2
2
n 2 1
1
1
1
2
3 1
=
+
+
+ ···+
n
n
n
n
n n
n n
Here we are computing the limit of a
sequence {Rn }. Sequences are
discussed in detail in Section 9.1.
However the concept is very similar to
a limit at infinity except that in writing
lim we restrict n to be a positive
1 1 2
· (1 + 22 + 32 + · · · + n2)
n n2
=
1 2
(1 + 22 + 32 + · · · + n2)
n3
=
1 n(n + 1)(2n + 1) (n + 1)(2n + 1)
·
=
n3
6
6n2
Factor out common terms.
Simplify.
Summation formula; simplify.
Now consider the limit
(n + 1)(2n + 1)
6n2
1 n+1
2n + 1
= lim
n→∞ 6
n
n
1
1
1
= lim
1+
2+
n→∞ 6
n
n
lim Rn = lim
n→∞
n→∞
integer. In particular, we know that
1
lim = 0
n→∞ n
When we write lim Rn = 31 , we mean
n→∞
=
n→∞
1
3
that we cna make Rn as close to as
desired by taking n sufficiently large.
=
Write out the summation.
Rewrite as a product.
Rewrite each fraction as a sum; simplify.
1
1
·1·2 =
6
3
Evaluate the limit.
1
This is an amazing result. It says that the exact area of the region S is .
3
1
Note that it can be shown that the left Riemann sums also approach , that is,
3
1
lim Ln =
n→∞
3
Figures 5.18 and 5.19 illustrate these results.
1
1
1
(a) Right Riemann sum, n = 10.
1
1
(b) Right Rieman sum, n = 30.
1
(c) Right Riemann sum, n = 50.
F IGURE 5.18
Notice that each right Riemann sum is an overestimate because f (x) = x2 is increasing.
20 CHAPTER 5 Integration
1
1
1
1
(a) Left Riemann sum, n = 10.
1
1
(b) Left Rieman sum, n = 30.
(c) Left Riemann sum, n = 50.
F IGURE 5.19
Notice that each left Riemann sum is an underestimate because f (x) = x2 is increasing.
As n increases, both Ln an Rn become better and better approximations to the area of S.
Therefore, we define the area A to be the limit of these Riemann sums, that is,
A = lim Rn = lim Ln =
n→∞
n→∞
1
3
Here is a formal definition for the area of a region S.
Definition 1
The area of the region S that is bounded above by the graph of a continuous function
y = f (x), the x-axis, and the lines x = a and x = b is the limit of the sum of the areas
of approximating rectangles.
n
∑ f (xi ) ∆x
x→∞
A = lim Rn = lim [ f (x1 ) ∆x + f (x2 ) ∆x + · · · + f (xn ) ∆x] = lim
n→∞
n→∞
i=1
If f is continuous, then the limit in Definition 5.2 always exists. It can also be shown that
we get the same value if left endpoints are used.
n
∑ f (xi−1 ) ∆x
x→∞
A = lim Ln = lim [ f (x0 ) ∆x + f (x2 ) ∆x + · · · + f (xn−1 ) ∆x] = lim
n→∞
n→∞
(1)
i=1
In fact, instead of using left endpoints or right endpoints, we could take the height of the
ith rectangle to be the value of f at any number x∗i in the ith subinterval [xi−1 , xi ]. Figure 5.20 illustrates an example of approximating rectangles for arbitrary sample points
in each subinterval.
SECTION 5.2 Riemann Sums 21
F IGURE 5.20
Approximating rectangles using
arbitrary sample points in each
subinterval.
So, a more general expression for the area of S is
n→∞
n
∑ f (x∗i ) ∆x
x→∞
A = lim [ f (x∗1 ) ∆x + f (x∗2 ) ∆x + · · · + f (x∗n ) ∆x] lim
(2)
i=1
A Closer Look
When approximating the area of a region S, we could form lower or upper sums by
selecting the sample points x∗i so that f (x∗i ) is the minimum, or maximum, value of f
on the ith subinterval. The function in the previous examples, f (x) = x2 , happens to be
increasing on [0, 1]. Therefore, all the lower sums arise from left endpoints and upper
sums from right endpoints. But this is certainly not always the case.
It can be shown that an equivalent definition of area is the following: A is the unique
number that is smaller than all the upper sums and bigger than all the lower sums.
F IGURE 5.21
The lower sum is represented by
the short, darker rectangles, and
the upper sum is represented by
the tall rectangles.
That is, the upper and lower sums sandwich, or squeeze, the area of the region S.
Example 4 General Expression and an Estimate
Let A be the area of the region bounded above by the graph of f (x) = e−x , below by the
x-axis, and between the lines x = 0 and x = 2.
(a) Using right endpoints and n equal subintervals, find an expression for A as a limit.
(b) Estimate the area using a midpoint sum with n = 4 and n = 10 equal subintervals.
SOLUTION
2−0 2
=
(a) The width of each subinterval is ∆x =
n
n
2 4 6
2i
2n
The partition is P = 0, , , , . . . , , . . .
=2
n n n
n
n
The right Riemann sum is
22 CHAPTER 5 Integration
Rn = f (x1 ) ∆x + f (x2 ) ∆x + · · · + f (xn ) ∆x
= e−x1 ∆x + e−x2 ∆x + · · · + e−xn ∆x
−2/n 2
−4/n 2
−2n/n 2
=e
+e
+ ···+ e
n
n
n
The area is the limit of the right Riemann sum as n → ∞.
2 −2/n
(e
+ e−4/n + e−6/n + · · · + e−2n/n)
n→∞ n
A = lim Rn = lim
n→∞
2 n −2i/n
∑e
n→∞ n
i=1
= lim
It is difficult to evaluate this limit analytically, but technology is very useful here.
This suggests we need a more efficient way to find area and evaluate limits like this.
2−0
= 0.5
4
The subintervals are: [0, 0.5], [0.5, 1], [1, 1.5], [1.5, 2]
(b) For n = 4, ∆x =
The midpoints of the subintervals are:
x∗1 = 0.25, x∗2 = 0.75, x∗3 = 1.25, and x∗4 = 1.75
The midpoint sum is
4
M4 =
∑ f (x∗i ) ∆x
i=1
= f (0.25) ∆x + f (0.75) ∆x + f (1.25) ∆x + f (1.75) ∆x
= e−0.25 (0.5) + e−0.75(0.5) + e−1.25(0.5) + e−1.75(0.5)
The sample points are
midpoints.
Evaluate f ; use ∆x = 0.5.
1 −0.25
(e
+ e−0.75 + e−1.25 + e−1.75) = 0.8557
2
Figure 5.22 illustrates this sum and Figure 5.23 shows a technology solution.
=
1
1
2
F IGURE 5.22
An illustration of the midpoint sum with n = 4
equal subintervals.
F IGURE 5.23
A technology solution.
For n = 10, ∆x = 0.2.
The subintervals are: [0, 0.2], [0.2, 0.4], . . ., [1.8, 2]
The midpoints of the subintervals are:
x∗1 = 0.1, x∗2 = 0.3, x∗3 = 0.5, . . . , x∗10 = 1.9
The midpoint sum is
M10 = f (0.1) ∆x + f (0.3) ∆x + f (0.5) ∆x + · · ·+ f (1.9) ∆x
= 0.2(e−0.1 + e−0.3 + e−0.5 + · · · + e−1.9) = 0.8632
Figure 5.24 illustrates this sum and suggests this estimate is better than the estimate
with n = 4. Figure 5.25 shows a technology solution.
SECTION 5.2 Riemann Sums 23
1
1
2
F IGURE 5.24
An illustration of the midpoint sub with n = 10
equal subintervals.
F IGURE 5.25
A technology solution.
The Distance Problem
The concept of Riemann sums can also be applied to the distance problem: Find the
distance traveled by an object during a certain period if the velocity of the object is
known at all times. In some sense, this is a backward problem: We know the velocity but
want the distance traveled.
If the velocity of the object is constant, then the distance problem is easy to solve using
the formula
distance = velocity × time
But if the velocity varies, then finding the total distance traveled is a little more challenging. The good news is that we will be able to visualize the result and relate it back to the
area problem. Consider the following example.
Example 5 Broken Odometer
Suppose the odometer in a car is broken but we still want to estimate the distance driven
over a 30 second time interval The table below shows the speedometer reading for selected times, in this case every five seconds.
Time (s)
Velocity (m/h)
0
17
5
21
10
24
15
29
20
32
25
31
30
28
In order to have the time and the velocity in consistent units, let’s convert the velocity
readings to feet per second (1 mi/h = 5280/3600 ft/s):
Time (s)
Velocity (ft/s)
0
25
5
31
10
35
15
43
20
47
25
45
30
41
During any five second interval, the velocity doesn’t change very much, so we can estimate the distance traveled during a time interval by assuming that the velocity is constant.
During the first five seconds, assume the velocity of the car is 25 ft/s (the initial velocity).
Then the approximate distance traveled during the first five seconds is
25 ft/s × 5 s = 125 feet
Similarly, during the second time interval we will assume the velocity is 31 ft/s, the
velocity at time t = 5 s. The estimate for the distance traveled from t = 5 s to t = 10 s is
31 ft/s × 5 s = 155 feet
Use similar estimates for the remaining time intervals. An estimate for the total distance
traveled over the 30 second time interval is
(25 × 5) + (31 × 5) + (35 × 5) + (43 × 5) + (47 × 5) + (45 × 5) = 1130 ft
We could also have used the velocity at the end of each time interval instead of the
24 CHAPTER 5 Integration
velocity at the beginning as the assumed constant velocity. Then the estimate for total
distance traveled is
(31 × 5) + (35 × 5) + (43 × 5) + (47 × 5) + (45 × 5) + (41 × 5) = 1210 ft
A more accurate estimate could be obtained if we had velocity readings every two seconds, or even every second.
The calculations in the previous example sure look like Riemann sums. The connection is
even more apparent if we sketch a graph of the velocity function of the car. Figure 5.26
shows a graph with a smooth curve connecting the points of the velocity function and
rectangles with heights equal to the initial velocities in each time interval.
F IGURE 5.26
The odometer readings with
rectangles that represent the
distance traveled over each five
second interval.
The area of the first rectangle is 525 × 5 = 125, which is the estimate for the distance
traveled in the first five seconds. The area of each rectangle can be interpreted as a
distance because the height represents velocity and the width represents time. The sum
of the rectangle areas in Figure 5.26 is L6 = 1130, a left Riemann sum, which is the
initial estimate for the total distance traveled.
In general, suppose the velocity of an object at time t is given by v = f (t), where a ≤ t ≤ b
and f (t) ≥ 0 (so the object is moving in the positive direction). In addition, suppose we
know the velocity of the object a times a = t0 ,t1 ,t2 , . . . ,tn = b. If the times are equally
b−a
.
spaced, then ∆t =
n
Assume the velocity is approximately constant over each subinterval. During the first
time interval, the velocity of the object is approximately f (t0 ) and the distance traveled is
f (t0 ) ∆t. Similarly, the distance traveled during the second time interval is about f (t1 ) ∆t.
The total distance traveled during the time interval [a, b] is approximately
n
f (t0 ) ∆t + f (t1 ) ∆t + · · · + f (tn−1 ) ∆t = ∑ f (ti−1 ) ∆t
i=1
which is a left Riemann sum.
If we use the velocity at the right endpoints of each subinteval, the estimate for the total
distance is
n
f (t1 ) ∆t + f (t2 ) ∆t + · · · + f (tn ) ∆t = ∑ f (ti ) ∆t
i=1
which is a right Riemann sum.
The more frequently we measure the velocity, the more accurate the estimate becomes.
It seems reasonable that the exact distance d traveled is the limit of these Riemann sums.
n
n
lim ∑ f (ti ) ∆t
∑ f (ti−1 ) ∆t = n→∞
n→∞
d = lim
i=1
(3)
i=1
Later in this chapter we will present this result formally.
Because Equation 3 has the same form as the expressions for area, it follows that the
distance traveled is equal to the area under the graph of the velocity function.
Example 6 Distance and Unequal Subintervals
SECTION 5.2 Riemann Sums 25
A car stops at a toll booth and then proceeds onto the New Jersey Turnpike. The car’s
velocity, in m/s, at various times t, 0 ≤ t ≤ 30, is given in the table below.
Time (s)
Velocity (m/s)
0
0
2
2.2
5
4.5
8
9.0
10
13.4
15
16.5
25
20.0
30
22.0
Assume that the velocity is increasing over the time interval [0, 30] as the car speeds up
to enter the Turnpike from an on ramp. Find a lower and an upper estimate for the total
distance traveled by the car over the interval [0, 30].
SOLUTION
Since the velocity of the car is increasing over the interval [0, 30], the left Riemann sum
will produce the lower estimate and the right Riemann sum will produce the upper estimate of the total distance traveled.
Note that the subintervals are not all of equal width.
The left Riemann sum is
7
L7 =
∑ f (ti−1 ) ∆ti
Use a general left Riemann sum.
i=1
= (0)(2) + (2.2)(3) + (4.5)(3) + (9.0)(2) + (13.4)(5) + (16.5)(10) + (20.0)(5)
= 370.1
Use the data in the table.
The right Riemann sum is
7
R7 =
∑ f (ti ) ∆ti
Use a general right Riemann sum.
i=1
= (2.2)(2) + (4.5)(3) + (9.0)(3) + (13.4)(2) + (16.5)(5) + (20.0)(10) + (22.0)(5)
= 464.2
Use the data in the table.
A lower estimate for the distance traveled over [0, 30] is 370.1 meters and an upper estimate is 464.2 meters.
Other Applications
In Chapter 6 we will see that other quantities of interest in the natural and social sciences,
for example, work done by a variable force or cardiac output of the heart, can also be
interpreted and computed in a similar manner. So, let’s begin to think about a general
interpretation of a Riemann sum.
In the distance problems above, we are given the velocity of the car, or the rate of change
of distance, at certain times. A Riemann sum uses the rate function (velocity) to accumulate small changes in distance. The next example presents another application of this
very important concept of accumulation.
Example 7 Too Much Radon
Radon is a colorless, odorless gas that can seep into a home through the basement and
can cause health problems. Suppose radon is leaking into a basement at a rate of r(t)
liters per minute. Values of the rate at 10 minute intervals are given in the table below.
t (m)
r(t) (L/m)
0
10
20
30
40
50
60
0.01
0.02
0.03
0.04
0.03
0.02
0.05
Find a left Riemann sum and a right Riemann sum to estimate the total amount of radon
that seeped into the home over the interval [0, 60].
SOLUTION
26 CHAPTER 5 Integration
During any 10 minute interval, we assume the rate doesn’t change very much. So, we
can estimate the amount of radon that leaked into the basement during a time interval by
assuming the rate is constant.
During the first 10 minutes, assume the rate is 0.01 l/m. So, the approximate amount of
radon that leaked into the basement during the first 10 minutes is
0.01 L/m × 10 m = 0.1 L
Try to draw a graph of the rate
function with rectangles that represent
the leaked radon over each 10 minute
interval.
Continue in this manner to accumulate the total amount of radon that leaked into the
basement over the time interval [0, 60], that is, compute a left Riemann sum.
L6 = (0.01)(10) + (0.02)(10) + (0.03)(10) + (0.04)(10) + (0.03)(10) + (0.02)(10)
= 1.5 liters
We could also use the right endpoint of each subinterval as an estimate of (constant) rate.
This leads to an accumulation of the amount of radon that leaked into the basement using
a right Riemann sum.
R6 = (0.02)(10) + (0.03)(10) + (0.04)(10) + (0.03)(10) + (0.02)(10) + (0.05)(10)
= 1.9 liters
It is tempting to think that L6 is the lower estimate and R6 is the upper estimate for the
total amount of radon that leaked into the basement. However, this isn’t true in this
example. We do not know if r is increasing or decreasing over the entire interval [0, 60].
In fact, the table of values suggests that r is not increasing or decreasing over this interval.
SECTION 5.2 Riemann Sums 27
5.2 EXERCISES
8. The graph of the function f is given below.
Concepts and Vocabulary
1. True or False If f is a nonnegative, continuous, and increasing
function on an interval [a, b], then a left Riemann sum is an
underestimate of the area bounded by the graph of f , the x-axis,
and the lines x = a and x = b.
8
6
2. True or False If f is a nonnegative, continuous, and increasing
function on an interval [a, b], and if Ln , Mn , and Rn are the left,
right, and midpoint sums for n subintervals, then Ln ≤ Mn ≤ Rn .
3. True or False If g is a nonnegative, continuous, and increasing
function on an interval [a, b], and if Ln is the left Riemann sum
for n subintervals, then Ln ≤ L2n .
n
∑ ai and ∑ bi are two summations, then
4. True or False If
n
n
n
∑ ai · ∑ bi =
i=1
i=1
i=1
i=1
∑ (ai · bi ).
i=1
5. True or False If f is a linear function such that f is positive on
L n + Rn
[a, b], then Tn =
.
2
n
6. True or False If
n
n
k
∑ ai + ∑ bi
i=1
i=1
∑ ai and ∑ bi are two summations, then
i=1
!
=
i=1
m
n
∑ k · ai + ∑ k · bi .
i=1
i=1
Practice
Note: unless specified, all approximating rectangles are assumed to
have the same width.
7. The graph of the function f is given below.
5
4
3
2
1
2
6
8
10
(a) Use the graph of f and five rectangles to find a lower
estimate and an upper estimate for the area bounded by the
graph of f , the x-axis, and the lines x = 0 and x = 10. In
each case, sketch the rectangles to illustrate the estimate.
(b) Find estimates using ten rectangles in each case.
4
2
0
0
2
4
6
8
10
12
(a) Use six rectangles to find estimates of each type for the
area bounded by the graph of f , the x-axis, and the lines
x = 0 and x = 12.
(i) L6 (sample points are left endpoints)
(ii) R6 (sample points are right endpoints)
(iii) M6 (sample points are midpoints)
(b) Is L6 an underestimate or overestimate of the true area?
Explain your reasoning.
(c) Is R6 an underestimate or an overestimate of the true are?
Explain your reasoning.
(d) Which of the numbers, L6 , R6 , or M6 gives the best
estimate of the true area? Explain your reasoning.
1
9. Let f (x) = .
x
(a) Estimate the area of the region bounded by the graph of f ,
the x-axis, and between the lines x = 1 and x = 2 using
four rectangles and a right Riemann sum. Sketch the
graph of f and the rectangles. Is your estimate an
underestimate or an overestimate of the true area? Explain
your reasoning.
(b) Repeat part (a) using a left Riemann sum.
10. Let f (x) = sin x.
(a) Estimate the area of the region bounded by the graph of f ,
π
the x-axis, and between the lines x = 0 and x = using
2
four rectangles and a right Riemann sum. Sketch the
graph of f and the rectangles. Is your estimate an
underestimate or an overestimate of the true area? Explain
your reasoning.
(b) Repeat part (a) using a left Riemann sum.
11. Let f (x) = 1 + x2 .
(a) Estimate the area of the region bounded by the graph of f ,
the x-axis, and between the lines x = −1 and x = 2 using
three rectangles and a right Riemann sum.Then use six
rectangles to improve your estimate. Sketch the graph of f
and the approximating rectangles.
(b) Repeat part (a) using a left Riemann sum.
(c) Repeat part (a) using a midpoint sum.
(d) From your sketches in parts (a)-(c), which sum appears to
produce the best estimate? Explain your reasoning.
28 CHAPTER 5 Integration
AP 21. Suppose the function f is positive on the interval [6, 24].
12. Let f (x) = x − 2 ln x, 1 ≤ x ≤ 5.
Selected values of f (x) are given in the table below.
(a) Sketch a graph of the function f .
(b) Estimate the area of the region bounded by the graph of f ,
x
6 12 20 24
the x-axis, and between the lines x = 1 and x = 5 using
f (x) 5
7
4
7
four rectangles and a (i) right Riemann sum and (ii) a
midpoint sum. In each case sketch the graph and the
Let R be the region bounded by the graph of f , the x-axis, and
approximating rectangles.
the lines x = 6 and x = 24. Find a trapezoidal approximation
for the area of R using the subintervals defined by the table.
(c) Find the right Riemann sum and the midpoint sum using
eight rectangles.
6
AP 22. Suppose ∑ f (x∗i ) ∆xi represents a Riemann sum approximation
13. Find the upper and lower sums for f (x) = 2 + sin x, 0 ≤ x ≤ π ,
i=1
with n = 2, 4, and 8. Illustrate each case with a sketch similar
for the area under the graph of f (x) = x2 + 2 between the lines
to Figure 5.21.
x = 2 and x = 5. If a midpoint Riemann sum is used, find the
value of f (x∗3 ).
14. Find the upper and the lower sums for f (x) = 1 + x2 ,
−1 ≤ x ≤ 1, with n = 3 and 4. Illustrate each case with a sketchAP 23. Selected values of f (x) over the interval [2, 38] are given in the
similar to Figure 5.21.
table below.
AP 15. Suppose f is a function that is nonnegative and continuous on
the interval [0, 2]. Which of these represents R4 , the right
Riemann sum with 4 equal subintervals?
1
1
1
1
(A) f (0) · + f (0.5) · + f (1) · + f (1.5) ·
4
4
4
4
1
1
1
1
(B) f (0.5) · + f (1) · + f (1.5) · + f (2) ·
4
4
4
4
1
1
1
1
(C) f (0) · + f (0.5) · + f (1) · + f (1.5) ·
2
2
2
2
1
1
1
1
(D) f (0.5) · + f (1) · + f (1.5) · + f (2) ·
2
2
2
2
AP 16. If the area under the graph of y = sin x over the interval [0, π ] is
estimated by L4 , a left Riemann sum with 4 equal subintervals,
then the exact value of L4 is
√
√
√
1+ 2
π (1 + 2)
π 2
(A)
(B)
(C) 2
(D)
2
4
2
Use technology to find the right Riemann sum using subintervals of
equal width for n = 10, 30, 50, and 100. Use these estimates to
guess the value of the exact area.
17. The region bounded by the graph of f (x) = x4 , the x-axis, and
between the lines x = 0 and x = 1.
18. The region bounded by the graph of y = cos x, the x-axis, and
π
between the lines x = 0 and x = .
2
1
and consider the region bounded by the
1 + x2
graph of f , the x-axis, and the lines x = 0 and x = 1.
(a) Find the left and right Riemann sums for n = 10, 30, and
50.
(b) Illustrate the sums in part (a) by graphing the rectangles.
(c) Show that the exact area of the region lies between 0.780
and 0.791.
19. Let f (x) =
20. Let f (x) = ln x and consider the region bounded by the graph of
f , the x-axis, and the lines x = 1 and x = 4.
(a) Find the left and right Riemann sums for n = 10, 30, and
50.
(b) Illustrate the sums in part (a) by graphing the rectangles.
(c) Show that the exact area of the region lies between 2.50
and 2.59.
x
f (x)
2
7
8
9
14
18
20
20
26
16
32
10
38
8
Suppose this table is used to compute a midpoint Riemann sum
for f over the interval [2, 38] using as many subintervals as
possible. What is the value of this midpoint sum?
(A) 422
(B) 468
(C) 482
(D) 52
Applications and Extensions
24. The speed of a runner increased steadily during the first three
seconds of a race. Her speed at half-second intervals is given in
the following table.
t (s)
0
0.5
1.0
1.5
2.0
2.5
3.0
v (ft/s)
0
6.2
10.8
14.9
18.1
19.4
20.2
Find lower and upper estimates for the distance that she
traveled during these three seconds.
25. The table below shows the speedometer readings at 10-second
intervals during a 1-minute period for a car racing at the
Daytona International Speedway in Florida.
Time (s)
Velocity (mi/h)
0
10
20
30
40
50
60
182.9
168.0
106.6
99.8
124.5
176.1
175.6
(a) Estimate the distance the race car traveled during this time
period using the velocities at the beginning of each
subinterval.
(b) Find another estimate of the distance traveled by using the
velocities at the end of each subinterval.
(c) Are your estimates in parts (a) or (b) upper and lower
estimates? Explain your reasoning.
SECTION 5.2 Riemann Sums 29
26. Oil is leaking from a tank at a rate of r(t) liters per hour. The
rate is decreasing over time and the values of the rate at
two-hour intervals are given in the table below.
t (h)
r(t) (L/h)
0
2
4
6
8
10
8.7
7.6
6.8
6.2
5.7
5.3
30. In someone infected with measles, the virus level N (measured
in number of infected cells per mL of blood plasma) reaches a
peak density at about t = 12 days (when a rash appears) and
then decreases fairly rapidly as a result of immune response.
The area of the region bounded by the graph of N(t), the t-axis
and the lines t = 0 and t = 12, as shown in the figure below, is
equal to the total amount of infection needed to develop
symptoms (measured in density of infected cells × time).
Find lower and upper estimates for the total amount of oil that
leaked from the tank over the interval [0, 10].
27. On May 7, 1992, the space shuttle Endeavour was launched on
mission STS-49. The purpose of this mission was to install a
new perigee kick motor in an Intelsat communications satellite.
The table below, provided by NASA, gives the velocity data for
the shuttle between liftoff and the jettisoning of the solid rocket
boosters.
Event
Time (s)
Velocity (ft/s)
Launch
Begin roll maneuver
0
10
0
185
End roll maneuver
Throttle to 89%
Throttle to 67%
Throttle to 104%
Maximum dynamic pressure
15
20
32
59
62
319
447
742
1325
1445
125
4151
Solid rocket booster separation
0
12
f (t) = −t(t − 21)(t + 1)
Use this model and a midpoint sum with six equal subintervals
to estimate the total amount of infection needed to develop
symptoms of measles.
Source: J.M. Heffernan et al., “An In-Host Model of Acute Infection: Measles as
a Case Study,” Theoretical Population Biology 73 (2006): 134-47.
31. The table below shows the number of people per day who died
from SARS in Singapore at two-week intervals beginning on
March 1, 2003.
28. The graph of the velocity of a braking car is shown in the figure
below.
Date
Deaths per day
Date
Deaths per day
March 1
0.0079
April 26
0.5620
March 15
0.0638
May 10
0.4630
March 29
0.1944
May 24
0.2897
April 12
0.4435
(a) Estimate the number of people who died of SARS in
Singapore between March 1 and May 24, 2003 using a left
Riemann sum and a right Riemann sum.
(b) Explain how to interpret the number of SARS deaths as an
area under a curve.
40
20
0
0
21
The function N = f (t) has been modeled by
Use these data to estimate the height above the earth’s surface
of the Endeavour, 62 seconds after liftoff.
60
1000
1
2
3
4
5
6
Use this graph and six subintervals to estimate the distance
traveled by the car while the brakes are applied.
Source: A. Gummel et al., “Modeling Strategies for Controlling SARS
Outbreaks,” Proceedings of the Royal Society of London: Series B 271 (2004):
2223-32.
29. The graph of the velocity of a car accelerating from rest to a AP 32. If f is increasing, continuous, positive, and concave up on an
interval [a, b], and if Ln , Mn , Tn , and Rn are the left, midpoint,
speed of 120 km/h over a period of 30 seconds is shown in the
trapezoidal, and right sums for n subintervals, which of these is
figure below.
not true?
(A) Ln ≤ Mn ≤ Rn
(B) Ln ≤ Tn ≤ Mn
120
L n + Rn
(C) Mn <
(D) L2n > Ln
2
80
AP 33. If f is decreasing, continuous, positive, and concave down on
40
0
0
10
20
30
Use six subintervals to estimate the distance traveled by the car
during this time period.
an interval [a, b], and if Ln , Mn , Tn , and Rn are the left,
midpoint, trapezoidal, and right sums for n subintervals, which
of these is true?
(A) Ln ≤ Mn ≤ Tn ≤ Rn
(B) Ln ≥ Mn ≥ Tn ≥ Rn
(C) Ln ≥ Tn ≥ Mn ≥ Rn
(D) Ln ≥ Rn ≥ Mn ≥ Tn
Use Definition 5.2 to find an expression for the area of the region
bounded by the graph of f and above the x-axis over the interval
30 CHAPTER 5 Integration
44. Let f (x) = x3 .
(a) Use Definition 5.2 to find an expression as a limit for the
area of the region bounded by the graph of f and above
the x-axis between the lines x = 0 and x = 1.
(b) Evaluate the limit in part (a).
indicated. Do not evaluate the limit.
2x
34. f (x) = 2
, 1≤x≤3
x +1
√
35. f (x) = x2 + 1 + 2x, 4 ≤ x ≤ 7
√
36. f (x) = sin x, 0 ≤ x ≤ π
Extended Applications
Describe a region whose area is given by the limit.
r
n
n
3i
π
3
iπ
1+
37. lim ∑
38. lim ∑
tan
n→∞
n→∞
n
n
4n
4n
i=1
i=1
AP 39. The expression
n
lim
n→∞
∑
i=1
6
3+ i
n
2
! 6
+1
n
represents the exact area of a region R. Find the midpoint
Riemann sum approximation for R for a regular partition with 3
subintervals.
AP 40. Describe the plane region that corresponds to the Riemann sum
r
1
+
n
r
2
+
n
r
3
+··· +
n
r ! n
1
n
n
AP 41. Consider the area of the region bounded by the graph of
f (x) = x2 , above the x-axis, and between the lines x = 4 and
x = 6. If a regular partition with 100 subintervals is used, what
is the difference between the value of the left Riemann sum and
the value of the right Riemann sum? Find this difference
without actually calculating either of the Riemann sums.
AP 42. Selected values of f (x) are given in the table below.
x
f (x)
2
7
5
9
8
k
10
11
The trapezoidal sum associated with the function f and this
table is 61. What is the value of k?
(A) 4
(B) 5
(C) 6
(D) 7
AP 43. Describe a plane region that corresponds to the Riemann sum
2000
∑
i=1
1+
2
i
n
2
! 2
+3
n
45. Suppose f is continuous and increasing on the interval [a, b].
Let A be the area of a region bounded above by the graph of
y = f (x), below by the x-axis, between the lines x = a and
x = b. Let L − N and Rn be the let and right Riemann sums,
respectively.
(a) How are A, Ln , and Rn related?
(b) Show that
b−a
[ f (b) − f (a)]
n
Draw a diagram to illustrate this equation by showing that
the n rectangles representing Rn − Ln can be reassembled
to form a single rectangle whose area is the right side of
the equation.
(c) Explain why the following equation is true.
b−a
Rn − A <
[ f (b) − f (a)]
n
Rn − L n =
46. Suppose A is the area of the region bounded by the graph of
y = ex , the x-axis, and the lines x = 1 and x = 3. Use the
previous Exercises to find a value of n such that
Rn − A < 0.0001.
47. Let An be the area of a polygon with n equal sides inscribed in
a circle with radius r.
(a) By dividing the polygon into n congruent triangles with
2π
central angle
, show that
n
2π
1 2
nr
sin
=
n
2
n
(b) Show that lim An = π r2
n→∞
SECTION 5.3 The Definite integral 31
5.3 The Definite integral
Learning
Objectives
Essential
Knowledge
LO 3.2A(a)
EK 3.2A1
LO 3.2A(b)
EK 3.2A2
LO 32C
EK 3.2A3
EK 3.2C1
EK 3.2C2
Review
• The area and distance problems.
• Properties of summations and common summation formulas.
• Riemann sums.
Preview
• The definite integral: definition and evaluation
• Properties of the definite integral
In the previous section, we discovered that a limit of the form
n
lim [ f (x∗1 ) ∆x + f (x∗2 ) ∆x + · · · + f (x∗n ) ∆x]
∑ f (x∗i ) ∆x = n→∞
n→∞
lim
(4)
i=1
arises when we try to compute the area of a plane region. We also saw limits of this type
when we tried to find the distance traveled by an object. Surprisingly, this same type of
limit occurs in a wide variety of applications, even when f is not necessarily a positive
function. Later we will see limits of this form occur in finding lengths of curves, volumes
of solids, centers of mass, force due to water pressure, and work, as well as many other
quantities. This type of limit (of a Riemann sum) is so important that it has a special
name and notation.
Definition 1 Definite integral
If f is a function defined on a closed interval [a, b], let P be a partition of [a, b] defined
by
P = {a = x0 , x1 , x2 , x3 , . . . , xn = b}
Choose sample points x∗i in [xi−1 , xi ], the ith subinterval, and let ∆xi = xi − xi−1 . Then
the definite integral of f from a to b is
Z b
n
∑ f (x∗i ) ∆x∗i
n→∞
f (x) dx = lim
a
i=1
provided the limit exits and gives the same value for all possible choices of sample
points. If it does exist, we say that f is integrable on [a, b].
A Closer Look
(1) The precise meaning of the limit that defines the definite integral is as follows:
For every number ε > 0 there exists an integer N such that
Z b
n
∗
f (x) dx − ∑ f (xi ) ∆xi < ε
a
i=1
for every integer n > N and for every choice of x∗i in [xi−1 , xi ].
(2) The norm of a partition P is defined to be the maximum width of all subintervals
and is denoted ||P||. Therefore, ||P|| = max{∆xi }. As n → ∞ the width of the largest
subinterval goes to 0, that is ||P|| → 0. So, another way to define the definite integral
is
Z b
a
R
n
∑ f (x∗i ) ∆xi
||P||→0
f (x) dx = lim
i=1
(3) The symbol was introduced by Leibniz and is called a definite integral sign. It is
an elongated S and was chosen because an definite integral is a limit of sums.
32 CHAPTER 5 Integration
The function f (x) is called the integrand.
a and b are called the limits of integration; a is the lower limit and b is the upper
limit.
For now, the symbol dx has no specific meaning by itself; rather
Z b
f (x) dx is all
a
one symbol, representing the definite integral. The dx simply indicates that the independent variable is x.
The procedure of calculating an definite integral is called integration.
(4) The definite integral, denoted by
Z b
f (x) dx, is a number; it does not depend upon x.
a
We could use any letter in place of x without changing the meaning or value of the
definite integral:
Z b
f (x) dx =
a
Z b
f (t) dt =
a
Z b
f (r) dr
a
(5) Recall from the previous section, the sum
n
∑ f (x∗i ) ∆xi
i=1
is called a Riemann sum. So, Definition 1 says that the definite integral is the limit
of a Riemann sum and that an integrable function can be approximated to within any
desired degree of accuracy by a Riemann sum.
In the previous section we learned that if f is continuous and positive on the interval [a, b],
then a Riemann sum can be interpreted as a sum of areas of approximating rectangles.
For example, Figure 5.27 shows a midpoint sum with all subintervals of equal width. By
comparing Definition 1 with the definition of area in the previous section, we see that the
definite integral
Z b
f (x) dx can be interpreted as the area bounded of the region bounded
a
above by the graph of y = f (x), below by the x-axis, and between the lines x = a and
x = b (or simply, the area under the curve y = f (x) from a to b). See Figure 5.28.
F IGURE 5.27
If f (x) ≥ 0, a Riemann sum, in this case a
midpoint sum, is the sum of areas of rectangles.
F IGURE 5.28
If f (x) ≥ 0, the definite integral limba f (x) dx is the
area under the curve y = f (x) from a to b.
If f takes on both positive and negative values over the interval [a, b] then a Riemann
sum is the sum of the areas of rectangles that lie above the x-axis and the negatives of
the areas of the rectangles that lie below the x axis. This is because f (x∗i ) < 0 for some
subintervals and, therefore, the product f (x∗i ) ∆xi < 0. In Figure 5.29 this is the areas of
the blue rectangles minus the areas of the gold rectangles.
SECTION 5.3 The Definite integral 33
F IGURE 5.29
The Riemann sum here is the areas of the blue
rectangles minus the areas of the gold rectangles.
F IGURE 5.30
Z
b
f (x) dx is the net area.
a
If we take the limit of Riemann sums formed this way, the result is illustrated in Figure 5.30. Therefore, a definite integral can be interpreted as a net area, that is, a difference of areas:
Z b
a
f (x) dx = A1 − A2
where A1 is the area of the region above the x-axis and below the graph of f , and A2 is
the area of the region below the x-axis and above the graph of f .
Remember that in a Riemann sum, the subintervals may be of unequal width. There
are many practical applications in which the partition is not regular, but technology is
extremely helpful in calculating these sums quickly and accurately.
The definite integral is defined for an integrable function, but not all functions are integrable. The following theorem shows that the most commonly occurring functions are,
in fact, integrable. This theorem is proved in more advanced mathematics courses.
Theorem 1
If f is continuous on the interval [a, b], or if f has a finite number of jump discontinuities, then f is integrable on [a, b], that is, the definite integral
Z b
f (x) dx exists.
a
Figure 5.31 shows the graph of a function on the interval [0, 5] with a finite number of
jump discontinuities. Theorem 1 tells us this function is integrable, although it is a little
unclear (right now) how to find this definite integral
Z 5
0
f (x) dx
2
1
F IGURE 5.31
The graph of a function with a
finite number of jump
discontinuities. This function is
integrable over the interval [0, 5].
1
2
3
4
5
-1
-2
-3
If f is integrable on [a, b] then the limit in Definition 1 exists and gives the same value
no matter how we choose the sample points x∗i . Therefore, to simplify the calculation
(for now) we often take the sample points to be the right endpoints, that is, we use a right
Riemann sum with subintervals of equal width. Then x∗i = xi and the definition of an
definite integral simplifies as follows.
34 CHAPTER 5 Integration
Theorem 2
If f is integrable on [a, b], then
Z b
n
∑ f (xi ) ∆x
n→∞
f (x) dx = lim
a
i=1
b−a
∆x =
n
where
xi = a + i ∆x
and
Example 1 Interpret a Riemann Sum
Write an equivalent expression for
n
∑ (x3i + xi sin xi ) ∆x
n→∞
lim
i=1
as an definite integral on the interval [0, π ].
SOLUTION
Compare the given limit with the limit in Theorem 2.
Let f (x) = x3 + x sin x in the limit expression in Theorem 2.
This produces the expression in this example.
We are given that a = 0 and b = π .
n
Therefore, lim
n→∞
∑ (x3i + xi sin xi ) ∆x =
i=1
Z π
0
(x3 + x sin x) dx
When we apply the definite integral to different physical situations, it will be important to recognize limits of (Riemann) sums as definite integrals. When Leibniz chose
the notation for an definite integral, he selected the pieces as reminders of the limiting
process.
In general, when we write
n
∑ f (x∗i ) ∆xi =
n→∞
lim
i=1
R
we replace ∑ by , x∗i by x, and ∆x by x.
Z b
f (x) dx
a
Evaluating definite integrals
When we use the limit definition to evaluate a definite integral, we will need to use the
properties of summations and the summation formulas, as given in the previous section.
Some of the evaluations can be long, but most are very prescriptive.
Example 2 Evaluate a Definite Integral
Let f (x) = x3 − 6x.
(a) Let P be a regular partition of [0, 3] with 6 subintervals. Find R6 , a right Riemann
sum.
(b) Evaluate
Z 3
0
(x3 − 6x) dx
SOLUTION
(a) For n = 6, a = 0, and b = 3, the length of each subinterval is
b−a 3−0 1
=
=
∆x =
n
6
2
The partition is P = {0, 0.5, 1.0, 1.5, 2.0, 2.5, 3}.
The right Riemann sum is
SECTION 5.3 The Definite integral 35
6
R6 =
∑ f (xi ) ∆x
i=1
= f (0.5) ∆x + f (1.0) ∆x + f (1.5) ∆x + f (2.0) ∆x + f (2.5) ∆x + f (3.0) ∆x
Use the right endpoint of each subinterval.
1
(−2.875 − 5 − 5.625 − 4 + 0.625 + 9)
2
= −3.9375
=
Evaluate f ; use the value for ∆x.
Figure 5.32 illustrates this right Riemann sum.
5
1
F IGURE 5.32
An illustration of R6 , a right
Riemann sum.
2
3
-5
Notice that f is not a positive function. Therefore, the Riemann sum does not represent a sum of areas of rectangles. However, it does represent the sum of the areas
of the blue rectangles (above the x-axis) minus the sum of the areas of the gold
rectangles (below the x-axis), the net area.
In the sum, n is constant; i is the
variable. Therefore, the constant
passes through the summation
symbol.
3
n
b−a 3
(b) For a regular partition with n subintervals, ∆x =
=
n
n
3i
3n
3 6 9
The partition is P = 0, , , , . . . , , . . . ,
n n n
n
n
Use a right Riemann sum to evaluate the definite integral.
Z 3
n
n
3i 3
3
(x − 6x) dx = lim ∑ f (xi ) ∆x = lim ∑ f
n→∞
n→∞
n n
0
i=1
i=1
" #
3i 3
3i
3 n
−6
= lim ∑
n→∞ n
n
n
i=1
3 n 27 3 18
i
−
i
∑ 3
n→∞ n
n
i=1 n
"
#
81 n 3 54 n
= lim
∑ i − n2 ∑ i
n→∞ n4
i=1
i=1
Evaluate f ; constants pass
freely through summation
symbols.
= lim
!
81 n(n + 1) 2 54 n(n + 1)
− 2
n4
2
n
2
= lim
n→∞
#
1 2
1
81
− 27 1 +
1+
= lim
n→∞ 4
n
n
"
Write as two summations;
simplify.
Summation formulas.
Rearrange terms.
27
81
− 27 = − = −6.75
4
4
This definite integral cannot be interpreted as an area because f takes on both positive an negative values. However, it can be interpreted as the difference of areas
A1 − A2, or net area, where A1 and A2 are shown in Figure 5.33.
=
36 CHAPTER 5 Integration
5
1
2
3
F IGURE 5.33
R3 3
0 (x − 6x) dx = A1 − A2 = −6.75.
-5
Figure 5.34 illustrates an approximation to this definite integral by showing the positive and negative terms in the right Riemann sum, Rn , for n = 40. The values in the
table show the Riemann sums approaching the exact value of the definite integral,
−6.75, as n → ∞.
5
1
2
3
-5
F IGURE 5.34
An illustration of R40 .
n
40
100
500
1000
5000
Rn
−6.3998
−6.6130
−6.7229
−6.7365
−6.7473
TABLE 5.4
As n increases, Rn → −6.75.
This example reinforces the notion that we need a simpler method to evaluate definite
integrals. We’ll find one in the next section.
Example 3 Technology to the Rescue
Set up, but do not evaluate, an expression for the definite integral
sums. Use technology to evaluate this expression.
Z 3
ex dx as a limit of
1
SOLUTION
Let f (x) = ex , a = 1 and b = 3.
b−a 3−1 2
=
=
For a regular partition with n subintervals, ∆x =
n
n
n
2
4
6
2n
The partition is P = 1, 1 + , 1 + , 1 + , . . . , 1 +
=3
n
n
n
n
2i
n
Use a right Riemann sum to find an expression for the definite integral.
The right endpoint of the ith subinterval is xi = 1 +
Z 3
1
n
ex dx = lim
n→∞
∑ f (xi ) ∆x
Limit of a right Riemann sum.
i=1
2i 2
= lim ∑ f 1 +
n→∞
n n
i=1
n
2 n 1+2i/n
∑e
n→∞ n
i=1
= lim
Use values for xi and ∆x.
Evaluate f ; constant passes freely through summation.
We cannot use the summation formulas to evaluate this expression. However, we can use
technology to investigate the limit numerically. Figures 5.35 and 5.36 show two different
SECTION 5.3 The Definite integral 37
approaches to evaluating the sum for various values of n.
F IGURE 5.35
Sum evaluation on the Home
Screen.
F IGURE 5.36
Sum evaluation using the Table
feature.
If we use a computer algebra system to evaluate the sum and simplify, we obtain
n
A computer algebra system can be
used to find a closed form expression
for this sum because it is a geometric
series. The limit could also be found
using l’Hospital’s Rule.
∑ e1+2i/n =
i=1
e1+2/n (e2 − 1)
e2/n − 1
We can also use technology to evaluate the limit:
Z 3
1
2 e1+2/n(e2 − 1)
·
= e3 − e = 17.3673
n→∞ n
e2/n − 1
ex dx = lim
Figure 5.37 illustrates this definite integral, and Figures 5.38 and 5.39 show the technology results.
20
10
1
2
3
F IGURE 5.37
Since f (x) = ex is positive, the definite integral
represents the area under the curve.
F IGURE 5.38
A closed form expression for the sum.
F IGURE 5.39
The value of the definite integral.
We still need a quicker method for evaluating this type of definite integral.
Example 4 The Key to a Deep Connection
It is important to connect, or associate, a Riemann sum with an appropriate integral.
Write the definite integral equivalent to the limit
n
2
2i
·
lim ∑ ln 1 +
n→∞
n
n
i=1
SOLUTION
The constant divided by n at the end of the expression is a good candidate for ∆x.
2
n
Look for an expression of the form a + i∆x.
Therefore, ∆x =
Use the argument of the natural logarithm function: xi = a + i∆x = 1 +
2i
n
38 CHAPTER 5 Integration
b−a 2
= =⇒ b − a = 2 =⇒ b − 1 = 2 =⇒ b = 3
n
n
We can use this information to rewrite the limit expression:
n
n
2i
2
lim ∑ ln 1 +
· = lim ∑ ln(xi ) · ∆x
n→∞
n
n n→∞ i=1
i=1
∆x =
Since we know the values of a and b:
n
lim
n→∞
∑ ln(xi ) · ∆x =
i=1
Z 3
ln x dx
1
Example 5 The Definite Integral as an Area
Evaluate the following definite integrals by interpreting each in terms of areas.
Z 3
Z 1p
(x − 1) dx
(b)
(a)
1 − x2 dx
0
0
SOLUTION
√
(a) Since f (x) = 1 − x2 ≥ 0, we can interpret this interval as the area under the graph
of f and above the x-axis from 0 to 1.
The graph of f is a quarter circle with radius 1, as shown in Figure 5.40.
Therefore, the definite integral is one quarter of the area of a circle of radius 1.
Z 1p
π
1
1 − x2 dx = π (1)2 =
4
4
0
2
1
1
1
1
F IGURE 5.40
R √
The definite integral 01 1 − x2 dx is the area of a
quarter circle.
2
3
-1
F IGURE 5.41
R
The definite integral 03 (x − 1) dx is the net area,
A1 − A2 .
(b) The graph of y = x − 1 is a line with slope 1 as shown in Figure 5.41.
We can evaluate the definite integral as the difference of the areas of the two triangles.
Z 3
1
1
3
(x − 1) dx = A1 − A2 = · 2 · 2 − · 1 · 1 =
2
2
2
0
Properties of the Definite integral
In the definition of the definite integral,
Z b
a
f (x) dx, we implicitly assume that a < b.
However, this definition as a limit of Riemann sums makes sense even if a > b. For a
SECTION 5.3 The Definite integral 39
regular partition, we can write
∆x =
b−a
a−b
=−
n
n
and, therefore,
Z a
b
f (x) dx = −
Z b
f (x) dx
a
If a = b, then ∆x =, and so
Z a
f (x) dx = 0
a
Here are some basic properties of definite integrals that will help us to evaluate definite
integrals. We assume f and g are continuous functions.
Properties of the Definite Integral
(1)
Z b
c dx = c(b − a),
where c is any constant
Z b
[ f (x) + g(x)] dx =
Z b
Z b
c f (x) dx = c
Z b
[ f (x) − g(x)] dx =
a
(2)
a
g(x) dx
f (x) dx,
where c is any constant
a
a
(4)
Z b
Z b
a
a
a
(3)
f (x) dx +
Z b
a
f (x) dx −
Z b
g(x) dx
a
A Closer Look
(1) In words, property 1 says that the definite integral of a constant function f (x) = c is
the constant times the length of the interval. If c > 0 and a < b, then c(b − a) is the
area of the rectangle bounded above by f (x) = c, below by the x-axis, on the left by
x = a, and on the right by x = b. See Figure 5.42.
F IGURE 5.42
a c dx = c(b − a).
Rb
(2) In words, property 2 says that the definite integral of a sum is the sum of the definite
integrals. For positive functions, this can be interpreted as the area under f + g is
the area under f plus the area under g. Figure 5.43 illustrates this concept. The
corresponding vertical line segments have equal height.
Property 2 follows from the definition of an definite integral and the fact that the
limit of a sum is the sum of the limits.
40 CHAPTER 5 Integration
F IGURE 5.43
Z
b
[ f (x) + g(x)] dx =
Zab
f (x) dx +
a
Z b
Z b
g(x) dx.
a
n
Definition of the definite
integral.
∑ [ f (x∗i ) + g(x∗i )]∆xi
n→∞
[ f (x) + g(x)] dx = lim
a
i=1
= lim
n→∞
"
n
∑
f (x∗i ) ∆xi +
i=1
n
∑
g(x∗i ) ∆xi
i]1
n
n
i=1
i=1
#
lim ∑ g(x∗i ) ∆xi
∑ f (x∗i ) ∆xi + n→∞
n→∞
= lim
Z b
=
f (x) dx +
Z b
g(x) dx
a
a
Write as separate
summations.
Limit of a sum is the sum of
the limits.
Definition of the definite
integral.
(3) Property 3 can be proved in a similar manner and says that the definite integral of a
constant times a functions is the constant times the definite integral of the function.
That is, constants pass freely through definite integral signs.
(4) Property 4 is proved by writing f − g = f + (−g) and using Properties 2 and 3 with
c = −1. In words, the definite integral of a difference is the corresponding difference
of the definite integrals.
Example 6 Property Value
Z
Z b
b
1
1
x2 dx = (b3 − a3).
x dx = (b2 − a2 ) and
2
3
a
a
Use the properties of definite integrals to evaluate
Assume
(a)
Z 2
−1
(3x + 5) dx
(b)
Z 1
0
(4 + 3x2) dx
SOLUTION
(a)
Z 2
−1
(3x + 5) dx =
Z 2
−1
=3
3x dx +
Z 2
−1
x dx +
Z 2
5 dx
Property 2.
Z 2
5 dx
Property 3.
−1
−1
1
= 3 · [22 − (−1)2 ] + 5[2 − (−1)]
2
=
9
39
+ 15 =
2
2
Given formula; Property 1.
SECTION 5.3 The Definite integral 41
(b)
Z 1
0
(4 + 3x2) dx =
Z 1
0
4 dx + 3
Z 1
x2 dx
Properties 2 and 3.
0
1
= 4(1 − 0) + 3 · (13 − 03)
3
= 4+3·
Property 1; given formula.
1
=5
3
The next property shows how to combine definite integrals of the same function over
adjacent intervals.
Properties of the definite integral (Continued)
Suppose that all of the following definite integrals exist.
(5)
Z b
f (x) dx =
Z c
f (x) dx +
f (x) dx
c
a
a
Z b
A Closer Look
The general proof of this property is lengthy. However, there is a very reasonable geometric interpretation if f (x) ≥ 0 and a < c < b. In Figure 5.44, the area under the graph
of y = f (x) from a to c plus the area from c to b is equal to the total area from a to b.
F IGURE 5.44
Geometric interpretation of
Property 5.
Example 7 Interval Shuffle
Suppose
Z 10
f (x) = 17 and
Z 8
f (x) dx = 12, find
0
0
Z 10
f (x) dx.
8
SOLUTION
Use Property 5.
Z 10
0
f (x) dx =
Z 8
0
f (x) dx +
Z 10
8
f (x) dx
Solve for the unknown definite integral.
Z 10
8
f (x) dx =
Z 10
0
f (x) dx −
Z 8
0
f (x) dx = 17 − 12 = 5
Properties 1-5 are true whether a < b, a = b, or a > b. The following properties, in which
we compare the magnitude of functions and definite integrals, are true only if a ≤ b.
42 CHAPTER 5 Integration
Comparison Properties of the definite integral
Suppose the following definite integrals exist and a ≤ b.
(6) If f (x) ≥ 0 for a ≤ x ≤ b, then
Z b
f (x) dx ≥ 0.
a
(7) If f (x) ≥ g(x) for a ≤ x ≤ b, then
Z b
a
f (x) dx ≥
Z b
g(x) dx.
a
(8) If m ≤ f (x) ≤ M for a ≤ x ≤ b, then
m(b − a) ≤
Z b
a
f (x) dx ≤ M(b − a)
A Closer Look
(1) If f (x) ≥ 0, then
Z b
f (x) dx represents the area of the region bounded above by the
a
graph of y = f (x), below by the x-axis, and between the lines x = a and x = b. So,
the geometric interpretation of Property 6 is simply that areas are positive.
(2) Property 7 says that a bigger function has a bigger definite integral. This follows
from Property 6 and 4, and because f − g ≥ 0.
(3) Property 8 is illustrated in Figure 5.45 for the case where f (x) ≥ 0.
F IGURE 5.45
Geometric interpretation of
Property 8.
If f is continuous, take m and M to be the absolute minimum and maximum values
of f on the interval [a, b]. In this case, Property 8 says that the area under the graph
of f is greater than the area of the rectangle with height m, and less than the area of
the rectangle with height M.
Proof of Property 8
Since m ≤ f (x) ≤ M, use Property 7 to obtain
Z b
a
m dx ≤
Z b
a
f (x) dx ≤
Z b
M dx
a
Use Property 1 to evaluate the definite integrals on the left and right sides.
m(b − a) ≤
Z b
a
f (x) dx ≤ M(b − a)
Property 8 is useful for finding a very rough estimate of the magnitude of an definite
integral. If we can find reasonable values for M and m, and then we can easily bound the
definite integral.
Example 8 Definite Integral Estimate
Use Property 8 to estimate
Z 1
0
2
e−x dx.
SOLUTION
2
Because f (x) = e−x is a decreasing function on [0, 1], its absolute maximum value is
M = f (0) = 1 and its absolute minimum value is m = f (1) = e−1 .
SECTION 5.3 The Definite integral 43
Using Property 8,
e−1 (1 − 0) ≤
Z 1
0
2
e−x dx ≤ 1(1 − 0),
Since e−1 = 0.368,
0.368 ≤
Z 1
0
e−1 ≤
or
Z 1
0
2
e−x dx ≤ 1
2
e−x dx ≤ 1
Figure 5.46 illustrates this inequality.
1
F IGURE 5.46
R
2
The definite integral, 01 e−x dx,
is greater than the area of the
lower rectangle and less than the
area of the square.
1
44 CHAPTER 5 Integration
5.3 EXERCISES
11. The graph of a function g is shown in the figure below.
Concepts and Vocabulary
1. True or False The definite integral of a function f on a closed
interval [a, b], if it exists, is the limit of a Riemann sum of f on
that interval as the number of subintervals of [a, b] increases
without bound.
1
-2
2. True or False Suppose the function g is continuous and g ≤ 0
for all x in [a, b]. Then the area enclosed by the graph of g, the
x-axis, and the lines x = a and x = b is negative.
c
[ f (x) + g(x)] dx =
a
a
c · f (x) dx +
Z b
a
c · g(x) dx
Estimate
f (x) dx +
Z b
f (x) dx =
Z 4
−2
f (x) dx using six subintervals with a (a) right
10
−12
x
f (x)
14
−6
and upper estimates for
Z 30
x
f (x)
8. Find the midpoint Riemann sum for f (x) = x2 − 4, 0 ≤ x ≤ 3,
with n = 6 equal subintervals. What does the Riemann sum
represent? Illustrate with a diagram.
1
9. Consider the function f (x) = , for 1 ≤ x ≤ 2.
x
(a) Find the right Riemann sum with n = 4 equal
subinternvals. Draw a diagram to illustrate this Riemann
sum.
(b) Repeat part (a) with a midpoint Riemann sum.
10. The graph of a function f is shown in the figure below.
3
−3.4
4
−2.1
Use the table to estimate
1
5
6
7
8
9
-2
-3
Estimate
Z 10
0
f (x) dx using five subintervals with a (a) right
Riemann sum, (b) left Riemann sum, and (c) midpoint sum.
Z 9
6
0.3
7
0.9
8
1.4
9
1.8
f (x) dx using three equal
Use a midpoint Riemann sum with n equal subintervals to
approximate the definite integral.
14.
Z 9
√
sin x dx,
0
16.
Z 2
0
x
dx,
x+1
n=4
15.
Z 1p
0
n=5
17.
Z π
x3 + 1 dx,
x sin2 x dx,
0
n=5
n=4
18. Use technology to find the left and right Riemann sums for the
x
on the interval [0, 2] with n = 100 equal
function f (x) =
x+1
subintervals. Explain why these estimates show that
0.895 <
4
30
8
subintervals and a (a) right Riemann sum, (b) left Riemann
sum, and (c) midpoint sum. If the function is increasing, can
you say whether your estimates are less than or greater than the
exact value of the definite integral? Explain your reasoning.
3
2
5
−0.6
3
3π
7. Find the left Riemann sum for f (x) = cos x, 0 ≤ x ≤
, with
4
n = 6 equal subintervals. What does the Riemann sum
represent? Illustrate with a diagram.
3
26
3
13. Selected values of f (x) were obtained from an experiment and
are given in the table below.
6. Find the right Riemann sum for f (x) = x − 1, −6 ≤ x ≤ 4, with
five equal subintervals. Explain with the aid of a diagram, what
the Riemann sum represents.
2
22
1
f (x) dx.
10
Practice
1
18
−2
Suppose f is an increasing function. Use the table to fine lower
f (x) dx
even if a, b, and c are not in increasing order.
-1
-1
4
12. Selected values of f (x) are given in the table below.
a
c
a
Z b
3
Riemann sum, (b) left Riemann sum, and (c) midpoint sum.
5. True or False If f is integrable on all intervals that contain a,
b, and c, then
Z c
2
-2
4. True or False If f and g are integrable on [a, b], then
Z b
1
-1
3. True or False The function h is integrable on [a, b] if and only
if it is continuous on [a, b].
Z b
-1
Z 2
0
x
dx < 0.908
x+1
19. Use technology to construct aRtable of values of right Riemann
sums for the definite integral 0π sin x dx with n = 5, 10, 50, and
100 equal subintervals. What value to these numbers appear to
be approaching?
20. Use technology to construct a table of values of left and right
R
2
Riemann sums for the definite integral 02 e−x dx with n = 5,
10, 50, and 100 equal subintervals. Between what two numbers
must the value of the definite integral lie? Can you make a
R 2 −x2
similar statement for the definite integral −1
e dx? Explain
your reasoning.
SECTION 5.3 The Definite integral 45
Express each definite integral as a limit of a Riemann sum. Do not
evaluate the limit.
Z 3p
Z 5
1
37.
dx
x2 +
36.
4 + x2 dx
x
1
2
Express each limit as a definite integral on the given interval.
n
exi
∆x,
n→∞ ∑ 1 + xi
21. lim
[0, 1]
i=1
n
22. lim
n→∞
∑ xi
i=1
n
23.
p
1 + xi ∆x,
lim [5(x∗i )3 − 4x∗i ] ∆x,
n→∞
i=1
∑
x∗
n
24. lim
n→∞
∑ (x∗ )2i + 4 ∆x,
n
n→∞ ∑
i=1
(A)
(C)
Z 2
0
Z 4
38.
[2, 7]
sin 5x dx
[1, 3]
2+
2i
n
2
·
(2 + x)2 dx
(B)
2
(D)
x dx
2
Z 4
2
Z 2
(2 + x)2 dx
2
2x dx
-4
n
n→∞ ∑
(B) lim
i=1
n
(D) lim
n→∞
∑
i=1
"
"
1+
1+
i
n
2
i
n
2 #
#
+1
Evaluate each definite integral by interpreting it in terms of
areas.
1
n
(a)
1
n
Applications and Extensions
Z 2
0
(c)
Z 7
(4 − 2x) dx
28.
(x2 + x) dx
30.
−2
31.
Z 1
0
Z 4
1
Z 2
0
f (x) dx
(b)
f (x) dx
(d)
Z 9
0
f (x) dx
f (x) dx
4
(2x − x3 ) dx
3
2
1
AP 32. Which of the following definite integrals has a (net area) value
equal to zero?
Z 2 p
(A)
4 − x2 dx
−2
33. Let
0
(x2 − 4x + 2) dx
(x3 − 3x2 ) dx
(C)
Z 5
41. The function g is defined on the closed interval [0, 7]. The
graph of g consists of two line segments and a semicircle as
shown in the figure below.
Use the Theorem 2 to evaluate each definite integral.
Z 0
8
-2
5
29.
4
2
0
(x2 + 1) dx is equivalent to
"
#
n i 2
1
(A) lim ∑
+1
n→∞
n
n
i=1
#
" n
2
i 2
+1
(C) lim ∑
n→∞
n
n
i=1
2
x6 dx
2
2
is equivalent to
n
Z 2
Z 5
Z 10
40. The graph of f is shown in the figure below.
1
27.
39.
0
2
AP 26.
Z π
i
i=1
AP 25. lim
Express each definite integral as a limit of sums. Use a computer
algebra system to find both the sum and the limit.
[2, 5]
Z 2
−4
(x + 2) dx
(B)
Z 1
|x| dx
(D)
Z 0
(2x + 1) dx
f (x) = x2 − 3x.
−1
−1
34. Show that
a
35. Show that
Z b
a
1
x dx = (b2 − a2 ).
2
1
x2 dx = (b3 − a3 )
3
2
3
4
5
6
7
-2
-3
Use the graph to evaluate each definite integral.
(a)
R4
(a) Find an approximatin to the definite integral 0 f (x) dx
using a right Riemann sum with n = 8 equal subintervals.
(b) Draw a diagram to illustrate the approximation in part (a).
R
(c) Use Theorem 2 to evaluate 04 f (x) dx.
(d) Interpret the definite integral in part (c) as a difference of
areas and illustrate with a diagram.
Z b
1
-1
Z 2
(b)
g(x) dx
Z 6
(c)
g(x) dx
g(x) dx
0
2
0
Z 7
Evaluate each definite integral by interpreting it in terms of areas.
Z 9
Z 2
1
42.
(1 − x) dx
43.
x − 2 dx
3
0
−1
Z 3 p
Z 6
45.
9 − x2 dx
44.
(2x − 5) dx
−3
2
Z 0 p
1 + 9 − x2 dx
−3
Z 3 1 x dx
48.
−4 2
46.
47.
Z 5 −5
49.
Z 6
1
x−
p
25 − x2 dx
|x − 2| dx
46 CHAPTER 5 Integration
50.
Z 1
|2x − 1| dx
0
51.
Z 4 p
−1
63. The graph the function f is given in the figure below.
x2 − 4x + 4 dx
Evaluate each definite integral by interpreting it in terms of areas.
52. Evaluate
Z 1p
1 + x4 dx
1
Z π
53. Given that
0
3π
, what is
8
sin4 x dx =
Z b
Z 0
π
1
1
x dx = (b2 − a2 ). Use this fact
54. In Example 5 we assumed
2
a
and the properties of definite integrals to evaluate
Z 1
0
(5 − 6x2 ) dx.
55. Use the properties of definite integrals and the Example results
Z 3
in this section to evaluate
1
2
3
4
5
sin4 θ d θ ?
(2ex − 1) dx.
56. Use the Example results in this section to evaluate
Z 3
x+2
e
dx.
If F(x) =
Z x
2
f (x) dx, which of the following values is largest?
Explain your reasoning.
(A) F(0)
(B) F(1)
(D) F(3)
(C) F(2)
(E) F(4)
64. The figure below shows the graph of f . The areas of the regions
between the graph of f and the x-axis are labeled in the figure.
1
57. Use the properties of definite integrals, the Example results in
this section, and the fact that
Z π /2
0
Z π /2
0
cos x dx = 1 to evaluate
(2 cos x − 5x) dx.
58. Write the following expression as a single definite integral of
the form
Z b
a
f (x) dx.
Z 2
−2
f (x) dx +
Z 5
2
f (x) dx −
Z −1
−2
Determine the value of each definite integral or explain why the
value cannot be determined.
f (x) dx
(a)
Z e
0
59. If
60. If
Z 8
2
Z 9
0
f (x) dx = 7.3 and
f (x) dx = 37 and
0
(
2
Z 9
Z 9
61. Let f (x) =
Z 4
0
f (x) dx = 5.9, find
Z 8
4
f (x) dx.
if x < 3
g(x) dx = 16, find
x
if x ≥ 3
. Find
Z 5
0
Z 0
| f (x)| dx
Z e
f (|x|) dx
a
(e)
−e
[2 f (x) + 3g(x)] dx
3
(c)
(b)
Z d
(f)
Z −a
f (x) dx
b
Z 0
f (x) dx
(d) a
f (x) dx
−c
f (−x) dx
65. Each of the regions in the figure below bounded by the graph of
f and the x axis has area 3.
f (x) dx.
62. The graph the function f is given in the figure below.
-4
-2
2
4
3
2
Find the value of
1
-1
1
2
3
4
5
6
7
8
-1
-2
List the following quantities in increasing order, from smallest
to largest, and explain your reasoning.
(A)
Z 8
0
(D)
Z 8
4
f (x) dx
f (x) dx
(B)
Z 3
0
(E)
f (x) dx
f ′ (1)
(C)
Z 8
3
f (x) dx
Z 2
−4
[ f (x) + 2x + 5] dx
66. What is the smallest positive value of M for which the value of
Z 0
Z M
π π cos
cos
x dx will equal the value of
x dx?
5
5
−2.5
5
(A) 2.5
(B) 5
(C) 7.5
(D) 12.5
AP 67. If
Z 6
2
(A) 9
f (x) dx = 7, what is the value of
(B) 14
Z 6
2
(C) 15
[ f (x) + 2] dx?
(D) 17
SECTION 5.3 The Definite integral 47
Z 3
g(x) dx = 1 and
AP 68. Given the function g such that
1
Z 2
Z 3
2
g(x) dx = −3, what is the value of
(A) 3
(B) 3.5
Use the properties of definite integrals and the Example results to
verify each inequality.
g(x) dx?
1
(C) 4
(D) 4.5
69. Suppose f has absolute minimum value m and absolute
maximum value M. Between what two values must
Z 2
0
0
71.
(x − 4x + 4) dx ≥ 0
0
72. 2 ≤
73.
1 + x2 dx ≤
−1
π
≤
12
Z π /3
π /6
sin x dx ≤
1 + x dx
Z 1
x3 dx
√
3π
12
Z 5
1
dx
x+4
77.
Z 2
(x3 − 3x + 3) dx
79.
Z 2π
75.
0
0
76.
Z π /3
78.
Z 2
π /4
0
tan x dx
xe−x dx
0
x sin x dx ≤
and
Z 2
1
Z 2
1
arctan x dx,
0
π
Z 2
1
reasoning.
Z 0.5
0
cos(x2 ) dx or
√
cos x dx is larger? Explain your reasoning.
Hint: −| f (x)| ≤ f (x) ≤ | f (x)|
(b) Use the result of part (a) to show that
Z
Z
2π
2π
≤
f
(x)
sin
2x
dx
| f (x)| dx
0
0
Express each limit as a definite integral.
n
85. lim
n→∞
i4
Hint: Consider f (x) = x4 .
∑ n5
i=1
n
(x − 2 sin x) dx
π2
8
1
n→∞ ∑ 1 + (i/n)2
86. lim
i=1
√
arctan x dx,
arctan(sin x) dx has the largest value? Explain your
84. Suppose f is continuous on [a, b].
Zb
Z b
| f (x)| dx
f (x) dx ≤
(a) Show that a
a
Use Property 8 to estimate the value of each definite integral.
74.
Z π /2
0
√
1 + x2 dx ≤ 2 2
Z 1 p
81.
82. Which of the definite integrals
Z 0.5
Z 1√
0
26
3
83. Which of the definite integrals
2
Z 1p
1
x4 + 1 dx ≥
f (x) dx
Use the properties of definite integrals to verify the inequality
without evaluating the definite integrals.
70.
Z 3p
Extended Applications
lie? Which property of definite integrals allows you to make
this conclusion?
Z 4
80.
48 CHAPTER 5 Integration
5.4 The Fundamental Theorem of Calculus
Learning
Objectives
Essential
Knowledge
LO 1.1C
LO 2.1A
LO 3.3A
LO 3.3B(a)
EK 1.1C2
EK 2.1A3
EK 3.3A1
EK 3.2A2
EK 3.3A3
EK 3.3B1
LO 3.3B(b)
EK 3.3B2
EK 3.3B3
Review
• The area interpretation of a definite integral.
• Basic antiderivatives.
Preview
• The connection between differential and integral calculus.
• A simpler way to evaluate a definite integral.
The Fundamental Theorem of Calculus (FTC) is the most important idea in calculus.
It establishes a connection between the two branches of calculus: differential calculus
and integral calculus. Differential calculus arose from the tangent line problem and integral calculus arose from the seemingly unrelated area problem. Newton’s mentor at
Cambridge, Isaac Barrow (1630-1677), discovered that these two problems are actually
closely related. In fact, he realized that differentiation and integration are inverse processes.
The Fundamental Theorem of Calculus is an amazing result and provides the precise
inverse relationship between the derivative and the integral. It was Newton and Leibniz
who exploited this relationship and used it to develop calculus into a systematic mathematical method. In particular, they realized that the FTC enabled them to compute areas
and integrals very easily without having to find the limit of a Riemann sum.
The first part of the FTC deals with functions defined by an equation of the form
g(x) =
Z x
f (t) dt
(5)
a
where f is a continuous function on [a, b] and x varies between a and b. Note that
g depends only on x, which appears as the upper limit in the integral. If x is a fixed
number,Zthen the integral in Equation 5 is a definite number. If we let x vary, then the
x
f (t) dt also varies and, therefore, defines a function of x, denoted by g(x).
number
a
If f is a positive function, then g(x) can be interpreted as the area under the graph of f
(and above the x-axis) from a to x, where x can vary from a to b. We can interpret g as
an area so far function. See Figure 5.47.
F IGURE 5.47
A visualization of the area so far
function, g.
Example 1 Find the Area So Far
The graph of the function f is given in the figure below.
SECTION 5.4 The Fundamental Theorem of Calculus
49
2
1
-1
1
2
3
4
5
-1
-2
F IGURE 5.48
Graph of y = f (t).
Let g(x) =
Z x
0
-3
f (t) dt and find the values g(0), g(1), g(2), g(3), g(4), and g(5). Use
these values to sketch a rough graph of g.
SOLUTION
g(0) =
Z 0
0
f (t) dt = 0
By definition, ∆x = 0; no area under one value.
g(1) =
Z 1
f (t) dt =
1
·1·2 = 1
2
g(2) =
Z 2
f (t) dt =
Z 1
0
0
0
f (t) dt +
Area of a triangle.
Z 2
f (t) dt
= 1 + (1 · 2) = 3
g(3) = g(2) +
Z 3
2
Adjacent intervals; integral property.
1
g(1) plus the area of a rectangle.
f (t) dt = 3 + 1.3 = 4.3
An estimate of the area under the graph of f from 2 to 3 is 1.3.
F IGURE 5.49
A visualization of values of the area so far function.
For t > 3, f (t) is negative. Therefore, we need to start subtracting areas.
g(4) = g(3) +
g(5) = g(4) +
Z 4
3
Z 5
4
f (t) dt = 4.3 + (−1.3) = 3.0
f (t) dt = 3 + (−1.3) = 1.7
Using these values, a sketch of the graph of g is given in Figure 5.50.
50 CHAPTER 5 Integration
4
3
2
F IGURE 5.50
A sketch of g, the area so far
function.
1
1
2
3
4
5
Note that because f (t) is positive for t < 3, we add area for t < 3, and so g is increasing
up to x = 3, where it attains a maximum value. For x > 3, g is decreasing because f (t) is
negative.
Instead of estimating values and then constructing a rough sketch, it would be advantageous and practical to be able to determine a formula for the area so far function. Recall
Z b
1
x dx = (b2 − a2 ).
the result presented in the last section:
2
a
If we let f (t) = t and a = 0, then,
g(x) =
Z x
0
x2
1
t dt = (x2 − 02) =
2
2
Notice that g′ (x) = x, that is, g′ = f or the derivative of the area so far function is f .
This suggests that if g is defined as the integral of f , as in Equation 5, then g is an
antiderivative of f , well, at least in this case.
This also appears to be the case in Example 1. If we sketch the derivative of the function
g shown in Figure 5.50 by estimating slopes of tangents, then we get a graph like the one
in Figure 5.48. This also suggests that g′ = f .
To see why this amazing result might be true
Z in general, consider a continuous function
x
f such that f (x) ≥ 0. The function g(x) =
f (t) dt can be interpreted as the area of the
a
region bounded above by the graph of f , below by the x-axis, from a to x; the area so far
function. See Figure 5.47.
Let’s try to compute g′ (x) from the definition of a derivative. For h > 0, the expression
g(x + h) − g(x) is obtained by subtracting areas. So, this difference is represented by the
area under the graph of f from x to x + h, the blue area in Figure 5.51.
F IGURE 5.51
A visualization of the expression
g(x + h) − g(x).
For small values of h, Figure 5.51 suggest that this area (in blue) is approximately equal
to the area of the rectangle with height f (x) and width h:
g(x + h) − g(x)
≈ f (x)
h
Therefore, intuitively, we expect that
g(x + h) − g(x) ≈ h f (x) =⇒
g′ (x) = lim
h→0
g(x + h) − g(x)
= f (x)
h
SECTION 5.4 The Fundamental Theorem of Calculus
51
This is in fact true, even when f is not necessarily positive, and is the first part of the
Fundamental Theorem of Calculus.
The Fundamental Theorem of Calculus, Part 1
If f is a continuous function on [a, b], then the function g defined by
g(x) =
Z x
f (t) dt
a≤x≤b
a
is continuous on [a, b] and differentiable on (a, b), and g′ (x) = f (x).
A Closer Look
(1) We will abbreviate the name of this theorem as FTC1. In words, it says that the
derivative of definite integral with respect to its upper limit is the integrand evaluated
at the upper limit.
Z x
d
f (t) dt = f (x)
(2) Other notation:
dx a
Loosely speaking, this expression says that if we first integrate f and then differentiate the result, we get back the original function f . That is, integration and differentiation are inverse operations, what one does, the other undoes.
PROOF
If x and x + h are in (a, b), then
g(x + h) − g(x) =
=
=
Z x+h
f (t) dt −
Z
f (t) dt +
a
x
f (t) dt
Definition of g.
a
Z x+h
x
a
Z x+h
Z x
Z
f (t) dt −
x
f (t) dt
Poperty 5.
a
f (t) dt
Simplify.
x
For h 6= 0,
g(x + h) − g(x) 1
=
h
h
Z x+h
f (t) dt
(6)
x
For now, let’s assume that h > 0. Since f is continuous on [x, x + h], the Extreme Value
Theorem says that there are numbers u and v in [x, x + h] such that f (u) = m and f (v) =
M, where m and M are the absolute minimum and maximum values of f on [a, x + h].
See Figure 5.52.
F IGURE 5.52
Consider the absolute extreme
values on the interval [x, x + h].
Use Property 8 of definite integrals:
Rewrite this expression using f :
mh ≤
f (u)h ≤
Since h > 0, divide this inequality by h:
Z x+h
x
Z x+h
x
f (u) ≤
f (t) dt ≤ Mh
f (t) dt ≤ f (v)h
1
h
Z x+h
x
f (t) dt ≤ f (v)
52 CHAPTER 5 Integration
Use Equation 6 to replace the middle part of this inequality:
f (u) ≤
g(x + h) − g(x)
≤ f (v)
h
(7)
For h < 0, we can argue in a similar manner to obtain this same inequality.
As h → 0, both u → x and v → x since u and v lie between x and x + h.
Therefore, since f is continuous at x:
lim f (u) = lim f (u) = f (x)
u→x
h→0
lim f (v) = lim f (v) = f (x)
and
v→x
h→0
Using Equation 7 and the Squeeze Theorem,
g′ (x) = lim
h→0
g(x + h) − g(x)
= f (x)
h
(8)
If x = a or b, then Equation 8 can be interpreted as a one-sided limit. We can then show
that g is continuous on [a, b].
Example 2 FTC, Part 1
Find the derivative of each function.
Z xp
Z
(a) g(x) =
1 + t 2 dt
(b) h(x) =
0
x
1
sin t
dt
1 + t2
SOLUTION
√
(a) f (t) = 1 + t 2 is continuous. Use the Fundamental Theorem of Calculus, Part 1.
Z x p
p
d
1 + t 2 dt = 1 + x2
g′ (x) =
dx 0
sint
is continuous. Use the FTC1.
1 + t2
Z x
d
sint
sin x
′
h (x) =
dt =
2
dx 1 1 + t
1 + x2
(b) f (t) =
Example 3 Fresnel Function
Admittedly, the expression g(x) =
Z x
f (t) dt seems like a strange way to define a func-
a
The Fresnel function is an example in
which the definition is not in closed
form. For a given value x, the Fresnel
funciton cannot be evaluated (exactly)
in a finite number of standard
operations.
tion. However, functions of this form are common in physics, chemistry, and statistics.
For example, the Fresnel function
2
Z x
πt
dt
sin
S(x) =
2
0
named after the French physicist Augustin Fresnel (1788-1827) is used in the study of
optics and recently has been applied to the design of highways.
The Fundamental Theorem of Calculus Part 1 tells us how to differentiate the Fresnel
function:
Z x 2 2
πt
πx
d
′
dt = sin
sin
S (x) =
dx 0
2
2
This means we can use all the methods of differential calculus to analyze the function S.
2
Z x
πx
f (t) dt.
and the Fresnel function S(x) =
Figure 5.53 shows the graph of f (x) = sin
2
0
This figure suggests that S(x) is indeed the area under the graph of f from 0 to x (until
x ≈ 1.4 when S(x) becomes a difference of areas). Figure 5.54 shows a larger part of the
graph of S.
SECTION 5.4 The Fundamental Theorem of Calculus
53
1.0
0.5
0.5
1
2
-5 -4 -3 -2 -1
-0.5
1
2
3
4
-0.5
-1.0
-1.0
F IGURE 5.53
2
Graph of f (x) = sin π2x and
2
Z x
πt
sin
S(x) =
dt.
2
0
F IGURE 5.54
Graph of the Fresnel function.
Start with the graph of S in Figure 5.53 and think about what its derivative should look
like. It seems reasonable that S′ (x) = f (x). For example, S is increasing when f (x) >
0 and decreasing when f (x) < 0. These graphs provide a visual confirmation of the
Fundamental Theorem of Calculus Part 1.
Example 4 FTC1 and The Chain Rule
"Z 4
#
x
d
sect dt .
Find
dx 1
SOLUTION
Because the upper bound is a function of x, we need to use the Chain Rule together with
the FTC1.
"Z 4
#
Z u
x
d
d
sect dt
sect dt =
Let u = x4 .
dx 1
dx 1
Z u
d
du
=
sect dt
Chain Rule.
du 1
dx
= sec u
du
dx
FTC1.
= sec(x4 ) · 4x3
Use u = x4 .
In the previous section we evaluated several definite integrals using the definition involving Riemann sums. This process can be long and tedious. The second part of the
Fundamental Theorem of Calculus provides a much simpler method for the evaluation
of definite integrals.
The Fundamental Theorem of Calculus, Part 2
If f is continuous on [a, b], then
Z b
a
f (x) dx = F(b) − F(a)
where F is any antiderivative of f , that is, a function such that F ′ = f .
A Closer Look
(1) We will abbreviate this theorem as FTC2.
(2) This theorem says that the value of
Z b
a
f (x) dx can be obtained by finding an an-
tiderivative F of the integrand f , then, subtracting, in the proper order, the values of
F at the endpoints of [a, b].
54 CHAPTER 5 Integration
PROOF
Let g(x) =
Z x
f (t) dt.
a
From the FTC1, g′ (x) = f (x). Therefore, g is an antiderivative of f .
If F is any other antiderivative of f on [a, b], we know that F and g differ by a constant:
F(x) = g(x) + C
(9)
for a < x < b. Since both F and g are continuous on [a, b], if we take the limits of both
sides of Equation 9 (as x → a+ and x → b− ) then this equation is also true for x = a and
x = b. Therefore, F(x) = g(x) + C for all x in [a, b].
Evaluate g at x = a: g(a) =
Z a
f (t) dt = 0
a
Use Equation 9 with x = b and x = a:
F(b) − F(a) = [g(b) + C] − [g(a) + C]
= g(b) − g(a) = g(b) =
Z b
f (t) dt
a
The FTC2 is an amazing result. The definite integral, which was defined by a methodical
but sometimes complicated procedure involving Riemann sums, can now be evaluated
by simply evaluating an antiderivative, F, at only two points, a and b.
The FTC2 is certainly reasonable if we interpret it in physical terms. If v(t) is the velocity
of an object and s(t) is its position at time t, then v(t) = s′ (t), so s is an antiderivative of
v. In Section 5.2 we considered an object that always moves in the positive direction and
made the guess that the area under the velocity curve is equal to the distance traveled. In
symbols:
Z b
a
v(t) dt = s(b) − s(a)
This is exactly what the FTC2 says in this context.
Example 5 FTC2
Evaluate the definite integral
Z 3
ex dx
1
SOLUTION
The function f (x) = ex is continuous everywhere and F(x) = ex is an antiderivative.
Compare this calculation with the
more complicated one in Section 5.2.
Use the FTC2:
Z 3
1
If we use F(x) = ex +C, the constant
C just cancels out in the subtraction.
ex dx = F(3) − F(1) = e3 − e
Note that the FTC2 says we can use any antiderivative F of f . So, we may as well use
the simplest one, that is, F(x) = ex , instead of ex + 7 or ex + C.
We often use the notation
ib
F(x) = F(b) − F(a)
a
So the equation of the FTC2 can be written as
Z b
ib
f (x) dx = F(x)
where
a
a
b
h
ib
Other common notations are F(x) and F(x) .
a
F′ = f
a
Example 6 Area Under a Curve
Find the area of the region R bounded above the graph of y = x2 , below by the x-axis,
and between the lines x = 0 and x = 1.
SECTION 5.4 The Fundamental Theorem of Calculus
55
SOLUTION
Figure 5.55 shows the region R.
1.0
0.5
F IGURE 5.55
Find the area of the region R.
0.5
We can represent the area of the region R with a definite integral:
1.0
Z 1
x2 dx
0
1
An antiderivative of f (x) = x2 is F(x) = x3 .
3
Use the FTC2:
1
Z 1
13 03
1
x3
=
−
=
x2 dx =
Area of R =
3 0
3
3
3
0
Example 7 Remembering Common Antiderivatives
Evaluate
Z 6
dx
3
x
SOLUTION
The given integral is an abbreviation for
Z 6
1
3
x
dx
1
is F(x) = ln |x|.
x
Since 3 ≤ x ≤ 6, we can use F(x) = ln x. Use the FTC2.
Z 6
i6
1
6
dx = ln x = ln 6 − ln3 = ln = ln 2
3
3
3 x
Example 8 Area Under The Cosine Curve
An antiderivative of f (x) =
Find the area of the region R bounded above by the graph of y = cos x, below by the
π
x-axis, and between the lines x = 0 and x = b, where 0 ≤ b ≤ .
2
SOLUTION
An antiderivative of f (x) = cos x is F(x) = sin x. Use the FTC2.
Z b
ib
cos x dx = sin x = sin b − sin0 = sin b
0
0
π
π
is
If b = , then we have shown that the area under the cosine curve from 0 to
2
π 2
= 1 (an amazing result).
sin
2
56 CHAPTER 5 Integration
F IGURE 5.56
The area of the region bounded
by the graph of y = cos x, the
x-axis, between x = 0 and
x = π /2 is exactly 1!
-1
Example 9 Don’t Forget The Assumptions
Why is the following calculation wrong?
3
Z 3
1
1
x−1
4
= − −1 = −
dx =
2
−1 −1
3
3
−1 x
SOLUTION
We know this result must be wrong because the answer is negative but f (x) =
and Property 6 of integrals says that
Z b
a
1
≥0
x2
f (x) dx ≥ 0 when f (x) ≥ 0.
The FTC2 applies only to continuous functions.
1
is not continuous at x = 0 ∈ [−1, 3].
x2
Z 3
1
dx using the FTC2.
Therefore we cannot evaluate
2
−1 x
This definite integral does not exist.
f (x) =
Differentiation and Integration as Inverse Processes
We often write the two parts of the FTC together.
The Fundamental Theorem of Calculus
Suppose f is continuous
on [a, b].
Z x
f (t) dt, then g′ (x) = f (x).
(1) If g(x) =
a
(2)
Z b
a
f (x) dx = F(b) − F(a), where F is any antiderivative of f , that is, F ′ = f .
Earlier in this section we wrote
d
dx
Z
x
a
f (t) dt = f (x)
which says that if we integrate f and then differentiate the result, we end up at the original
function f .
In the FTC2, since F ′ (x) = f (x), we can write
Z b
a
F ′ (x) dx = F(b) − F(a)
This notation suggests that if we start with a function F, first differentiate and then integrate, we end up at the original function F, but in the form F(b) − F(a). Taken together,
these two parts of the FTC say that differentiation and integration are inverse processes.
What one does, the other undoes.
SECTION 5.4 The Fundamental Theorem of Calculus
57
The Fundamental Theorem of Calculus is absolutely the most important theorem in calculus and, indeed, it ranks as one of the great accomplishments of the human mind. Before it was discovered, from the time of Eudoxus and Archimedes to the time of Galileo
and Fermat, problems of finding areas, volumes, and lengths of curves were extremely
difficult. But now, armed with the systematic method that Newton and Leibniz developed
out of the Fundamental Theorem of Calculus, many of these challenging problems are
readily accessible.
58 CHAPTER 5 Integration
5.4 EXERCISES
Concepts and Vocabulary
7. Let g(x) =
1. Explain in your own words what is meant by the statement that
“differentiation and integration are inverse processes.”
Z x
0
f (t) dt, where f is the function whose graph is
shown below.
2. True or False If F is any antiderivative of f , where f is
continuous on [a, b], then
Z b
a
4
3
2
f (x) dx = F(b) − F(a)
1
3. True or False If f is continuous on an interval containing a,
then
#
"Z 2
x
d
f (t) dt = f (x2 )
dx a
4. True or False If
Z b
a
Z b
a
f (x) dx = F(b) − F(a) and
f (x) dx = G(b) − G(a) then F ′ (x) = G′ (x).
5. True or False If f is continuous on its entire domain, then
Z x
Z x
d
d
f (t) dt =
f (t) dt
dx a
dx b
-1
1
2
3
4
5
6
7
-1
-2
-3
(a) Evaluate g(0), g(1), g(2), g(3), and g(6).
(b) On what interval is g increasing?
(c) Where does g have a maximum value? Explain your
reasoning.
(d) Sketch a rough graph of g.
8. Let g(x) =
Z x
0
f (t) dt, where f is the function whose graph is
shown below.
for any a and b in the domain of f .
Practice
6. Let g(x) =
3
Z x
0
f (t) dt, where f is the function whose graph is
2
shown below.
1
1
3
2
3
4
5
6
-1
2
-2
1
-3
1
2
3
4
5
6
7
-1
-2
(a) Evaluate g(x) for x = 0, 1, 2, 3, 4, 5, and 6.
(b) Estimate g(7).
(c) Where does g have a maximum value? Where doe it have
a minimum value? Explain your reasoning.
(d) Sketch a rough graph of g.
-4
Evaluate g(0) and g(6).
Estimate g(x) for x = 1, 2, 3, 4, and 5.
On what interval is g increasing?
Where does g have a maximum value? Explain your
reasoning.
(e) Sketch a rough graph of g.
(f) Use the graph in part (e) to sketch the graph of g′ 9x).
Compare with the graph of f .
(a)
(b)
(c)
(d)
Sketch the area represented by g(x). Then find g′ (x) in two ways:
(a): by using Part 1 of the Fundamental Theorem of Calculus and (b)
by evaluating the integral using Part 2 and then differentiating.
9. g(x) =
Z x
1
t 2 dt
10. g(x) =
Z x
0
(2 + sin t) dt
Use Part 1 of the Fundamental Theorem of Calculus t find the
derivative of each function.
Z x
Z xp
12. g(x) =
ln(1 + t 2 ) dt
11. g(x) =
t + t 3 dt
1
0
Z u √
Z s
t
13. g(s) = (t − t 2 )8 dt
14. h(u) =
dt
0 t +1
5
SECTION 5.4 The Fundamental Theorem of Calculus
15.
Z 0√
1 + sec t dt
x
Hint:
1 + sec t dt = −
Z 2
t 3 sint dt
17. h(x) =
Z ex
lnt dt
19. g(x) =
Z 3x+2
21. g(x) =
Z π /4
16. R(y) =
Z x√
0
(A) k2 − k
1 + sec t dt
AP 50. Let g(x) =
y
18. h(x) =
1
1
√
"Z
t
dt
1 + t3
θ tan θ d θ
Z √x
1
z2
dz
4
z +1
20. g(x) =
Z x4
22. g(x) =
Z 1 p
x
−x2
(B)
(C)
2x
1 + x2
26.
(x2 + 2x − 4) dx
Z 2
4
3
2
t3 − t2 + t
5
4
5
0
28.
Z 9
√
25.
−1
dt
27.
(D)
1
1 + x4
Z π
π /6
sin θ d θ
32.
Z 1
34.
Z 4
2 + x2
0
√
36.
Z π /2
38.
Z 1
π /6
40.
v4
Z 1
y2
46.
Z 1/√2
1/2
47.
Z π
0
48.
Z 2
√
4
1 − x2
f (x) dx
35.
Z 2
(3u − 2)(u + 1) du
Z π /3
39.
Z 3
dy
csc2 θ d θ
Z 18 r
3
z
Z √3
45.
Z 4
√
1/ 3
2
3
4
5
-2
(2 sin x − ex ) dx
43.
-3
(a) Find G′ (2).
(b) Find all value of x for which G has a relative maximum.
Justify your answer.
52. Let h(x) =
Z x2
0
f (t) dt where f is the function with domain
[0, 10] and whose graph is shown below.
dz
8
dx
1 + x2
2s ds
0
5
dx
where
f (x) =
(
f (x) dx where
(
2
if − 2łx ≤ 0
f (x) =
2
4−x
if 0 < x ≤ 2
−2
π /4
1
-1
√
(4 − t) t dt
1
Z 3 3
y − 2y2 − y
f (t) dt where f is the function whose graph is
-1
Z 4
41.
(xe + ex ) dx
1
e dx
37.
(D) 9.5
shown below.
33.
0
dv
0
1
−1
(1 + r)3 dr
Z 2 3
v + 3v6
0
44.
dx
csct cott dt
1
42.
x
Z x2
x−2/3 dx
0
3
What is the value of g(0) + g(1) + g(2) + g(3)?
(A) 4.5
(B) 5
(C) 6
2
Z 5
2
-1
Z 8
−5
(u + 2)(u − 3) du
1
0
31.
x100 dx
1
(1 − 8v3 + 16v7 ) dv
1
1
30.
2
Z 1
0
29.
x dx
shown below.
51. Let G(x) =
Z 1
(D) ln k
f (t) dt where f is the function whose graph is
-1
4x
1 + x4
(C) ln |k2 − k|
1
Evaluate each definite integral.
1
0
1
1
−
k2 k
1 + t 2 dt
Applications and Extensions
24.
Z x
cos2 θ d θ
sin x
dx =
(B)
0
#
1
dt =
1 + t2
x2
(A) 0
Z 3
x
k
Z 0√
x
d
AP 23.
dx
Z k2
1
AP 49. For k > 0,
sin x
cos x
if 0 ≤ x <
if
π
2
59
π
2
≤x≤π
10
(a) Find h′ (x).
(b) Find all values of x, if any, for which h as a relative
minimum. Justify your answer.
60 CHAPTER 5 Integration
53. Consider the function A defined by A(k) =
where 0 ≤ k ≤ 5.
Z k
0
x·
p
25 − x2 dx,
56. The graph of f (t) = |2t − 10| − 4 is shown in the figure below.
dA
(a) For what value of k on the interval 0 ≤ k ≤ 5 is
a
dk
maximum? Justify your answer.
(b) The line x = k moves right so that k increases 2 units per
dk
= 2. How fast is A changing at the
second, that is,
dt
instant k = 3?
8
6
4
2
54. Let A be the region in the first quadrant bounded by the axes,
x+7
the graph of y = √
, and the line x = k, as shown in the
x2 + 16
figure below.
-2
2
4
6
8
10
-2
-4
Let F(x) =
Z x
1
f (t) dt. Find all values of m such that F(m) = 0.
Sketch the region enclosed by the given graphs and calculate its area.
√
57. y = x, y = 0, x = 4
58. y = x3 ,
If k increases at the rate of 5 units per minute, how fast is the
area of A changing when k = 3?
55. Let R be the region in the first quadrant bounded by the axes,
π sin
x
2 , and the line x = k, as shown in the
the graph of y =
x
figure below.
y = 0,
59. y = 4 − x2 ,
60.
x=1
y=0
y = 2x − x2 ,
y=0
Use a graph to give a rough estimate of the area of the region that
lies beneath the graph of the function given, above the x-axis, and
between the indicated interval.
√
61. y = 3 x, 0 ≤ x ≤ 27
62. y = x−4 ,
1≤x≤6
63. y = sin x,
0≤x≤π
2
64. y = sec x,
0≤x≤
π
3
Evaluate the integral and interpret it as a difference of areas.
Illustrate with a sketch.
The vertical line x = k moves at a rate of 0.5 units per second.
How fast is the area of R increasing when k = 1.5?
√
√
√
√
2
2
2
2
(B)
(C)
(D)
(A)
3
4
6
8
65.
Z 2
−1
x3 dx
66.
Z 2π
π /6
cos x dx
Explain what is wrong in each equation.
67.
Z 1
−2
68.
x−4 dx =
Z 2
4
2
dx = − 2
x3
x
−1
69.
70.
x−3
−3
Z π
π /3
Z π
0
1
=−
2
=
−2
−1
3
8
3
2
sec θ tan θ d θ = sec θ
iπ
sec2 x dx = tan x = 0
iπ
π /3
= −3
0
Find the derivative of each function.
71. g(x) =
Hint:
Z 3x 2
u −1
2x
u2 + 1
Z 3x 2
u −1
2x
u2 + 1
du
du =
Z 0 2
u −1
2x
u2 + 1
du +
Z 3x 2
u −1
0
u2 + 1
du
SECTION 5.4 The Fundamental Theorem of Calculus
72. g(x) =
Z 1+2x
74. F(x) =
Z 2x
t sint dt
1−2x
√
x
76. Let f (x) =
0
75. g(x) =
Z sin x
t2
e dt
x
arctant dt
Z x
73. F(x) =
Z x2
cos x
ln(1 + 2v) dv
(1 − t 2 ) dt. Find the intervals on which the graph
of f is increasing.
t2
dt. Find the intervals on which the
2
0 t +t +2
graph of g is concave down.
Z xp
7 + r2 dr
78. Let F be the function defined by F(x) = −4x +
77. Let g(x) =
Z x
2
for all x. Which of the following statements is true?
(A) F(2) = 0.
(B) F has a critical point at x = 3.
(C) F is increasing for x > 0.
(D) The graph of F has no point of inflection.
79. The continuous function g is defined on the interval
−4 ≤ x ≤ 10. The graph of g consists of three line segments
and a semicircle, as shown in the figure below.
4
2
-4
-2
2
4
6
8
10
81. Let F(x) =
-6
(a) What is the average value of g for −2 ≤ x ≤ 4?
(b) What is the average rate of change of g for −2 ≤ x ≤ 4?
(c) Suppose J is the function such that J ′ (x) = g(x), and
J(−3) = 5. Find J(8).
Z x
g(t) dt. Find the absolute minimum value
Z x
g(t) dt. Find the x-coordinates of all points
0
of F on the interval −4 ≤ x ≤ 10. Justify your answer.
0
of inflection on the graph of F.
(f) Let H(x) =
80. Let F(x) =
Z x
1
Z x2
0
2
83. The error function defined by
Z x
2
2
erf(x) = √
e−t dt
π 0
is used in probability, statistics, and engineering.
Z b
2
1√
π [erf(b) − erf(a)]
(a) Show that
e−t dt =
2
a
2
(b) Show that the function y = ex erf(x) satisfies the
2
differential equation y′ = 2xy + √
π
84. The Fresnel function S was defined in Example 3 and graphed
in Figures 5.53 and 5.54.
(a) At what values of x does this function have local
maximum values? Justify your answer.
(b) Find the intervals on which the graph of this function is
concave up.
(c) Use technology to solve the following equation:
2
Z x
πt
sin
dt = 0.2
2
0
85. The sine integral function is defined as
Si(x) =
-4
(e) Let F(x) =
2
et dt. Find an equation of the tangent line to the
graph of y = F(x) at the point with x-coordinate 2.
Z sin x p
Z y
82. Let f (x) =
f (x) dx. Find
1 + t 2 dt and g(y) =
0
3 Z
π 4
f ′ (x) dx = 17,
. If f (1) = 12, f ′ is continuous, and
g′′
6
1
what is the value of f (4)?
-2
(d) Let F(x) =
Z x
g(t) dt. Find H ′ (3) and H ′′ (3).
f (t) dt, where f is the function whose graph is
Z x
sint
0
t
dt
and is important in electrical engineering. Note that the
sint
is not defined when t = 0. However, its
integrand f (t) =
t
limit is 1 as t → 0. So we define f (0) = 1 and this makes f a
continuous function everywhere.
(a) Sketch the graph of Si.
(b) At what values of x does this function have local
maximum values? Justify your answer.
(c) Find the coordinates of the first inflection point to the right
of the origin.
(d) Does the graph of this function have horizontal
asymptotes? Explain your reasoning.
(e) Use technology to solve the following equation:
Z x
sint
dt = 1
t
0
Let g(x) =
shown below.
Z x
0
f (t) dt, where f is the function whose graph is shown.
(a) At what values of x do the local maximum and minimum
values of g occur?
(b) Where does g attain its absolute maximum value? Find the
intervals on which the graph of g is concave down.
-3
-2
-1
1
2
Find the intervals on which the graph of F is concave down.
61
(c) Sketch the graph of g.
62 CHAPTER 5 Integration
96. Find a function f and a number a such that
86.
3
6+
2
Z x
f (t)
t2
0
√
dt = 2 x
for all x > 0
1
2
4
6
97. The area of the region labeled B is three times the area of the
region labeled A. Find an expression for b in terms of a.
8
-1
-2
-3
87.
1
3
5
7
9
-
-
Evaluate each limit by interpreting the sum as a Riemann sum for a
function defined on [0, 1].
n i4
i
88. lim ∑
−
5
n→∞
n2
i=1 n
r
r
r
r !
n
1
1
2
3
n
+
+
+··· +
89. lim ∑
n→∞
n
n
n
n
i=1 n
Extended Applications
90. Justify Equation 7 for the case h < 0.
91. If f is continuous and g and h are differentiable functions, find
a formula for
Z h(x)
d
f (t) dt
dx g(x)
p
92. (a) Show that 1 ≤ 1 + x3 ≤ 1 + x3 for x ≥ 0.
Z 1p
(b) Show that 1 ≤
1 + x3 dx ≤ 1.25
0
2
93. (a) Show that cos(x ) ≥ cos x for 0 ≤ x ≤ 1.
Z π /6
1
cos(x2 ) dx ≥
(b) Deduce that
2
0
94. Show that
0≤
Z 10
5
x2
x4 + x2 + 1
dx ≤ 0.1
by comparing the integrand to a simpler function.
95. Let
and g(x) =

0



 x
f (x) =

2−x



0
Z x
0
if x < 0
if 0 ≤ x ≤ 1
if 1 < x ≤ 2
if x > 2
f (t) dt
(a) Find an expression for g(x), as a piecewise defined
function.
(b) Sketch the graphs of f and g.
(c) Where is f differentiable? Where is g differentiable?
98. A manufacturing company owns a major piece of equipment
that depreciates at the (continuous) rate f = f (t), where t is the
time measured in months since its last overhaul. Because a
fixed cost A is incurred each time the machine is overhauled,
the company wants to determine the optimal time T (in
months) between overhauls.
(a) Explain why
Z t
0
f (s) ds represents the loss in value of the
machine over the period of time t since the last overhaul.
(b) Let C = C(t) be given by
Z t
1
C(t) =
f (s) ds
A+
t
0
What does C represent and why would the company want
to minimize C?
(c) Show that C has a minimum value at the numbers t = T
where C(T ) = f (T ).
99. A high-tech company purchases a new computing system
whose initial value is V . The system will depreciate at the rate
f = f (t) and will accumulate maintenance costs at the rate
g = g(t), where t is the time measured in months. The company
wants to determine the optimal time to replace the system.
(a) Let
Z
1 t
C(t) =
[ f (s) + g(s)] ds
t 0
Show that the critical numbers of C occur at the numbers t
where C(t) = f (t) + g(t).
(b) Suppose that

 V − V t if 0 < t ≤ 30
15 450
f (t) =

0
if t > 30
V t2
for t > 0.
12,900
Determine
Z the length of time T for the total depreciation
and g(t) =
t
D(t) =
0
f (s) ds to equal the initial value V .
(c) Determine the absolute minimum of C on (o, T ].
(d) Sketch the graph sof C and f + g on the same coordinate
axes, and verify the result in part (a) in this case.
SECTION 5.5 Indefinite Integrals
63
5.5 Indefinite Integrals
Learning
Objectives
Essential
Knowledge
LO 3.1A
LO 3.3B(a)
EK 3.1A1
EK 3.1A2
EK 3.3B1
LO 3.3B(b)
EK 3.3B2
LO 3.4A
LO 3.4C
LO 3.4E
EK 3.3B3
EK 3.4A1
EK 3.4A2
EK 3.4C1
EK 3.4E1
Review
• Formulas for antiderivatives.
• The Fundamental Theorem of Calculus.
Preview
• Calculation and notation for a general antiderivative.
• The integral of a rate of change is the net change.
The Fundamental Theorem of Calculus is a very powerful method for evaluating the
definite integral of a function, assuming that we can find an antiderivative of the function.
In this section we introduce a notation for antiderivatives, review the basic formulas for
antiderivatives, and use them to evaluate definite integrals. We also apply the FTC2 to
science and engineering problems, and in general to problems involving a rate of change.
Indefinite Integrals
Both parts of the Fundamental Theorem of Calculus establish a connection between antiderivatives and definite integrals. Part 1 says that if f is continuous, then the function
Z x
f (t) dt is an antiderivative of f , and Part 2 says that
Z b
f (x) dx can be found by
a
a
simply evaluating F(b) − F(a), where F is an antiderivative of f .
The FTC suggests that we need a convenient notation for antiderivatives, so that they
are easy to work with in connection with the FTC. Because ofZthe relationship between
antiderivatives and integrals given by the FTC, the notation
f (x) dx is traditionally
used for an antiderivative of f and is called an indefinite integral. Therefore,
Z
f (x) dx = F(x)
means F ′ (x) = f (x)
For example, using this notation we can now write
Z
x3
x dx = + C
3
2
because
d x3
+ C = x2
dx 3
An indefinite integral is a very general antiderivative, representing an entire family of
functions (one antiderivative for each value of the constant (C).
It is important to carefully distinguish between definite and indefinite integrals. A definite integral, for example,
example,
Z
Z b
f (x) dx, is a number. However, an indefinite integral, for
a
f (x) dx, is a function (or family of functions). The connection between these
two is given by the FTC2: If f is continuous on [a, b], then
Z
Z b
ib
f (x) dx = f (x) dx
a
a
We can only use the Fundamental Theorem of Calculus effectively and efficiently if we
have a toolbox full, or supply, of antiderivatives of functions. Therefore, Table 5.5 shows
a summary of antidifferentiation formulas from Section 5.1, together with a few more,
using the notation of indefinite integrals. Any formula in this table can be verified by
differentiating the function on the right side to obtain the integrand. For example,
Z
sec2 x dx = tan x + C because
d
(tan x + C) = sec2 x
dx
64 CHAPTER 5 Integration
Z
Z
Z
Z
Z
Z
Z
Z
c f (x) dx = c
Z
Z
f (x) dx
[ f (x) + g(x)] dx =
Z
f (x) dx +
Z
g(x) dx
k dx = kx + C
xn dx =
xn+1
+C
n+1
Z
1
dx = ln |x| + C
x
Z
bx
+C
bx dx =
ln b
(n 6= −1)
ex dx = ex + C
Z
sin x dx = − cos x + C
Z
sec2 x dx = tan x + C
Z
sec x tan x dx = sec x + C
Z
1
dx = tan−1 x + C
2
x +1
cos x dx = sin x + C
csc2 x dx = − cotx + C
csc x cotx dx = − csc x + C
1
√
dx = sin−1 x + C
1 − x2
TABLE 5.5
Table of Indefinite Integrals
Recall that the most general antiderivative on a given interval is obtained by adding a
constant to a particular antiderivative. We will use the convention that when a formula
for a general antiderivative is given, it is valid only on an interval. Therefore, it is
customary to write, for example,
Z
1
1
dx = − + C
x2
x
with the understanding that this expression is valid on the interval (0, ∞) or on the interval (−∞, 0). This is true despite the fact that the general antiderivative of the function
1
f (x) = 2 , x 6= 0, is
x
 1

 − + C1 if x < 0
x
F(x) =

1
 − + C if x > 0
2
x
Example 1 Indefinite Integral
Find the general indefinite integral
Z
(10x4 − 2 sec2 x) dx
SOLUTION
Use Table 5.5 and the convention discussed above.
A formula in Table 5.5 allows us to
split this integral into two separate
integrals; integrate term by term.
Then remember that constants pass
freely through integral symbols.
Z
(10x4 − 2 sec2 x) dx = 10
= 10
Z
x4 dx − 2
Z
sec2 x dx
x5
− 2 tan x + C
5
= 2x5 − 2 tan x + C
It’s easy to check the answer here. The derivative must be the integrand. The graph of
the indefinite integral is shown in Figure 5.57 for several values of C. In this example,
the value of C is the y-intercept.
SECTION 5.5 Indefinite Integrals
65
2
-1
F IGURE 5.57
The graph of the indefinite
integral for several values of C.
1
-2
-4
Example 2 Indefinite Integral
Z
cos θ
dθ
sin2 θ
SOLUTION
Evaluate
This indefinite integral isn’t immediately apparent in Table 5.5. Use trigonometric identities to rewrite the integrand, then use Table 5.5.
Z
Z cos θ
cos θ
1
dθ
dθ =
sin θ
sin θ
sin2 θ
=
Z
csc θ cot θ d θ = − csc θ + C
Example 3 Definite Integral
Evaluate
Z 3
0
(x3 − 6x) dx.
SOLUTION
Use the FTC2 and Table 5.5.
3
Z 3
x4
x2
(x3 − 6x) dx =
−6
4
2 0
0
1 4
1 4
2
2
=
·3 −3·3 −
·0 −3·0
4
4
=
81
27
− 27 − 0 = −
4
4
Table 5.5.
FTC2.
Simplify.
Example 4 Definite Integral and Interpretation
Z 2
3
3
2x − 6x + 2
Find
dx and interpret the result in terms of areas.
x +1
0
SOLUTION
Use the FTC2 and Table 5.5.
2
Z 2
3
x4
x2
3
−1
2x − 6x + 2
dx = 2 − 6 + 3 tan x
x +1
4
2
0
0
2
1
= x4 − 3x2 + 3 tan−1 x
2
0
1 4
2
−1
(2 ) − 3(2 ) + 3 tan 2 − 0
=
2
= −4 + 3 tan−1 2
This is the exact value of the definite integral. We can use technology to obtain an
66 CHAPTER 5 Integration
approximation, as shown in Figure 5.58.
4
2
2
-2
F IGURE 5.58
An approximation to the definite
integral.
F IGURE 5.59
Graphical interpretation of the definite integral:
net area.
Figure 5.59 shows the graph of the integrand. The value of the definite integral can be
interpreted as a net area: the sum of the areas labeled with a plus sign minus the area
labeled with a minus sign.
Example 5 Definite Integral
√
Z 9 2
2t + t 2 t − 1
Evaluate
dt
t2
1
SOLUTION
Rewrite the integrand by carrying out the division. Then use Table 5.5.
√
Z 9 2
Z 9
2t + t 2 t − 1
(2 + t 1/2 − t −2 ) dt
dt
=
t2
1
1
9
3/2
−1
t
t 
2 3/2 1 9
= 2t +
−
 = 2t + t +
3
−1
3
t 1
2
1
2
2
1
1
= 2 · 9 + · 93/2 +
− 2 · 1 + · 13/2 +
3
9
3
1
2
292
1
= 18 + 18 + − 2 + + 1 =
9
3
9
Divide.
Table 5.5.
FTC2.
Simplify.
Note that some CAS machines are able to compute and return this exact answer. See
Figure 5.60.
F IGURE 5.60
The exact value of the definite
integral using technology.
Applications
The Fundamental Theorem of Calculus Part 2 says that if f is continuous on [a, b] and F
is any antiderivative of f , then
SECTION 5.5 Indefinite Integrals
Z b
67
f (x) dx = F(b) − F(a)
a
Since F is an antiderivative of f , then F ′ = f , so the conclusion of the FTC2 can be
rewritten as
Z b
a
F ′ (x) dx = F(b) − F(a)
We have learned that F ′ (x) represents the rate of change of y = F(x) with respect to
x, and F(b) − F(a) is the change in y as x changes from a to b. Remember however,
that y could, for example, increase, then decrease, and then increase again. Therefore,
F(b) − F(a) really represents the net change in y. Using this interpretation, the FTC2
can be restated in terms of the net change.
Net Change Theorem
The definite integral of a rate of change (F ′ ) is the net change (in the original function
F):
Z b
a
F ′ (x) dx = F(b) − F(a)
A Closer Look
(a) Another way to interpret the definite integral
Z b
a
F ′ (x) dx: this represents an accu-
mulation of the change in F over the interval [a, b].
(b) We can rearrange the terms in the Net Change Theorem to provide an alternate interpretation and practical approach for solving many problems.
=
F(b)
|{z}
End amount
F(a)
| {z}
+
Start amount
Z b
F ′ (x) dx
| {z }
Net change
a
(c) This principle has many uses and can be applied to all rates of change in the natural
and social science previously discussed.
Here are some examples of the Net Change Theorem in action.
(1) Let V ′ (t) represent the rate at which water flows into a reservoir at time t. Then V (t)
is the volume of water in the reservoir at time t and
Z t2
t1
V ′ (t) dt = V (t2 ) − V (t1 )
is the net change in the amount of water in the reservoir between time t1 and time t2 .
(2) If C(t) is the concentration of the product of a chemical reaction at time t, then the
rate of reaction is the derivative C′ (t). Then
Z t2
t1
C′ (t) dt = C(t2 ) − C(t1 )
is the change in the concentration of the product from time t1 to time t2 .
(3) If the mass of a rod measured from the left end to a point x is m(x), then the linear
density is ρ (x) = m′ (x). Therefore,
Z b
a
ρ (x) dx = m(b) − m(a)
is the mass of the segment of the rod that lies between x = a and x = b.
(4) If the rate of growth of a population is P′ (t), then
68 CHAPTER 5 Integration
Z t2
t1
P′ (t) dt = P(t2 ) − P(t1)
is the net change in population during the time period from t1 to t2 . The population
increases when births occur and decreases when deaths happen. The net change
takes into account both actions: births and deaths.
(5) If C(x) is the cost of producing x units of a commodity, then the marginal cost is the
derivative C′ (x). Therefore,
Z x2
x1
C′ (x) dx = C(x2 ) − C(x1 )
is the increase in cost when production is increased from x1 units to x2 units; the net
change in cost.
(6) If a particle moves along a horizontal line so that its position at time t is given by
s(t), then its velocity is v(t) = s′ (t), and
Z t2
t1
v(t) dt = s(t2 ) − s(t1 )
(10)
is the net change in position, or displacement of the particle over the time period
from t1 to t2 . We have already seen examples in which this was true for a particle
moving only in a positive direction However, now we can interpret Equation 10 as
the net change in position of the particle over the interval [t1 ,t2 ].
(7) If we want the total distance traveled by an object over a time interval, then, theoretically, we have to consider separately the intervals where v(t) ≥ 0 (the particle is
moving to the right) and the intervals where v(t) ≤ 0 (the particle is moving to the
left). However, we can simplify this process if we consider the function |v(t)|, the
speed of the object. Therefore,
Z t2
t1
|v(t)| dt = total distance traveled
(11)
Figure 5.61 shows how both displacement and total distance traveled can be interpreted in terms of areas under a velocity curve.
F IGURE 5.61
The displacement and total
distance traveled can be
visualized in terms of the area
under the graph of a velocity
function.
displacement =
Z t2
t1
v(t) = A1 − A2 + A3
total distance traveled =
Z t2
t1
|v(t)| dt = A1 + A2 + A3
(8) The acceleration of an object is a(t) = v′ (t). Therefore,
Z t2
t1
a(t) dt = v(t2 ) − v(t1)
is the net change in velocity over the time interval [t1 ,t2 ].
SECTION 5.5 Indefinite Integrals
69
Example 6 Displacement versus Total Distanced Traveled
A particle moves along a horizontal line so that its velocity at time t, t ≥ 0, is given by
v(t) = t 2 − t − 6, where v is measured in meters per second.
(a) Find the displacement of the particle during the time interval [1, 4].
(b) Find the total distance traveled during this same time period.
SOLUTION
(a) Use Equation 10 to find the displacement.
s(4) − s(1) =
=
Z 4
1
v(t) dt =
Z 4
t3 t2
− − 6t
3
2
1
(t 2 − t − 6) dt
Net Change Theorem.
4
Table 5.5.
1
3
43 42
1
12
9
=
− −6·4 −
− −6·1 = −
3
2
3
2
2
FTC2.
This means that the particle’s net change in position over the time interval [1, 4] is
4.5 meters to the left.
(b) In order to find the total distance traveled, we need to integrate |v(t)|.
Therefore, we need to find the intervals on which v(t) ≥ 0 and those on which
v(t) ≤ 0.
v(t) = t 2 − t − 6 = (t − 3)(t + 2)
v(t) ≤ 0 on the interval [1, 3] and v(t) ≥ 0 on [3, 4].
Use Equation 11 to find the total distance traveled.
Z 4
1
Z 4
v(t) dt
Split the integral into two parts,
one where v(t) ≤ 0 and one
where v(t) ≥ 0.
Z 4
Use the expression for v(t).
|v(t)| dt =
Z 3
[−v(t)] dt +
=
Z 3
−(t 2 − t − 6) dt +
1
1
3
3
(t 2 − t − 6) dt
3
3 3
4
t
t
t2
t2
= − + + 6t +
− − 6t
3
2
3
2
1
3
Table 5.5.
61
FTC2; simplify.
= 10.167
6
Figure 5.62 shows a technology solution for the total distance traveled.
=
F IGURE 5.62
A technology solution: the
absolute value of the velocity is
the integrand.
Example 7 End Velocity
A particle moves
along a horizontal line so that its acceleration at time t, t ≥ 0, is given
√
by a(t) = t t 4 + t, where t is measured in seconds and s is in feet. If its velocity at time
t = 2 is 5, what is its velocity at time t = 4?
SOLUTION
70 CHAPTER 5 Integration
The velocity at time t = 2 is 5: v(2) = 5.
We need the velocity at time t = 4: v(4).
Use a rearrangement of terms in the Net Change Theorem.
v(4) = v(2) +
= 5+
Z 4
a(t) dt
Net Change Theorem rearranged.
2
Z 4 p
t
t 4 + t dt
Use given information.
2
= 5 + 60.989 = 65.989
Use technology to obtain a numerical solution.
Figure 5.63 shows a technology solution.
F IGURE 5.63
A technology solution in two
steps: The start amount plus the
net change (in velocity).
Example 8 Power Consumption
On a certain day in September in the city of San Francisco, the rate, in megawatts per
hour, at which power is consumed is given P(t), where t is measured in hours past midnight. The graph of P is shown in the figure below. Estimate the total power used on that
day.
SOLUTION
Since P is the rate of power consumption and P(t) ≥ 0, we need to find
total accumulation of power used over the 24 hour period.
Z 24
P(t) dt, the
0
There is no formula for P(t), so we cannot find a closed form expression for the integral.
Therefore, we will approximate the value of the definite integral using a midpoint Riemann sum with 12 equal subintervals.
Z 24
0
P(t) dt ≈ [P(1) + P(3) + P(5) + · · ·+ P(21) + P(23)]∆t
= (440 + 400 + 420 + · · ·+ 670 + 550)(2)
Midpoint Riemann sum.
Read values from the graph.
= 15,840
The total power consumed over the time interval [0, 24] was approximately 15,840
SECTION 5.5 Indefinite Integrals
megawatts.
71
A Note on Units
In problems involving net change, it is important to report the correct units. In Example
7, the definite integral
Z 24
P(t) dt is defined as the limit of the sum of terms of the form
0
P(ti∗ ) ∆t. The rate, P(ti∗ ), is measured in megawatts per hour and ∆t is measured in hours.
megawatts
Therefore, the product is measured in megawatts:
· (hour) = megawatts
hour
In general, the unit of measurement for the definite integral
the unit for f (x) and the unit for x.
Z b
a
f (x) dx is the product of
72 CHAPTER 5 Integration
5.5 EXERCISES
Concepts and Vocabulary
29.
1. True or False Every indefinite integral has unique numerical
value.
Z 2
(2x − 3)(4x2 + 1) dx
30.
Z π
(5ex + 3 sin x) dx
32.
0
31.
0
2. True or False If F and G are any two antiderivatives of f (x),
ib
ib
then F(x) = G(x) .
a
3. True or False
a
If C′ (x)
represents the marginal cost of
Z b
producing x units of a commodity, then
a
4. True or False If a particle moves along a horizontal line with
velocity v(t), then
35.
Z 1
0
37.
√
u
39.
Z 2
x
Z 1
0
t1
41.
2
−
2 x
43.
Verify that each formula is correct by differentiation.
√
Z
1 + x2
1
√
5.
+C
dx = −
2
2
x
x 1+x
6.
7.
8.
Z
Z
Z
1
1
cos x dx = x + sin 2x +C
2
4
47.
tan2 x dx = tan x − x +C
49.
√
x a + bx dx =
Find each general indefinite integral.
(x1.3 + 7x2.5 ) dx
Z 2
3
11.
5 + x2 + x3 dx
3
4
9.
Z
(u + 4)(2u + 1) du
√
Z
1+ x+x
15.
dx
x
13.
17.
19.
21.
Z
Z
Z
1+r
dr
r
(2 + tan2 θ ) d θ
2t (1 + 5t ) dt
10.
Z p
4
x5 dx
Z 22.
Z
5
3
sin 2x
dx
sin x
Evaluate each definite integral.
Z 3
−2
27.
(x2 − 3) dx
Z 0 1 4 1 3
t + t − t dt
−2
2
4
26.
Z 2
(4x3 − 3x2 + 2x) dx
Z 3
(1 + 6w2 − 10w4 ) dw
1
28.
0
38.
Z 1
2
dt =
t2
2
(A) t 3 + +C
t
2
3
(C) t − +C
t
−
x2
4
x3
dx
4
dp
1
+
p2
0
Z 4√
y−y
36.
dy
y2
1
40.
(5x − 5x ) dx
Z π /4
sec θ tan θ d θ
0
42.
Z π /3
sin θ + sin θ tan2 θ
sec2 θ
Z 5
1
2ex +
44.
dx
x
1
dθ
0
48.
(x − 2|x|) dx
Z Z 1
dt
Z 2
(x − 1)3
x2
1
Z 2
0
50.
|2x − 1| dx
Z 3π /2
0
dx
| sin x| dx
3t 2 +
2
u − 2u − u +
12.
du
Z 9
7
x+1
Z √
√ dx =
AP 52.
x
4
14.
t(t 2 + 3t + 2) dt
44
19
Z (B)
(A)
1
2
3
2
x +1+ 2
16.
dx
x +1
Z π /4
Z AP 53.
(cos x − sin x) dx =
1+r 2
0
18.
dr
√
r
√
2
Z
(B) 2 − 1
(A)
2
20. sect(sect + tant) dt
6
Find each general indefinite integral. Illustrate by graphing several
members of the family in the same viewing window.
Z Z
1
cos x + x dx
23.
24. (ex − 2x2 ) dx
2
25.
Z 2
Z 2
1
34.
46.
t4 − 1
t(1 − t)2 dt
1
dr
√
1 − r2
Z 1/√3 2
t −1
−1
AP 51.
Z
√
dt
3 2
t
0
2
(3bx − 2a)(a + bx)3/2 +C
15b2
dθ
Z 8
2+t
Z √3/2
−1
0
cos2 θ
0
2
dx
Z π /4
1 + cos2 θ
1
45.
(x10 + 10x ) dx
0
Practice
du
√
√
x( 3 x + 4 x) dx
1
v(t) dt represents the total distance
traveled by the particle over the time interval [t1 ,t2 ].
Z 4
4 + 6u
1
C′ (x) dx represents
the cost of producing b − a additional units of that commodity.
Z t2
33.
Z 1
(B) 6t −
4
+C
t3
(D) t 3 + 2 ln |t| +C
(C)
1456
15
(D)
1
2
(C)
√
2+1
√
(D) 2 2
Applications and Extensions
54. Use technology to estimate the x-intercepts of the graph of
f (x) = 1 − 2x − 5x4 . Use this information to estimate the area
of the region that lies under the graph of f and above the x-axis.
55. Use technology to estimate the x-intercepts of the graph of
f (x) = (x2 + 1)−1 − x4 . Use this information to estimate the
area of the region that lies under the graph of f and above the
x-axis.
SECTION 5.5 Indefinite Integrals
56. Let R be the region that is bounded on the left by the y-axis and
the right by the graph of x = 2y − y2 ; the shaded region in the
figure below.
73
63. Suppose R′ (t) is a marginal revenue function, the derivative of
the revenue function R(x), where x is the number of units sold.
Explain the meaning of
Z 5000
1000
R′ (x) dx.
64. Suppose f (x) is the slope of a mountain trail at a distance of x
miles from the start of the trail. Explain the meaning of
Z 5
2
3
f (x) dx.
65. If x is measured in meters and f (x) is measured in newtons,
what are the units for
1
Z 100
f (x) dx?
0
66. If the units for x are feet and the units for a(x) are pounds per
da
foot, what are the units for
? What are the units for
dx
Z
8
a(x) dx?
2
1
The area of R is given by the definite integral
Z 2
0
Find the area of the region R.
(2y − y2 ) dy
57. Let R be the region bounded above by the line y = 1, below by
√
the graph of y = 4 x, the y-axis, and the line x = 1, as shown in
the figure below.
AP 67. Which of the following definite integrals are undefined?
Z 1
Z 2
(A)
(C)
−1
1
dx
1 + x2
Z 1/2
−1/2
√
1
1 − x2
0
(D)
dx
Z 2
1
1
x2
1
dx
(x − 1)2
The velocity function (in meters per second) is given for a particle
moving along a line. Find (a) the displacement and (b) the total
distance traveled by the particle during the given time interval.
68. v(t) = 3t − 5,
1
(B)
0≤t ≤3
69. v(t) = t 2 − 2t − 3,
2≤t ≤4
The acceleration function (in m/s2 ) and the initial velocity are given
for a particle moving along a line. Find (a) the velocity at time t and
(b) the distance traveled during the given time interval.
70. a(t) = t + 4,
71. a(t) = 2t + 3,
1
v(0) = 5,
0 ≤ t ≤ 10
v(0) = −4,
0≤t ≤3
Find the area of the region R by writing x as a function of y and
AP 72. A particle moves along a horizontal line so that its velocity at
integrating with respect to y.
2
time t, t ≥ 0, is given by v(t) = 2et /10 , and the position of the
′
58. Suppose w (t) is the rate of growth of a child in pounds per
particle is given by s(t). It is known that s(2) = 5. Find the
Z 10
′
position
of the particle at time t = 5.
year. Explain the meaning of
w (t) dt.
5
59. Suppose r′ (t) is the rate at which water flows over a dam in
cubic feet per second. Explain the meaning of
Z 60
0
′
r (t) dt.
60. Suppose r(t) is the rate at which water flows into a tank and
d(t) is the rate at which water flows out of the tank at time t,
both measured in cubic feet per hour. Explain the meaning of
Z 6
3
[r(t) − d(t)] dt.
61. Suppose r′ (t) is the rate at which oil leaks from a tank in
gallons per minute at time t. Explain the meaning of
Z 120
r(t) dt.
0
62. A honeybee population starts with 100 bees and increases at a
rate of n′ (t) bees per week. Explain the meaning of
100 +
Z 15
0
n′ (t) dt.
AP 73. A particle moves along a horizontal line so that its velocity at
time t is given by v(t) = 2t − 3, in meters per second. The
distance traveled by the particle in meters over the time interval
from t = 1 to t = 4 is
(A) 6
(B) 8.25
(C) 6.5
(D) 2.25
74. The linear density of a rod of length 4 meters is given by
√
ρ (x) = 9 + 2 x measured in kilograms per meter, where x is
measured in meters from one end of the rod. Find the total
mass of the rod.
75. Water flows from the bottom of a storage tank at a rate of
r(t) = 200 − 4t liters per minute, where 0 ≤ t ≤ 50. Find the
amount of water that flows from the tank during the first 10
minutes.
74 CHAPTER 5 Integration
76. The velocity of a car was read from its speedometer at
10-second intervals and the values are given in the table below.
t (s)
v (mi/h)
t (s)
v (mi/h)
0
10
20
30
40
50
0
38
52
58
55
51
60
70
80
90
100
56
53
50
47
45
Use a midpoint Riemann sum to estimate the distance traveled
by the car over the interval [0, 100].
77. Suppose that a volcano is erupting and r(t) is the rate at which
solid materials are spewed into the atmosphere, where r(t) is
measured in tonnes per second and t is measured in seconds.
Selected values for r(t) are given in the table below and r(t) is
increasing over the interval [0, 6].
t
r(t)
0
2
1
10
2
24
3
36
4
46
5
54
6
60
(a) Find upper and lower estimates for the total quantity Q(6)
of erupted materials after six seconds.
(b) Use a midpoint Riemann sum to estimate Q(6).
78. The marginal cost of manufacturing x yards of a certain fabric
is
C′ (x) = 3 − 0.01x + 0.000006x2
in dollars per yard. Find the increase in cost if the production
level is raised from 2000 yards to 4000 yards.
81. Lake Lanier in Georgia, USA, is a reservoir created by Buford
Dam on the Chattahoochee River. The table below shows the
rate of inflow of water, in cubic feet per second, as measured
every morning at 7:30 AM by the US Army Corps of Engineers.
Day
Inflow rate
(ft3 /s)
July 18
5275
July 19
6401
July 20
2554
July 21
4249
July 22
3016
July 23
3821
July 24
2462
July 25
2628
July 26
3003
Use a midpoint Riemann sum to estimate the amount of water
that flowed into Lake Lanier from July 18th, 2013, at 7:30 AM
to July 26th at 7:30 AM.
82. A bacteria population is 4000 at time t = 0 and its rate of
growth is 1000 · 2t bacteria per hour at time t hours. What is the
population after one hour?
83. The function D is the data throughput, the rate at which data
flows through an Internet provider’s TI data line, measured in
megabits per second. A graph of the function D for midnight to
8:00 AM is shown in the figure below.
79. Water flows into and out of a storage tank. A graph of the rate
of change r(t) of the volume of water in the tank, in liters per
day, is hown in the figure below.
Use a midpoint Riemann sum to estimate the total amount of
data transmitted during that time period.
If the amount of water in the tank at time t = 0 is 25,000 L, use
a midpoint Riemann sum to estimate the amount of water in the
tank four days later.
80. The graph of the acceleration a(t) of a care, measured in ft/s2 ,
is shown in the figure below.
Use a midpoint Riemann sum to estimate the increase in the
velocity of the car during the six-second time interval.
SECTION 5.5 Indefinite Integrals
84. On December 9, 2004 in Ontario, Canada, the rate, in
megawatts per hour, at which power was consumed is given by
P(t), where t is measure in hours past midnight. The graph of P
is shown in the figure below.
Estimate the total power used on that day.
75
76 CHAPTER 5 Integration
5.6 The Method of Substitution
Learning
Objectives
Essential
Knowledge
LO 2.2A
EK 2.2A1
EK 2.2A2
Ek 2.2A3
EK 3.1A1
EK 3.1A2
EK 3.3B5
LO 3.1A
LO 3.3B(b)
Review
• Basic antidifferentiation formulas.
• The Chain Rule.
Preview
• The Substitution Rule.
Because the Fundamental Theorem of Calculus allows us to evaluate a definite integral
very quickly, it is important to be able to find antiderivatives. The basic antidifferentiation
formulas are useful, but don’t help us evaluate an integral like
Z
p
(12)
2x 1 + x2 dx
To find the indefinite integral, we use the problem-solving strategy of introducing something extra. Here the something extra is a new variable; we change from the variable
x to a new variable u (from the x world to the u world). In other words, we transform,
or reduce, the given indefinite integral into an equivalent expression in another variable,
preferably into a form found in the Table of Integrals.
It’s not clear yet why this is a good
choice for u.
Here is how this process works. Define u in terms of the variable x; here let u = 1 + x2
(the expression under the root sign). In general, given a function u = g(x), we call du and
dx differentials and the relationship between them is given by du = g′ (x) dx. Therefore,
du
the differential of u is du = 2x dx. Solve for the differential dx: dx =
2x
Use these expressions to transform the integral from the variable x to the variable u.
Z
Z
p
√
Use the expression for u.
2x 1 + x2 dx = 2x u dx
=
Z
=
2 3/2
u +C
3
=
2
(1 + x2)3/2 + C
3
√ du Z √
2x u
u du
=
2x
Use the expression for dx; simplfiy.
Basic antidifferentiation formula.
Use the expression for u to rewrite the solution in
terms of the original variable, x.
We can check this answer by using the Chain Rule to differentiate.
p
2 3
d 2
2 3/2
(1 + x ) + C = · (1 + x2)1/2 · 2x = 2x 1 + x2
dx 3
3 2
A Closer Look
(1) The expression for u in terms of the variable x is the connection, or bridge, between
the two integrals. It takes practice to cleverly define u in terms of x so that the
transformed integral is indeed reduced.
(2) The final answer in an indefinite integral problem, like the one above, is written in
terms of the original variable. Use the connection between u and x to return to the
original variable.
(3) The constant C associated with the antiderivative in u’s is actually different from the
constant C associated with the antiderivative in x’s. Well, we have to allow for the
possibility that it is different. However, we usually use the same symbol, C, for the
constant in both worlds.
(4) As we change variables, or transform the integral form all x’s to all u’s, there is often
an intermediate step that involves both variables.
SECTION 5.6 The Method of Substitution 77
In
Z general, this method works whenever we have an integral that we can write in the form
f (g(x)) g′ (x) dx. If F ′ = f , then
Z
F ′ (g(x)) g′ (x) dx = F(g(x)) + C
(13)
because, by the Chain Rule,
d
[F[g(x))] = F ′ (g(x)) g′ (x)
dx
If we make the change of variable or substitution u = g(x), then from Equation 13 we
have
Z
F ′ (g(x)) g′ (x) dx = F(g(x)) + C = F(u) + C =
or writing F ′ = f , we get
Z
f (g(x)) g′ (t) dx =
This proves the following rule.
Z
Z
F ′ (u) du
f (u) du
The Substitution Rule
If u = g(x) is a differentiable function whose range is an interval I and f is continuous
on I, then
Z
f (g(x)) g′ (x) dx =
Z
f (u) du
A Closer Look
(1) The Substitution Rule for integration was proved (above) using the Chain Rule for
differentiation.
(2) If u = g(x), then du = g′ (x) dx, so a way to remember and use the Substitution Rule
is to think of dx and du as differentials.
(3) The Substitution Rule says: It is permissible to operate with dx and du after the
integral signs as if they were differentials. Practically this means we can treat dx and
du as if they are variables that we can multiple and divide by, etc.
(4) When using the Substitution Rule, a good choice for u = g(x) is one in which the
derivative, g′ (x), occurs in the integral, except for a constant factor.
Example 1 From the x World to the u World, and Back
Find
Z
x3 cos(x4 + 2) dx
SOLUTION
This isn’t a basic antidifferentiation formula. However, notice that the derivative of the
inner function, x4 + 2, is 4x3 , and this occurs in the integrand, except for the constant.
Therefore, a reasonable substitution is u = x4 + 2.
Then, du = 4x3 dx =⇒ dx =
du
4x3
78 CHAPTER 5 Integration
Z
x3 cos(x4 + 2) dx =
Z
x3 cos u
=
Z
1
1
cosu du =
4
4
=
1
sin u + C
4
=
1
sin(x4 + 2) + C
4
Remember, we can check the answer
by differentiation.
du
4x3
Change variables.
Z
cos u du
Simplify; constants pass freely through
integral symbols.
Basic antidifferentiatoin formula.
Final answer in terms of the original
variable x. The important concept behind the Substitution Rule is to replace a relatively complicated
integral by a simpler, equivalent integral. This is done by transforming the original integral in x Zto a reduced integral in u. In Example 1 we replaced
Z a relatively complicated
1
3
4
integral, x cos(x + 2) dx, by a much simpler integral,
cos u du.
4
A reminder that the challenge in using the Substitution Rule is to select an appropriate
substitution, that is, a good function u = g(x). One helpful strategy is to choose u to
be some function in the integrand whose derivative also occurs (except for a constant
factor). This was the case in Example 1. If that strategy fails, try choosing u to be some
complicated part of the integrand, perhaps the inner function in a composite function.
Finding the right substitution is an art. It is not unusual to start with a bad guess for u.
If that happens, then the resulting integral, in u, may be even more difficult, or it may be
impossible to transform the entire integrand into the variable u. So, if your first guess
doesn’t work, just try another substitution.
Example 2 Substitution
Z √
Evaluate
2x + 1dx
SOLUTION
Let u = 2x + 1 =⇒ du = 2 dx =⇒ dx =
Z √
2x + 1dx =
=
=
Common Error
In using the Substitution Rule, the
antiderivative is left in terms of u.
When using the Substitution Rule, the
final answer must be in terms of the
original variable.
Z
1
du
2
Z
√ 1
1
u1/2 du
u · du =
2
2
Change variables; simplify.
1
1 u3/2
+ C = u3/2 + C
·
2 3
3
2
Basic antidifferentiation formula.
1
(2x + 1)3/2 + C
3
Final answer in terms of x.
There may be other substitutions that also work.
√
√
dx
=⇒ dx = 2x + 1du = u du
Let u = 2x + 1 =⇒ du = √
2x + 1
Z √
2x + 1dx =
=
Z
u · u du =
Z
u2 du
Change variables; simplify.
1
u3
+ C = (2x + 1)3/2 + C
3
3
Final answer in terms of x.
Example 3 Substitution Too
Z
x
√
Find
dx
1 − 4x2
SOLUTION
Let u = 1 − 4x2 =⇒ du = −8x dx =⇒ dx = −
du
8x
SECTION 5.6 The Method of Substitution 79
Z
x
√
dx =
1 − 4x2
Z
du
1
x
√ ·−
=−
u
8x
8
Z
u−1/2 du
Change variables; simplify.
1 √
= − (2 u) + C
8
1p
=−
1 − 4x2 + C
4
Basic antidifferentiation formula.
Final answer in terms of x.
We can check this answer by differentiation. However, in this case, let’s verify the solux
,
tion graphically. Figure 5.64 shows the graph of both the integrand, f (x) = √
1 − 4x2
p
1
and the indefinite integral, g(x) = −
1 − 4x2 (let C = 0).
4
0.5
-1.0
-0.5
0.5
-0.5
F IGURE 5.64
A graph of both the integrand
and the definite integral.
-1.0
Notice that g(x) decreases when f (x) is negative, increases when f (x) is positive, and
has it minimum value when f (x) = 0. So, the graphical evidence suggests that g is an
antiderivative of f .
Example 4 Substitution Involving an Exponential Function
Find
Z
e5x dx
SOLUTION
Let u = 5x =⇒ du = 5 dx =⇒ dx =
Z
e5x dx =
=
Z
eu ·
du 1
=
5
5
Z
du
5
eu du
Change variables, simplify.
1 u
1
e + C = e5x + C
5
5
Basic formula,final answer in terms of x.
With some practice and experience, it may be possible to evaluate integrals like those
in Example 1-4 without making an explicit substitution. Look for the pattern: The integrand is the product of the derivative of an outer function and the derivative of the inner
function.
Here is an alternate way to solve Example 1.
Z
x3 cos(x4 + 2) dx =
Z
cos(x4 + 2) · x3 dx =
1
4
Z
cos(x4 + 2) · (4x3) dx
The derivative of the inner function, x4 + 2, is 4x3 . We need a 4 in the integrand. Multiply by 1 in a convenient
1
form, and since constants pass freely through integral symbols, place the 4 and in appropriate places.
4
Z
1
1
d
cos(x4 + 2) dx · (x4 + 2) dx = sin(x4 + 2) + C
4
dx
4
Similarly, the solution to Example 4 could be written as:
=
Z
e5x dx =
1
5
Z
1
5 · e5x dx = e5x + C
5
80 CHAPTER 5 Integration
It is advisable to use an explicit substitution, especially in more complicated problems.
This makes it easier to manage constants and the expressions in x and in u.
Example 5 Reduced Integral, Really
Z p
Find
1 + x2 x5 dx.
SOLUTION
The substitution here isn’t so easy to spot. It’s subtle.
Write x5 = x4 · x and then use a typical substitution approach.
Let u = 1 + x2 =⇒ du = 2x dx =⇒ dx =
du
2x
Also, x2 = u − 1 =⇒ x4 = (u − 1)2
Z p
Z p
1 + x2 x5 dx =
1 + x2 x4 · x dx
Factor x5 as x4 · x.
=
Z
√
du
u(u − 1)2 x
2x
Change variables.
=
1
2
Z
√ 2
u(u − 2u + 1) du
Simplify; expand
binomial.
=
1
2
Z
(u5/2 − 2u3/2 + u1/2) du
1
=
2
=
Distribute.
2 7/2
2 5/2 2 3/2
u −2· u + u
+C
7
5
3
Basic
antidifferentiation
formula.
2
1
1
(1 + x2 )7/2 − (1 + x2)5/2 + (1 + x2)3/2 + C
7
5
3
Final answer in terms
of x. Example 6 Hidden Substitution
Find
Z
tan x dx.
SOLUTION
There doesn’t seem to be any possible substitution here.
Write tangent in terms of sine and cosine.
Z
tan x dx =
Z
sin x
dx
cos x
This suggests the substitution u = cos x =⇒ du = − sin x dx =⇒ dx = −
Z
tan x dx =
Z
=−
sin x
dx =
cos x
Z
Z
sin x
u
du
−
sin x
1
du = − ln |u| + C
u
du
sin x
Change variables.
Simplify; basic antidifferentiation formula.
= − ln | cos x| + C
Final answer in terms of x.
The previous Example presents an important result, and can be added to our list of
known, or basic, antidifferentiation formulas.
Z
tan x dx = − ln | cos x| + C
= ln(| cos x|−1 ) = ln
= ln | sec x| + C
1
| cos x|
(14)
SECTION 5.6 The Method of Substitution 81
Definite Integrals
There are two possible methods to evaluate a definite integral when the method of substitution is used. One method is to evaluate the indefinite integral fist, and then use the
Fundamental Theorem. That is, temporarily ignore the limits of integration, evaluate the
indefinite integral in terms of the original variable, and then use the FTC2 to find the final
answer (in the x world).
Here’s how this method works using the result from Example 2.
4
Z √
Z 4√
Find the indefinite integral first.
2x + 1dx =
2x + 1dx
0
0
=
=
1
(2x + 1)3/2
3
4
Change variables, integrate, return
to the variable x.
0
26
1 3/2 1 3/2 1
(9) − (1) = (27 − 1) =
3
3
3
3
Use the FTC2 in terms of x.
Another method, which is usually preferable, is to transform the limits of integration
also, that is, write, and use, the appropriate bounds in terms of the new variable.
This rule says that when using the
method of substitution in a definite
integral, we need to transform
everything in terms of the new
variable u, not only x and dx, but also
the limits of integration. The new
limits of integration are the values of
u that correspond to x = a and x = b.
The Substitution Rule for Definite Integrals
If g′ is continuous on [a, b] and f is continuous on the range of u = g(x), then
Z b
a
f (g(x)) g′ (x) dx =
Z g(b)
f (u) du
g(a)
PROOF
Let F be an antiderivative of f .
Then F(g(x)) is an antiderivative of f (g(x)) g′ (x).
Use the Fundamental Theorem of Calculus Part 2.
Z b
ib
f (g(x) g′ (x) dx = F(g(x)) = F(g(b)) − F(g(a))
a
a
Apply the FTC2 a second time.
Z g(b)
ig(b)
f (u) du = F(u)
= F(g(b)) − F(g(a))
g(a)
g(a)
A Closer Look
(1) This substitution rule, a change of variables in definite integrals, requires a change
of corresponding limits of integration via the equation u = g(x), which relates the
variables:
x = a =⇒ u = g(a);
x = b =⇒ u = g(b)
(2) If the Substitution Rule for Definite Integrals is used, the notation must identify the
correct limits of integration associated with the given variable.
Here are some other examples of appropriate notation.
82 CHAPTER 5 Integration
Z b
a
=
The differential dx indicates the variable of integration is x. Therefore, the
limits of integration are implicitly understood to mean x = a and x = b.
f (g(x)) g′ (x) dx
Z x=b
The differential du indicates the variable of integration is u. The limits of
integration are labeled to identify their association with the variable x.
f (u) du
x=a
Using the FTC2, F ′ = f ; u is the variable in the antiderivative so we still
need the labels x = a and x = b.
x=b
= F(u) x=a
b
= F(g(x)) a
The variable x is in the antiderivative so the labels x = a and x = b are no
longer required.
Example 7 Definite Integral and Substitution
Evaluate
Z 3
1
(x2 + 1)3 x dx
SOLUTION
Common Error
Change variables but keep the limits
of integration associated with x.
A change of variables in definite
integrals requires a change of
corresponding limits of integration.
The derivative of the inner function, x2 + 1, occurs in the integrand, except for a constant.
Therefore, a reasonable choice to change variables is u = x2 + 1.
Here is all the information, the consequences of this choice for u, we will need to change
variables from x to u.
u = x2 + 1
du = 2x dx
du
dx =
2x
Z 3
1
x = 1 : u = 12 + 1 = 2
x = 3 : u = 32 + 1 = 10
(x2 + 1)3 x dx =
Z 10
2
u3 x
du 1
=
2x
2
Z 10
u3 du
2
Change variables;
remember the limits of integration.
10
1 u4
·
2 4 2
1 104 24
= 1248
−
=
2
4
4
=
Basic antidifferentiation formula.
FTC2; simplify.
Remember, by using the Substitution Rule for Definite Integrals, we do not have to return
to the variable x after integrating. We simply evaluate the expression in terms of u, using
the appropriate values of u.
Example 8 Definite Integral and Substitution
Evaluate
Z 2
1
SOLUTION
dx
(3 − 5x)2
The inner function in the denominator is a good choice for u.
u = 3 − 5x
du = −5 dx
1
dx = − du
5
x = 1 : u = 3 − 5 · 1 = −2
x = 2 : u = 3 − 5 · 2 = −7
SECTION 5.6 The Method of Substitution 83
Z 2
1
Z −7
1
Z
1
1 −7 du
−
du
=
−
2
5
7 −2 u2
−2 u
1 −7
1 −7
1
=
=− −
5
u −2
5u −2
1
1 1
1
=
− +
=
5
7 2
14
dx
=
(3 − 5x)2
Change variables; notice that the bounds
become reversed in terms of u. That’s
OK. Continue with the FTC.
Basic antidifferentiation formula;
simplify.
FTC2.
Example 9 Definite Integral and an Interpretation
Find
Z e
ln x
x
SOLUTION
dx
1
The choice for u isn’t too clear here because there doesn’t appear to be an inner function.
However, the derivative of ln x also occurs in the integrand.
u = ln x
1
du = dx
x
dx = x du
Z e
ln x
1
x
x = 1 : u = ln 1 = 0
x = e : u = ln e = 1
dx =
=
=
Z 1
u
x
2 1
0
u
2
· x du =
Z 1
u du
Change variables.
0
Basic antidifferentiaton forumla.
0
12 02 1
−
=
2
2
2
FTC2; simlify.
ln x
Since the function f (x) =
is positive for x > 1, the definite integral represents the
x
area of the shaded region in Figure 5.65.
0.50
F IGURE 5.65
The definite integral represents
the area of the region bounded
ln x
,
above by the graph of y =
x
below by the x-axis, and between
the lines x = 1 and x = e.
Symmetry
The next theorem uses the Substitution Rule for Definite Integrals to simplify the calculation of integrals of functions that exhibit certain symmetry properties.
Integrals of Symmetric Functions
Suppose f is continuous on [−a, a]. Z
(a) If f is even, f (−x) = f (x), then
a
−a
(b) If f is odd, f (−x) = − f (x), then
Z a
f (x) dx = 2
−a
Z a
0
f (x) dx = 0.
f (x) dx.
84 CHAPTER 5 Integration
PROOF
Split the integral into two parts.
Z a
−a
f (x) dx =
Z 0
f (x) dx +
−a
Z a
0
f (x) dx = −
Z −a
0
f (x) dx +
Z a
0
f (x) dx
(15)
Use the substitution u = −x in the first integral on the far right side.
u = −x
x = 0 : u = −(0) = 0
du = −dx
x = −a : u = −(−a) = a
dx = −du
−
Z −a
f (x) dx = −
0
Z a
f (−u)(−du) =
0
Z a
0
f (−u) du
Use this result to write rewrite Equation 15
Z a
−a
f (x) dx =
Z a
0
f (−u) du +
Z a
0
f (x) dx
(16)
(a) If f is even then f (−u) = f (u). Equation 16 becomes
Z a
−a
f (x) dx =
Z a
0
f (u) du +
Z a
0
f (x) dx = 2
Z a
0
f (x) dx
(b) If f is odd, then f (−u) = − f (u). Equation 16 becomes
Z a
−a
f (x) dx = −
Z a
0
f (u) du +
Z a
0
f (x) dx = 0
Figures 5.66 and 5.67 illustrates the results in the theorem for Integrals of Symmetric
Functions.
F IGURE 5.66
If
Z
Z f is an even function,
a
a
−a
F IGURE 5.67
Z
If f is an odd function,
f (x) dx = 2
0
a
−a
f (x) dx.
f (x) dx = 0.
If f is positive and even, part (a) says that the area under the graph of y = f (x) from −a
to a is twice the area from 0 to a because of symmetry. Remember that an integral of the
form
Z b
f (x) dx represents net area: the area above the x-axis and below the graph of
a
y = f (x) minus the area below the x-axis and above the curve. Part (b) says the integral
is 0 because the areas cancel.
Example 10 Symmetry Simplifies
Evaluate each definite integral.
(a)
Z 2
−2
(x6 + 1) dx
SOLUTION
(b)
Z 1
tan x
dx
2
4
−1 1 + x + x
SECTION 5.6 The Method of Substitution 85
(a) Let f (x) = x6 + 1
f (−x) = (−x)6 + 1 = x6 + 1 = f (x)
Therefore, f is an even function.
Z 2
−2
6
(x + 1) dx = 2
Z 2
(x6 + 1) dx
0
2
1 7
x +x
7
0
7
128
284
0
=2
+2 −
+0 =
7
7
7
=2
(b) Let f (x) =
f (−x) =
Integral of a symmetric
function.
Integrate term by term; basic
antidifferentiation formulas.
FTC2; simplify.
tan x
1 + x2 + x4
− tan x
tan(−x)
=
= − f (x)
1 + (−x)2 + (−x)4
1 + x2 + x4
Therefore, f is an odd function.
Z 1
tan x
dx = 0
1
+
x2 + x4
−1
86 CHAPTER 5 Integration
5.6 EXERCISES
Concepts and Vocabulary
1. True
or False If u = g(x), then
Z
f ′ (g(x)) · g′ (x) dx = f (u) +C.
2. True or False If u = x2 + 1, then
to
Z
Z
p
x x2 + 1 dx is equivalent
u du.
Z 4
1 4
to
u du.
Z 1
0
Z 2
4. True or False
−2
Z 2 p
p
x3 x4 + 1 dx = 2
x3 x4 + 1 dx
0
Evaluate each indefinite integral by using the indicated substitution.
7.
8.
Z
Z
Z
2
x e−x dx,
u = −x2
p
x3 + 1 dx,
sin2 θ cos θ d θ ,
17.
19.
Z
Z
Z
21.
Z
23.
Z
25.
Z
27.
29.
31.
33.
Z
Z
Z
Z
cos(1 + 5t) dt
38.
Z
39.
(1 − 2x)9 dx
cos
πt dt
2
sin x cos x dx
Z
Z
44.
cot x dx
46.
Z
Z
42.
sin 2x
dx
1 + cos2 x
dx
√
1 − x2 sin−1 x
Z
1+x
49.
dx
1 + x2
47.
40.
x(2x + 5)8 dx
48.
50.
52.
Z
Z
Z
Z
Z
Z
Z
x3
p
x2 + 1 dx
Z
Z
√
u du
(B)
1√
u du
4
(D)
Z
Z r
p
x 2x2 + 1 dx is
u−1 √
u du
2
Z p
2u2 + 1 du
AP 54. Which of the following indefinite integrals cannot be
12.
14.
16.
20.
Z
Z
Z
Z
3
sint
√
24.
(ln x)2
dx
x
26.
Z
sec θ tan θ d θ
28.
√
ex 1 + ex dx
30.
(x2 + 1)(x3 + 3x)4 dx
32.
34.
Z
Z
Z
Z
(C)
1 + cos t dt
AP 55.
(A)
y2 (4 − y)2/3 dy
z2
z3 + 1
dz
sin x sin(cos x) dx
√
x x + 2 dx
dx
(a 6= 0)
ax + b
ecost sint dt
ln x
dx
x
(D)
Z
(C)
26
3
x
√
dx
1−x
2x + 1 dx =
8
3
(B) 8
(D)
39
2
Evaluate each indefinite integral. Illustrate and check that your
answer is reasonable by graphing both the integrand and its
antiderivative in the same viewing window (use C = 0).
56.
Z
x(x2 − 1)3 dx
58. ecos x sin x dx
Evaluate each definite integral.
Z 1
πt 60.
dt
cos
2
0
Z 1√
3
62.
1 + 7x dx
57.
64.
Z π /6
sint
66.
cos2 t
Z 2 1/x
e
1
x2
dx
dt
Z
tan2 θ sec2 θ d θ
59. sin x cos4 x dx
61.
Z 1
0
63.
0
0
sec2 x
dx
tan x
Z
Z 4√
0
sec2 2θ d θ
e−5r dr
√
Z
sin x
√ dx
22.
x
a + bx2
√
dx
3ax + bx3
transformed by u substitution into one that has a closed form
antiderivative?
Z
Z
p
2
(A) x2 2x3 − 1 dx
(B) ex dx
x2 ex dx
Z
Z
5t sin(5t ) dt
Z √
(C)
cos3 θ sin θ d θ
3
cot x csc2 x dx
(A)
18.
eu
du
(1 − eu )2
Z √
equivalent to
dx
5 − 3x
2
45.
u = sin θ
Evaluate each indefinite integral.
Z p
11. x 1 − x2 dx
15.
Z
u = x3 + 1
x3
9.
dx, u = x4 − 5
4
x −5
Z √
10.
2t + 1 dt, u = 2t + 1
Z
37.
x
dx
x2 + 4
π cos
x dx
x2
2t
dt
2t + 3
dt
√
2
cos t 1 + tant
sin x
dx
1 + cos2 x
cos(lnt)
dt
t
x
dx
1 + x4
√
x2 2 + x dx
AP 53. If the substitution u = 2x2 + 1 is used, then
Z
13.
Z
51.
u = 2x
cos 2x dx,
x2
36.
43.
Practice
6.
(arctan x)2
dx
x2 + 1
(3x + 1)4 dx is equivalent
3
1
5.
Z
41.
3. True or False If u = 3x + 1, then
Z
35.
Z 3
0
(3t − 1)50 dt
dx
5x + 1
65.
Z 2π /3
67.
Z 1
π /3
0
csc2
2
t x e−x dx
2
dt
SECTION 5.6 The Method of Substitution 87
68.
70.
Z π /4
−π /4
Z 13
(x3 + x4 tan x) dx
0
74.
Z 2 √
76.
Z e4
78.
Z 1 z
e +1
1
e
0
80.
Z 1
0
71.
cos x sin(sin x) dx
86. Which of the following areas are equal? Why?
Z a p
a2 − x2 dx
x
0
(a > 0)
x x − 1 dx
73.
75.
Z 4
x
√
dx
1 + 2x
77.
Z 2
(x − 1)e(x−1) dx
79.
Z T /2
−π /2
0
dz
Z π /3
0
dx
√
x ln x
ez + z
Z π /2
0
dx
p
3
(1 + 2x)2
Z a p
72.
x x2 + a2 dx
0
69.
0
x4 sin x dx
1
2
sin
2π t
−α
T
dt
0.5
1.0
0.5
1.0
5
dx
√
(1 + x)4
Applications and Extensions
√
81. Verify that f (x) = sin 3 x is an odd function and use that fact to
show that
Z 3
√
sin 3 x dx ≤ 1
0≤
1
−2
Use a graph to provide a rough estimate of the area of the region that
lies under each curve. Then find the exact area.
√
82. y = 2x + 1, 0 ≤ x ≤ 1
83. y = 2 sin x − sin 2x,
84. Evaluate
Z 2
−2
2
1
0≤x≤π
√
(x + 3) 4 − x dx by writing it as a sum of two
integrals and interpreting one of those integrals in terms of an
area.
Z 1 p
x 1 − x4 dx by making a substitution and
85. Evaluate
0
interpreting the resulting integral in terms of area.
87. A model for the basal metabolism rate, in kcal/h, of a young
pit
, where t is the time in hours
man is R(t) = 85 − 0.18 cos
12
measuref Zrom 5:00 AM. What is the total basal metabolsim of
24
this man,
0
R(t) dt, over a 24-hour period?
88. An oil storage tank ruptures at time t = 0 and oil leaks from the
tank at a rate of r(t) = 100e−0.01t liters per minute. How much
oil leaks out during the first hour?
89. A bacteria population starts with 400 bacteria and grows at a
rate of r(t) = (450.268)e1.12567t bacteria per hour. How many
bacteria will there be after three hours?
90. Breathing is cyclic and a full respiratory cycle from the
beginning of inhalation to the end of exhalation takes about 5
seconds. The maximum rate of air flow into the lungs is about
0.5 L/s. Thisexplains,
in part, why the function
1
2π t
f (t) = sin
has often been used to model the rate of
2
5
air flow into the lungs. Use this model to find the volume of
inhaled air in the lungs at time t.
88 CHAPTER 5 Integration
91. The rate of growth of a fish population was modeled by the
equation
60,000e−0.6t
G(t) =
(1 + 5e−0.6t )2
where t is measured in years and g in kilograms per year. If the
biomass was 25,))) kg in year 2000, what is the predicted
biomass for the year 2020?
92. Dialysis treatment removes uera and other waste products from
a patient’s blood by diverting some of the bloodflow externally
through a machine called a dialyzer. The rate at which urea is
removed from the blood (in mg/min) is often well desribed by
the equation
r
u(t) = C0 e−rt/V
V
where r is the rate of flow of blood through the dialyzer (in
mL/min), V is the volume of the patient’s blood (in mg), and C0
is the amount of urea in the blood (in mg) at time t = 0.
Evaluate the integral
Z 30
u(t) dt and interpret your answer in
0
the context of this problem.
93. Alabama Instruments Company has set up a production line to
manufacture a new calculator. The rate of production of these
calculators after t weeks is
dx
100
calculators/week
= 5000 1 −
dt
(t + 10)2
(Notice that the production approaches 5000 per week as time
goes on, but the initial production is lower because of the
workers’ unfamiliarity with the new assembly techniques.)
Find the number of calculators produced from the beginning of
the third week to the end of the fourth week.
94. If f is continuous and
Z 4
0
95. If f is continuous and
Z 9
0
f (x) dx = 10, find
f (x) dx = 4, find
Z 2
0
Z 3
0
f (2x) dx.
Extended Applications
96. If f is continuous for all real numbers, show that
Z b
x f (x ) dx.
−b
a
f (x) dx
For the case where f (x) ≥ 0 and 0 < a < b, draw a diagram to
intrepret this equation geometrically as an equality of areas.
97. If f is continuous for all real numbers, show that
Z b
f (x + c) dx =
Z b+c
f (x) dx
a+c
a
For the case where f (x) ≥ 0, draw a diagram to interpret this
equation geometrically as an equality of areas.
98. If a and b are positive numbers, show that
Z 1
0
xa (1 − x)b dx =
Z 1
0
xb (1 − x)a dx
99. If f is continuous on [0, π ], use the substitution u = x − π to
show that
Z π
Z
π π
x f (sin x) dx =
f (sin x) dx
2 0
0
100. Use Exercises 92 to evaluate the integral
Z π
0
x sin x
dx
1 + cos2 x
101. (a) If f is continuous, show that
Z π /2
0
f (cos x) dx =
(b) Use part (a) to evaluate
2
Z −a
f (−x) dx =
Z π /2
0
Z π /2
0
f (sin x) dx
cos2 x dx and
Z π /2
0
sin2 x dx.
CHAPTER 5 Review
89
5 REVIEW
Concepts and Vocabulary
1. (a) Write an expression for a Riemann sum of a function f .
Explain the meaning of the the notation that you use.
(b) If f (x) ≥ 0, what is the geometric interpretation of a
Riemann sum? Illustrate with a diagram.
(c) If f (x) takes on both positive and negative values, what is
the geometric interpretation of a Riemann sum? Illustrate
with a diagram.
5. (a) State the Net Change Theorem.
(b) If r(t) is the rateZ at which water flows into a storage
f (x) dx if f (x) ≥ 0?
Z b
f (x) dx if f (x) takes on both positive and negative
a
r(t) dt represent?
t1
6. Suppose a particle moves back and forth along a horizontal line
so that its velocity at time t is given by v(t), measured in feet
per second, and acceleration is given by a(t).
(a) Explain the meaning of
Z 120
v(t) dt.
60
2. (a) Write the definition of the definite integral of a continuous
function from a to b.
(b) What is the geometric interpretation of the definite integral
Z b
t2
tank,what does
(b) Explain the meaning of
(c) Explain the meaning of
Z 120
60
Z 120
|v(t)| dt.
a(t) dt.
60
(c) What is the geometric interpretation of the definite integral
7. (a) Explain the meaning of the indefinite integral
Z b
values? Illustrate with a diagram.
f (x) dx and the indefinite integral
a
4. State both parts of the Fundamental Theorem of Calculus.
True-False Quiz
Determine whether each statement is true or false. If it is true,
explain why. If it is false, explain why or give an example that
disproves the statement.
1. If f and g are continuous on [a, b], then
Z b
[ f (x) + g(x)] dx =
Z b
f (x) dx +
g(x) dx
a
a
a
Z b
2. If f and g are continuous on [a, b], then
Z b
Z b
Z b
g(x) dx
f (x) dx
[ f (x) g(x)] dx =
a
a
a
3. If f is continuous on [a, b], then
Z b
5 f (x) dx = 5
Z b
7. If f and g are continuous and f (x) ≥ g(x) for a ≤ x ≤ b, then
Z b
a
a
Z b
10.
a
6. If f ′ is continuous on [1, 3], then
Z 5
−5
Z 3
0
14. If
a
a
Z 3
1
f ′ (v) dv = f (3) − f (1).
g(x) dx
a
(ax2 + bx + c) dx = 2
Z 5
0
(ax2 + c) dx
12. All continuous functions have antiderivatives.
f (x) dx
5. If f is continuous on [a, b] and f (x) ≥ 0, then
s
Z b
Z bp
f (x) dx
f (x) dx =
Z b
11. All continuous functions have derivatives.
13.
f (x) dx
f (x) dx ≥
8. If f and g are differentiable and f (x) ≥ g(x) for a < x < b, then
f ′ (x) ≥ g′ (x) for a < x < b.
Z 1 sin x
x5 − 6x9 +
9.
dx = 0
(1 + x4 )2
−1
2
ex dx =
Z 1
0
x f (x) dx = x
f (x) dx.
9. Explain in your own words how to use the Substitution Rule.
4. If f is continuous on [a, b], then
Z b
Z
8. Explain in your own words what is meant by the statement,
“differentiation and integration are inverse processes.”
a
a
f (x) dx.
(b) Explain the connection between the definite integral
a
3. Explain in your own words a left, right, and midpoint Riemann
sum.
Z
Z 5
0
2
ex dx +
Z 3
2
ex dx
5
f (x) dx = 0, then f (x) = 0 for 0 ≤ x ≤ 1.
15. If f is continuous o [a, b], then
Z b
d
f (x) dx = f (x)
dx a
16. The definite integral |int02 (x − x3 ) dx represents the area under
the curve y = x − x3 from 0 to 2.
17.
Z 1
1
−2 x4
dx = −
3
8
18. If f has a discontinuity at 0, then
Z 1
−1
f (x) dx does not exist.
90 CHAPTER 5 Integration
7. The figure below shows the graphs of f , f ′ and
Exercises
1. Use the graph of the function f , shown in the figure below, to
find the Riemann sum with six subintervals. Take the sample
points to be (a) left endpoints and (b) right endpoints. In each
case draw a diagram and explain what the Riemann sum
represents.
Z x
0
f (t) dt.
Identify each graph, and justify your choices.
2. Let f (x) = x2 − x for 0 ≤ x ≤ x.
(a) Find the left Riemann sum and the right Riemann sum
with n = 4 eqaul subintervals. Explain, with the aid of a
diagram, what each Riemann sum represents.
(b) Use the definition of a definite integral (with right
endpoints) to calculate the value of the integral
Z 2
0
8. Evaluate each expression.
Z 1
d arctan x (a)
e
dx
dx
0
Z x
d
earctan t dt
(c)
dx 0
Z
d
dx
(b)
1
0
earctan x dx
9. The graph of f consists of three line segments as shown in the
figure below.
2
1
(x2 − x) dx
(c) Use the Fundamental Theorem of calculus to check your
answer to part (b).
(d) Draw a diagram to explain the geometric meaning of the
definite integral in part (b).
1
2
3
4
5
-1
-2
3. Evaluate
Z 1
0
x+
p
1 − x2 dx
If g(x) =
by interpreting it in terms of areas.
n
lim
n→∞
10.
∑ sin xi ∆x
i=1
as a definite integral on the interval [0, π ] and then evaluate the
integral.
5. If
0
CAS
f (x) dx = 10 and
6. (a) Write
0
f (t) dt, find g(4), g′ (4), and g′′ (4).
Evaluate the integral if it exists.
4. Express
Z 6
-3
Z x
Z 5
1
Z 4
0
f (x) dx = 7, find
Z 6
4
f (x) dx.
12.
1
Z 1
0
14.
16.
5
(x + 2x ) dx as a limit of a right Riemann sum.
Use a computer algebra system to evaluate the sum and to
compute the limit.
(b) Use the Fundamental Theorem of Calculus to check your
answer to part (a).
Z 2
18.
1
Z 1
0
Z 5
Z 1
0
22.
11.
(1 − x9 ) dx
13.
Z 9√
u − 2u2
1
20.
(8x3 + 3x2 ) dx
u
du
28.
Z
15.
Z 1
(x4 − 8x + 7) dx
(1 − x)9 dx
Z 1
√
4
0
y(y2 + 1)5 dy
17.
dt
(t − 4)2
19.
v2 cos(v3 ) dv
21.
Z π /4 4
t tant
Z
0
0
Z 2
y2
0
Z 1
2
u + 1 du
q
1 + y3 dy
sin(3π t) dt
0
dt
−π /4 2 + cos t
Z 1−x 2
24.
dx
x
26.
Z T
Z 1
sin x
−1
1 + x2
ex
dx
1 + e2x
23.
Z 1
25.
Z 10
0
1
x+2
√
dx
x2 + 4x
27.
sin π t cos π t dt
29.
Z
Z
dx
x
dx
x2 − 4
csc2 x
dx
1 + cot x
sin x cos(cos x) dx
CHAPTER 5 Review
30.
Z
√
e x
√ dx
x
Z
32.
Z
tan x ln(cos x) dx
34.
Z
x3
dx
1 + x4
36.
38.
Z
sin(ln x)
31.
dx
x
Z
x
33. √
dx
1 − x4
sec θ tan θ
dθ
1 + sec θ
Z 3
0
Z π /4
37.
0
2
|x − 4| dx
(A) f (x2 )
approximate
(1 + tan t)3 sec2 t dt
40.
cos x
√
dx
1 + sin x
41.
used to approximate
1
43. Sketch the graph of the function f (x) = cos2 x sin x and use the
Z 2π
46. g(x) =
48. y =
cos(t 2 ) dt
47. g(x) =
dt
49. y =
0
Z x4
0
Z x t
e
√
x
t
Z 1√
t + sint dt
x
Z sin x
1 − t2
1 + t4
1
Z 3x+1
2x
dt
sin(t 4 ) dt
Z 3p
1
x2 + 3 dx
Z 5
51.
3
Z 1
54.
Z 1
0
0
x2 cos x dx ≤
1
3
ex cos x dx ≤ e − 1
Z π /2
sin x
55.
Z 1
x
π /4
0
dx ≤
√
2
2
x sin−1 x dx ≤
continuous over its domain such that a < b < c?
Z c
g(x) dx =
Z b
g(x) dx + k
a
(C)
Z b
g(x) dx
g(x) dx = (c + k)
Z b
g( f (x)) · f ′ (x) dx =
g(x) dx
c
Z f (b)
f (a)
g(u) du
18
24
30
36
42
7
−4
−7
3
4
5
Z x
g(t) dt, then the value of a is:
a
(B) 2
(C) 8
(D)
√
4
32
p
65. A particle moves along a horizontal line so that its velocity at
time t is given by v(t) = t 2 − t, where v is measured in meters
per second. Find (a) the displacement and (b) the total distance
traveled by the particle during the time interval [0, 5].
AP 66. A particle moves along the x-axis so that its velocity at time t is
π
4
given by v(t) = 3t 2 − 9t + 6. At time t = 0 the particle is
located at s(0) = 5.
(a) Find the total distance traveled by the particle during the
time interval [0, 2].
(b) Find the position function s(t) corresponding to the
motion of this particle.
AP 67. A particle moves along a horizontal line so that its acceleration
Z b
a
a
Z a
g(x) dx = −
a
Z b
c
Z c
a
(d)
g(x) dx +
a
a
(B) c
Z c
12
5
x3 + 6x and f (1) + 2, what is the value of f (5)?
(A) 5.789
(B) 24.672
(C) 9.789
(D) 26.672
AP 56. Which of the following is not true of a function y = g(x) that is
(A)
6
AP 64. If f (x) =
1
dx
x+1
53.
x
f ′ (x)
AP 63. If x3 − 8 =
Use the properties of integrals to verify each inequality.
52.
table below gives values of f ′ (x) for selected values of x.
(A) 0
Use Property 8 of integrals to estimate the value of each definit
integral.
50.
(D) 48.5
interval [a, b], such that A represents the area bounded by f , and
L, M, R, and T represent the left, right, midpoint, and
trapezoidal sums, respectively, then which of these is not true?
(A) L < M < R
(B) M < R < T
L+R
(C) T > A
(D) A >
2
Find the derivative of each function.
45. F(x) =
(C) 45
AP 62. If f is a function that is increasing and concave up on the
Evaluate the definite integral to check your estimate.
t2
dt
1 + t3
(2x2 + 1) dx. What is the value of this
Approximate the value of f (42) using a midpoint Riemann sum
using as many subintervals as possible.
f (x) dx.
0
Z x
(D) 4x · f (x2 )
AP 61. Suppose f is a differentiable function such that f (6) = 8. The
42. Use a graph to find an estimate of the area of the region
√
bounded by the graph of y = x x, the x axis, and the lines x = 0
and x = 4. Then find the exact area.
44. F(x) =
(C) x · f (x2 )
sin(x3 ) dx.
approximation?
(A) 40.5
(B) 44.5
√
dx
x2 + 1
graph to estimate the value of the definite integral
f (t) dt =
AP 60. Suppose a midpoint Riemann sum with 3 equal subintervals is
Z 4
x3
Z
−x2
approximate the area bounded by the graph of y = 8 − 0.5x2
and the x-axis over the interval [0, 4], then M =
(A) 21.5
(B) 27.25
(C) 18
(D) 24.375
Evaluate each indefinite integral. Illustrate and check that your
answer is reasonable by graphing both the function and its
antiderivative in the same viewing window (use C = 0).
Z
#
x2
AP 59. If M represents the midpoint sum with n = 4 used to
| x − 1| dx
0
Z 3
0
Z 4
√
39.
(B) 2x · f (x)
"Z
58. Use a midpoint Riemann sum with n = 6 equal subintervals to
1 + x4
dx
x3
Z
35.
d
AP 57. If f is an even function, then
dx
91
g(x) dx
function at time t is given by a(t) = 12 cos(3t). At t = 0, its
velocity is v(0) = 2, and its position is s(0) = 0.
(a) Find the function s(t) corresponding to the motion of this
particle.
(b) For 0 ≤ t ≤ π , find the values of t for which this particle is
at rest.
92 CHAPTER 5 Integration
AP 68. The net area enclosed by the graph of y = x3 on the interval
[1, 3] is:
(A) 19.5
(B) 20
(C) 20.5
(D) 28
73. Let r(t) be the rate at which the world’s oil is consumed, where
t is measured in years starting at t = 0 on January 1, 2000, and
r(t) is measured in barrels per year. Explain the meaning of
Z 8
r(t) dt.
0
0.3t
can be modeled by the function r(t) = 0.1 + √
, where r
1
AP 74. For 1 ≤ t ≤ 4, oil is leaking from a tank t the rate of r(t) = √
t2 + 5
t
is measured as the percent interest on an investment at any time
gallons per minute. How many gallons of oil have leaked out
t, 0 ≤ t ≤ 10. The amount of an investment can be thought of as
for 1 ≤ t ≤ 4?
the accumulation of interest rate over an interval of time.
(A) 1
(B) 2
(C) 3
(D) 4
(a) Find the interest rat at time t = 5 years.
Z
2
(b) Write an expression for r(t) dt, the antiderivative of r(t).AP 75. Suppose y = f (x) is a function such that d y = 6x − 4.
dx2
Suppose also that the line y = 6x − 4 is tangent to the graph of f
(c) The amount that an investment of $k grows over an
Z b
at the point where x = 1. Find the function f .
r(t) dt. Find
interval of time, [a, b] is represented by k
a
76. The velocity of a runner, measured in meters per second, is
the amount that $1000 will grow to over the first five
given by a function v(t), where time t is measured in seconds.
years.
Selected values for v(t) are given in the table below.
AP 70. The rate at which water flows into a tank is given by a
t (s)
v (m/s)
t (s)
v (m/s)
differentiable function R(t), where R(t) is measured in gallons
0
0
3.0
10.51
per hour. The table below gives values of R at selected values of
0.5
4.67
3.5
10.67
t.
1.0
7.34
4.0
10.76
t
0
6 12 18 24
1.5
8.86
4.5
10.81
R(t) 13 15 18 14 10
2.0
9.73
5.0
10.81
2.5
10.22
Use a right Riemann sum with 4 equal subintervals to
approximate the number of gallons of water that flowed into the
Use a midpoint Riemann sum to estimate the distance traveled
tank over the 24 hour period.
by the runner during these 5 seconds.
AP 71. Water flows out of a tank and into a barrel at a rate r(t) gallons
77. Suppose a population of honeybees increases at a rate of r(t)
per hour, where t represents the number of hours past 9:00 AM.
bees per week, where the graph of r is shown in the figure
The table below show values of the differentiable function r at
below.
selected values of t.
AP 69. The rate at which an investment grows over a period of ten years
0
3
t
V (t)
1
4
2
5
3
6
4
4
5
5
6
2
12000
(a) Use a midpoint Riemann sum and 3 equal subintervals to
approximate the value of the definite integral
Z 6
8000
r(t) dt.
4000
0
(b) Using correct units, interpret the meaning of this definite
integral in the context of this problem.
(c) If ther were initally 10 gallons of water in the barrel at
9:00 AM, how many gallons of water are in the tank at
3:00 PM?
AP 72. Let f be a function with domain [0, 5] and whose graph is
0
78. Let
f (x) =
2
Evaluate
1
8
12
16
20
24
Use a midpoint Riemann sum with six subintervals to estimate
the increase in the bee population during the first 24 weeks.
shown in the figure below.
Z 1
−3
(
−x − 1
√
− 1 − x2
if − 3 ≤ x ≤ 0
if 0 ≤ x ≤ 1
f (x) dx by interpreting the definite integral as a
difference of areas.
0
-1
4
1
2
3
4
5
79. If f is continuous and
-2
Z π /2
-3
-4
Consider the function h defined by h(x) =
0
Z x2
0
f (t) dt
(a) Find h′ (2).
(b) Find all values of xfor which h has a relative maximum.
Explain your reasoning.
Z 2
0
f (2 sin θ ) cos θ d θ .
f (x) dx = 6, evaluate
CHAPTER 5 Review
πt 2
dt, was introduced
2
0
in this chapter. Fresnel also used the function
2
Z x
πt
cos
C(x) =
dt
2
0
80. The Fresnel function, S(x) =
Z x
sin
in his theory of the diffraction of light waves.
(a) Find the intervals on which C is increasing.
(b) Find the intervals on which C is concave up.
(c) Use technology to solve the following equation:
2
Z x
πt
dt = 0.7
cos
2
0
(d) Plot the graphs of C and S in the same viewing window.
How are these graphs related?
81. Suppose that the temperature of a long, thin rod placed along
C
the x-axis is initially
if |x| ≤ a and 0 if |x| > a. It can be
2a
shown that if the heat diffusivity of the rod is k, then the
temperature of the rod at the point x at time t is
C
T (x,t) = √
a 4π kt
Z a
0
e−(x−u)
2
/(4kt)
du
To find the temperature distribution that results from an initial
hot spot concentrated at the origin, wee need to compute
lim T (x,t). Use l’Hospital’s Rule to find this limit.
a→0
82. If f is a continuous function such that
Z x
1
Z x
f (t) dt = (x − 1)e2x +
1
e−t f (t) dt
for all x, find an explicit formula for f (x).
83. Suppose h is a function such that h(1) = −2, h′ (1) = 2,
h′′ (1) = 3, h(2) = 6, h′ (2) = 5, h′′ (2) = 13, and h′′ is
continuous everywhere. Evaluate
Z 2
1
h′′ (u) du
84. If f ′ is continuous on [a, b], show that
2
Z b
a
1
h→0 h
85. Find lim
f (x) f ′ (x) dx = [ f (b)]2 − [ f (a)]2
Z 2+h p
1 + t 3 dt
2
86. If f is continuous on [0, 1], show that
Z 1
0
87. Evaluate
1
lim
n→∞ n
f (x) dx =
Z 1
0
f (1 − x) dx
" #
n 9
1 9
2 9
3 9
+
+
+···+
n
n
n
n
88. Suppose f is continuous, f (0) = 0, f (1) = 1, f ′ (x) > 0, and
Z 1
1
f (x) dx = . Find the value of the definite integral
3
Z0
1
0
f −1 (y) dy.
93