Calculus Chapter 6
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Chapter 6 Integration
• 6.1 Antiderivatives and the rules of integration
• 6.2 Integration by substitution
• 6.3 Area and the definite integral
• 6.4 Fundamental Theorem of Calculus
• 6.5 Evaluating definite integrals
• 6.6 Area between two curves
• 6.7 Applications of the definite integral to business
and economics
• 6.8 Volumes of solids of revolution
Calculus Chapter 6
6.1 Antiderivatives and the Rules of Integration
• In Chapter 2, we consider:
If we know the position, can we find its velocity?
• In Chapter 6 and 7, we consider the opposite
problem:
If we know the velocity function f of the maglev,
can we find its position function f ?
• Definition of Antiderivative: A function F is an
antiderivative of f on an interval I if F (x) = f (x)
for all x in I.
• Example 1: F (x) = 13 x3 − 2x2 + x − 1 is an
antiderivative of f (x) = x2 − 4x + 1 because
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Calculus Chapter 6
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F (x) = x2 − 4x + 1 = f (x)
• Example 2: Let F (x) = x, G(x) = x + 2, and
H(x) = x + C, where C is a constant. Then F, G,
and H are all antiderivative of f (x) = 1.
• Theorem 1: Let G is an antiderivative of a function
f . Then, every antiderivative of F of f must be of
the form F (x) = G(x) + C, where C is a constant.
• In Example 2, we see that there are infinitely many
antiderivatives of the function f (x) = 1. (Figure 6.2)
– Figure 6.2 shows the graphs of some of these
antiderivatives for selected values of C.
Calculus Chapter 6
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Indefinite Integral
• Antidifferentiation or Integration: (integral sign )
f (x)dx = F (x) + C
– Read ”the indefinite integral of f (x) with respect
to x equals F (x) plus C”
– The indefinite integral of f is the family of
functions given by F (x) + C, where F (x) = f (x).
– Function f called the integrand
– Constant C called a constant of integration
– If independent variable is t, we write f (t)dt. In
this sense, both t and x are ”dummy variables.”
Calculus Chapter 6
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Basic Integration Rules
• Rule 1: Indefinite integral of a constant
kdx = kx + C
where k a constant
– Check F (x) =
– Example 4 (b):
d
(kx
dx
2
+ C) = k
π dx = π 2 x + C
• Rule 2: Power Rule
1
x dx =
xn+1 + C
n+1
(n = −1)
n
– Check F (x) =
– Example 5 (c):
d
dx
1
n+1
x
n+1
+C =
n+1 n
x
n+1
= f (x)
Calculus Chapter 6
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1
x3/2
dx =
• Rule 6:
−1
x dx =
– Observe
x
d
dx
−3/2
1 −1/2
dx = 1 x
+C = −2x−1/2 + C
−2
1
dx = ln |x| + C
x
ln |x| =
1
x
(x = 0)
(Rule 3, Section 5.5)
• Rule 3: Integral of a constant multiple of a function:
cf (x)dx = c
– Check
d(cf (x))
dx
f (x)dx
(c, a constant)
(x)
= c dfdx
– Only a constant can be ”moved out” of an
integral sign. For example, it is incorrect to write
Calculus Chapter 6
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x2 dx = x2
1dx
In fact,
1 3
2
x dx = x +C,
3
x2
1dx = x2 (x+C) = x3 +Cx2
– Example 6 (b):
−2
−3x dx = −3
3
x dx = (−3)(−1)x +C = + C
x
−2
• Rule 4: Sum rule
[f (x) ± g(x)]dx =
– Example 7:
−1
f (x)dx ±
g(x)dx
Calculus Chapter 6
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=
(3x5 + 4x3/2 − 2x−1/2 )dx
3x5 dx +
= 3
x5 dx + 4
4x3/2 dx −
x3/2 dx − 2
2x−1/2 dx
x−1/2 dx
2 5/2
1 6
= (3)( )x + (4)( )x − (2)(2)x1/2 + C
6
5
1 6 8 5/2
x + x −4x1/2 + C
=
2
5
where we combine the three constants of
integration to obtain one constant C
• Rule 5: Exponential
ex dx = ex + C
Calculus Chapter 6
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Differential Equations
• Given the derivative of a function f , can we find the
function f ?
• As an example, suppose we are given a differential
equation
f (x) = 2x − 1.
– A differential equation is an equation that involves
the derivative or differential of an unknown
function. In this case, the unknown function is f .
– All the solutions are given by
f (x) =
f (x)dx =
(2x − 1)dx = x2 − x + C
Calculus Chapter 6
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where C is an arbitrary constant. It is called the
d
f (x) = f (x).
general solution. Note that dx
– Particular solution: Specify the value the function
must assume at a certain value of x. For example,
f (1) = 3, then f (1) = 1 − 1 + C = 3. Thus, the
particular solution is f (x) = x2 − x + 3.
• Initial Value Problem: Find a function satisfying
1. a differential equation
2. one or more initial conditions
• Example 10: Find the function f if it is known that
f (x) = 3x2 − 4x + 8,
f (1) = 9
Calculus Chapter 6
⇒ f (x) =
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f (x)dx =
(3x2 −4x+8)dx = x3 −2x2 +8x+C
– Because f (1) = 1 − 2 + 8 + C = 9 ⇒ C = 2.
– Therefore, the required function
f (x) = x3 − 2x2 + 8x + 2
• Example: dy = 3ex dx, y(0) = 1. Then
dy
= 3ex
dx
⇒ y(x) =
3ex dx = 3ex + C
– Because y(0) = 1 = 3 + C ⇒ C = −2
Calculus Chapter 6
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Applications
• Example 11: Velocity of the maglev is
v(t) = 8t, (0 ≤ t ≤ 30). Find the position function.
Assume that initially the maglev is located at the
origin of a coordinate line.
– Let s(t) denote the position. Then
s (t) = v(t) = 8t with initial condition s(0) = 0.
⇒ s(t) =
s (t)dt =
8tdt = 4t2 + C.
– Because s(0) = 4(0) + C = 0 ⇒ C = 0. Therefore,
the required position function is
s(t) = 4t2 (0 ≤ t ≤ 30).
Calculus Chapter 6
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6.2 Integration by Substitution
• How the method of substitution works
– Consider the indefinite integral
2(2x + 4)5 dx
– One way to do (brute force): Expand the
expression (2x + 4)5 and then integrate the
integrand term by term
– Alternative: Change of variable u = 2x + 4. Then
du
= 2 ⇒ du = 2dx
dx
∗ Then substitute
2(2x + 4)5 dx =
(2x + 4)5 (2dx) =
u5 du
Calculus Chapter 6
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∗ Power function
1 6
1
u du = u + C = (2x + 4)6 + C
6
6
5
∗ Verify
d 1
1
6
(2x + 4) + C = ·6(2x+4)5 (2) = 2(2x + 4)5
dx 6
6
• Method of Integration by Substitution
– Write f (x) = x5 and g(x) = 2x + 4. Then
g (x) = 2 and
(f ◦ g)(x) = f (g(x)) = [g(x)]5 = (2x + 4)5
Calculus Chapter 6
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Therefore,
2(2x + 4)5 dx =
– Let u = g(x), then
2(2x + 4)5 dx =
du
dx
f (g(x))g (x)dx
= g (x) ⇒ du = g (x)dx
f (g(x))g (x)dx =
f (u)du
where it is easier to integrate
√
• Example 2: Find 3 3x + 1dx
√
– Step 1: f (x) = x, g(x) = 3x + 1. Let u = 3x + 1
be the ”inside function” of the composite function
√
√
f (g(x)) = 3x + 1. Then f (u) = u
– Step 2: Compute du = 3dx
Calculus Chapter 6
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– Step 3: Making the substitution
√
√
√
3 3x + 1dx =
3x + 1(3dx) =
udu
– Step 4: Evaluate
√
udu =
u
1/2
2 3/2
du = u + C
3
– Step 5: Replacing u by 3x + 1, we obtain
√
2
3 3x + 1dx = (3x + 1)3/2 + C
3
• Example 3: Find
x2 (x3 + 1)3/2 dx
– (1) The integrand contains the composite function
(x3 + 1)3/2 with ”inside function” g(x) = x3 + 1.
Calculus Chapter 6
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So, let u = x3 + 1
– (2) Compute du = 3x2 dx
– (3) Substitute
x2 (x3 + 1)3/2 dx =
du
1
u3/2 ( ) =
u3/2 du
3
3
– (4) Evaluate
1 3/2
1 2 5/2
2 5/2
u du = · u + C = u + C
3
3 5
15
– (5) Finally
2
3
3/2
x (x + 1)
• Example 4: Find
2 3
dx = (x + 1)5/2 + C
15
e−3x dx. Let u = −3x so that
Calculus Chapter 6
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du = −3dx.
⇒
1
1 u
1 −3x
−3x
u
e dx = −
e du = − e + C = − e + C
3
3
3
• Example 5: Find
du = 6xdx.
⇒
⇒
x
dx.
3x2 +1
Let u = 3x2 + 1 so that
xdx
1 du
1
1
2
+1)+C
=
=
ln
|u|+C
=
ln(3x
2
3x + 1
6
u
6
6
• Example 6: Find
du = x1 dx
(ln x)2
dx.
2x
Let u = ln x so that
1 2
1 3
1
(ln x)2
dx =
u du = u + C = (ln x)3 + C
2x
2
6
6
Calculus Chapter 6
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Applications
• Example 8: Sale forecasts for a new computer will
grow at the rate of 2000 − 1500e−0.05t , (0 ≤ t ≤ 60)
units per month.
– N (t) : Total expected number of sold computers
⇒ N (t) = 2000 − 1500e−0.05t
⇒ N (t) =
(2000 − 1500e−0.05t )dt
1500 −0.05t
+C
= 2000t +
e
0.05
– Because N (0) = 30, 000 + C = 0 ⇒ C = −30, 000
– Sell in the first year N (12) = 10, 464
Calculus Chapter 6
6.3 Area and the Definite Integral
• Second fundamental problem in calculus: Calculate
the area of the region bounded by the graph of a
nonnegative function f , the x-axis, and the vertical
lines x = a and x = b (Figure 6.7). Called the area
under the graph of f on the interval [a, b]
• Example 1: Let f (x) = x2 on the interval [0, 1]
(Figure 6.8a).
– Divide the interval [0, 1] into four subintervals
[0, 14 ], [ 14 , 12 ], [ 12 , 34 ], [ 34 , 1] of equal length 14 .
– Construct 4 rectangles with these subintervals as
bases and with heights given by the values of the
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Calculus Chapter 6
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function at the midpoints 18 , 38 , 58 , 78 of each
subinterval. (Figure 6.8 b)
– Approximate area by 4 rectangles
1 1
1 3
1 5
1 7
A ≈
f( ) + f( ) + f( ) + f( )
4 8
4 8
4 8
4 8
3 2
5 2
7 2
21
1 1 2
( ) +( ) +( ) +( ) =
=
4 8
8
8
8
64
– Better approximation:
n Rectangles
4
16
100
200
Approximate A 0.328125 0.333008 0.333325 0.333331
– Our computations seem to suggest that the
approximations approach the number 13 as n gets
Calculus Chapter 6
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larger and larger. This result suggests that we
define the area of the region under the graph of
f (x) = x2 on [0, 1] to be 13 square unit.
1
– Note: 0 f (x)dx = 13 x3 |10 = 13 (1) − 13 (0) =
(Section 6.4)
1
3
• Example 2: Let f (x) = 16 − x2 on the interval [1, 3]
(Figure 6.10 a).
– Divide the interval [1, 3] into four subintervals
[1, 32 ], [ 32 , 2], [2, 52 ], [ 52 , 3] of equal length 12 .
– Pick the left end point of each subinterval to
evaluate f (x) (Figure 6.10 b). Approximate area
by 4 rectangles
Calculus Chapter 6
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1 3
1
1 5
101
1
A ≈ f (1) + f ( ) + f (2) + f ( ) =
2
2 2
2
2 2
4
– Better approximation:
n Rectangles
4
100
50,000
100,000
Approximate A 25.2500 23.4132 23.3335 23.3334
– The approximations approach the number 23 13 as
n gets larger and larger.
– Note:
3
x3 3
27
1
f
(x)dx
=
16x
−
|
=
(48
−
)
−
(16
−
)=
1
3 1
3
3
(Section 6.4)
70
3
Calculus Chapter 6
Defining Area – The General Case
• Consider continuous and nonnegative function f .
Want to compute the area A on an interval [a, b]
• Divide the interval [a, b] into n subintervals of equal
length Δx = b−a
n
• Pick n arbitrary representative points x1 , x2 , · · · , xn
from the first, second, · · · , and nth subintervals
(Figure 6.11b)
• Approximate area by a Riemann sum
A ≈ f (x1 )Δx + f (x2 )Δx + · · · + f (xn )Δx
• Area under the graph of a function: Let f be a
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Calculus Chapter 6
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nonnegative continuous function on [a, b]. Then, the
area of the region under the graph of f is
A = lim [f (x1 ) + f (x2 ) + · · · + f (xn )]Δx
n→∞
– The Riemann sum will approach a unique number
as n becomes arbitrarily large.
Calculus Chapter 6
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Definite Integral
• Definite integral: Let f be defined on [a, b]. If
lim [f (x1 )Δx + f (x2 )Δx + · · · + f (xn )Δx]
n→∞
exists for all choices of representative points
x1 , x2 , · · · , xn in the n subintervals of [a, b] of equal
width Δx = (b − a)/n, then this limit is called the
definite integral of f from a to b and is denoted by
b
a f (x)dx. Thus,
b
a
f (x)dx = n→∞
lim [f (x1 ) + f (x2 ) + · · · + f (xn )]Δx
The number a is the lower limit of integration, and
the number b is the upper limit of integration.
Calculus Chapter 6
– If f is nonnegative, then the definite integral gives
the area under the graph of f on [a, b].
– We will see later that the definite integral and the
antiderivative of a function f are related, so the
limit is denoted by the integral sign .
– Definite integral is a number, whereas the
indefinite integral f (x)dx represents a family of
functions (antiderivatives)
– If the limit exists, we say that f is integrable on
[a, b].
• Theorem: Let f be continuous on [a.b]. Then, f is
integrable on [a, b].
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Calculus Chapter 6
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b
• If f is continuous on [a, b], then a f (x)dx is equal to
the area of the region above [a, b] minus the area of
the region below [a, b] (Figure 6.15).
Calculus Chapter 6
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6.4 Fundamental Theorem of Calculus (FTC)
• Theorem 2 (FTC by Newton and Leibniz): Let f be
continuous on [a, b]. Then,
b
a
f (x)dx = F (b) − F (a)
where F is any antiderivative of f , F (x) = f (x).
• Notation: F (x)|ba = F (b) − F (a)
• The constant of integration ”dropped out.” If
F (x) + C denotes an antiderivative of some function
f , then
F (x) + C|ba = [F (b) + C] − [F (a) + C] = F (b) − F (a)
Calculus Chapter 6
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• Example 1: Consider f (x) = x on [1, 3]. (Figure 6.16)
– Area given by the definite integral of f from 1 to 3
A=
3
1
3
1 2
9
1
xdx = ( x + C) = ( +C)−( +C) = 4
2
2
1
2
– Elementary means: Areas R1 and R2 are
2(1) + 12 (2)(2) = 4
• Example 2: In Section 6.3, we conjectured that the
area of f (x) = x2 on [0, 1] was 13 (Figure 6.17).
Observe that f is nonnegative on [0, 1], so the area is
A=
0
1
1
1
1
1 3 1
2
x dx = x = (1) − (0) =
3
0
3
3
3
Calculus Chapter 6
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Evaluate Definite Integrals
• Example 4: Evaluate
1
3
1
2
2
x
(3x
+
e
)dx
1
(3x2 + ex )dx = x3 + ex |31 = (27 + e3 ) − (1 + e)
• Example 5: Evaluate
3
2
1
(
1 x −
1 1
( − 2 )dx = ln |x| +
x x
1 2
1
1
= (ln 2+ )−(ln 1+1) = ln 2 −
x1
2
2
• Group discussion: Consider
1. We have
1
−1
1
)dx
x2
1
1
1
1
dx
=
−
x2
x
1
−1 x2 dx
−1
= −1 −
−1
−1
= −2
2. Observe that f (x) is positive at x ∈ [−1, 1] where
it is defined. Therefore, one might expect that the
Calculus Chapter 6
definite integral has a positive value, if it exists.
3. Explain this apparent contradiction in the result
(1) and the observation (2).
4. More in Section 7.4
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Calculus Chapter 6
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Applications
• Example 6: Daily marginal cost function is
C (x) = 0.000006x2 − 0.006x + 4
where C (x) is measured in dollars per unit and x
denotes the number of units produced. Daily fixed
cost is $100.
– Daily total cost for producing the first 500 units?
∗ Total cost function C(x)
∗ Daily fixed cost is C(0) = 100
∗ Need to find C(500). Using the FTC
Calculus Chapter 6
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C(500) − C(0) =
0
=
=
500
500
0
C (x)dx
(0.000006x2 − 0.006x + 4)dx
500
0.000002x − 0.003x + 4x
3
2
0
= 1500
∗ Therefore, C(500) = 1500 + C(0) = 1600
– Daily total cost (201st through 400th)
400
C(400) − C(200) =
C (x)dx
200
=
400
3
2
0.000002x − 0.003x + 4x
= 552
200
Calculus Chapter 6
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– Since C (x) ≥ 0 for x ∈ (0, ∞), so C (x)dx are
the region areas R1 and R2 in Figure 6.19
• Riemann sum: Consider
2
2
2
n 1
1
S = n→∞
lim
+ 2 + ··· + 2
2
n
n
n n
– High school math:
n(n + 1)(2n + 1)
1 + 2 + ··· + n =
6
1
n(n + 1)(2n + 1)
S = n→∞
lim
=
3
6n
3
2
⇒
2
2
– By calculus:
∗ Width n1 : The first subinterval [0, n1 ], the second
Calculus Chapter 6
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subinterval [ n1 , n2 ]
∗ Height: (1/n)2 for the first subinterval, (2/n)2
for the second subinterval. Look like x2
1 2
∗ So S = 0 x dx = 13
• Riemann sum
9
S = lim 1 + 2 + · · · + n
n→∞
1
9
n10
=
0
1
1
x dx =
10
9
• Riemann sum
2
4
2n 2
S = lim (1 + ) + (1 + ) + · · · + (1 + )
n→∞
n
n
n n
2
2 2
x 1 + xdx = x + = 4
=
2 0
0
Calculus Chapter 6
Validity of the Fundamental Theorem of Calculus
• Let A(t) denote the area of the region R under the
graph of y = f (x) from x = a to x = t, where
a ≤ t ≤ b (Figure 6.20)
• If h is a small positive number, then A(t + h) is the
area from x = a to x = t + h. Therefore, the area
difference A(t + h) − A(t) (Figure 6.21) can be
approximated by (Figure 6.22)
A(t + h) − A(t)
A(t + h) − A(t) ≈ h · f (t) ⇒
≈ f (t)
h
where the approximations improve as h is taken to
be smaller and smaller.
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Calculus Chapter 6
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• Take the limit as h → 0
A(t + h) − A(t)
A (t) = lim
,
h→0
h
A (t) = f (t)
So the area function A is an antiderivative of f (x).
• By Theorem 1 of Section 6.1, A(x) must have the
form A(x) = F (x) + C where F (x) is any
antiderivative of f and C is an arbitrary constant.
• Observe that A(a) = 0 ⇒ C = −F (a)
• The area of the region is A(b) (Figure 6.23), or
b
a f (x)dx,
⇒
b
a
f (x)dx = A(b) = F (b) + C = F (b) − F (a)
Calculus Chapter 6
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6.5 Evaluating Definite Integrals
• Properties: If f and g be integrable functions, then
1.
2.
3.
4.
a
a
f (x)dx = 0
a
cf (x)dx = c
b
b
a [f (x)
b
a
b
a
f (x)dx
± g(x)]dx =
f (x)dx =
c
a
b
a
(c is a constant)
f (x)dx ±
f (x)dx +
b
c
b
a
g(x)dx
f (x)dx
– If f ≥ 0 and a < c < b, then we have the area
interpretation (Figure 6.24)
5.
b
a
a
f (x)dx + b f (x)dx = a f (x)dx = 0
b
a
⇒ a f (x)dx = − b f (x)dx
a
Calculus Chapter 6
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Fundamental Theorem of Calculus, Part II
• Reference: Smith and Minton, Calculus, Section 4.5.
• If f is continuous on [a, b] and F (x) =
F (x) = f (x) on [a, b].
– Proof:
F (x + h) − F (x) =
x+h
a
x+h
x
f (t)dt −
=
a
f (t)dt, then
x
f (t)dt +
a
a
a
f (t)dt
f (t)dt =
x
= f (z)h
where h > 0 and z ∈ (x, x + h). Therefore,
F (x + h) − F (x)
= lim+ f (z) = f (x).
lim+
h→0
h→0
h
x+h
x
f (t)dt
Calculus Chapter 6
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– If h < 0, we may prove in a similar fashion that
F (x + h) − F (x)
= lim− f (z) = f (x).
lim−
h→0
h→0
h
– The two preceding one-sided limits imply that
F (x + h) − F (x)
F (x) = lim
= f (x).
h→0
h
• For F (x) =
x
2
(t
−
2t
+
3)dt,
compute
F
(x).
1
– From Part II of the FTC
⇒ F (x) = f (x) = x2 − 2x + 3
– From Part I of the FTC
Calculus Chapter 6
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x
1 3
7
1 3
2
2
F (x) =
t − t + 3t =
x − x + 3x −
3
3
3
1
1
⇒ F (x) = · 3x2 − 2x + 3 − 0 = x2 − 2x + 3
3
• Chain Rule: If F (x) =
F (x) = f (u(x))u (x).
• If F (x) =
x2
2
u(x)
a
f (t)dt, then
et dt, compute F (x).
– Method (1): Let u(x) = x2 , so that
u(x) t
F (x) = 2 e dt. From the chain rule,
u(x) du
F (x) = e
– Method (2):
dx
u(x)
=e
(2x) = 2xe
x2
Calculus Chapter 6
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F (x) =
t x
e
2
2
x2
=e
− e2
x2
⇒ F (x) = e
· (2x)
– Method (3):
F (x) =
=
=
=
F (x + h) − F (x)
lim
h→0
h
(x+h)2 t
x2 t
e dt − 2 e dt
2
lim
h→0
h
(x+h)2 t
e dt
x2
lim
h→0
h
2
2
(x
+
h)
−
x
2
ex lim
h→0
h
Calculus Chapter 6
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x2
= e
• If F (x) =
x2 √
2x
· (2x)
t2 + 1dt, compute F (x).
– Rewrite the integral
F (x) =
√
0
2x
= −
t2
2x
0
+ 1dt +
√
√
x2
0
t2 + 1dt +
0
t2 + 1dt
x2
√
t2 + 1dt
– From the chain rule,
d
d 2
2
2
2
F (x) = − (2x) + 1 (2x) + (x ) + 1 (x )
dx √
dx
√
= −2 4x2 + 1 + 2x x4 + 1
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• Find an equation of the tangent line at x = 0:
y(x) =
0
x
−t2 +1
e
dt
– For x = 0, then y(0) = 0.
– Slope at x = 0: By y (x) = e−x
2 +1
⇒ y (0) = e.
– Tangent line at x = 0: y = ex
• If f (x) =
2
x −t2
dt ,
0 e
then f (x) = 2e
• Exercise: Find the derivative f (x)
– f (x) =
– f (x) =
x3
0
0
(t2 − 3t + 2)dt
2
−x2
−t
(e
+ 1)dt
2
x
x
0
−t2
e
dt
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Method of Substitution for Definite Integrals
4 √
• Example 1: Evaluate 0 x 9 + x2 dx
– Method 1: Consider indefinite integral
√
I = x 9 + x2 dx
Make the substitution u = 9 + x2 , so that
du = 2xdx. Then
√
1 3/2
1
1
I=
udu = u + C = (9 + x2 )3/2 + C
2
3
3
⇒
0
4
√
4
1
2
2 3/2 2
x 9 + x dx = (9 + x ) = 32
3
3
0
– Method 2: Changing the limits of integration. Let
Calculus Chapter 6
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u = 9 + x2 .
∗ If x = 0 (the lower limit), then u = 9 + 0 = 9.
∗ If x = 4 (the upper limit), then u = 9 + 42 = 25.
∗ Thus, the integration interval over u becomes
[9, 25]. Then
25
1 3/2
2
I = ( u + C) = 32
3
3
9
2
• Example 2: Evaluate
0
2
xe2x dx
– Let u = 2x2 so that du = 4xdx. When
x = 0, u = 0, and when x = 2, u = 8. Then
0
2
xe
2x2
dx =
0
8
8
1 u 1 8
1 u
e du = e = (e − 1)
4
4
0
4
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• Example 3: Evaluate
1
x2
0 x3 +1 dx
– Let u = x3 + 1 so that du = 3x2 dx. When
x = 0, u = 1, and when x = 1, u = 2. Then
0
1
2
x2
1
1 21
1
ln 2
dx =
du = ln |u| = (ln 2−ln 1) =
3
x +1
3 1 u
3
3
3
1
• Let f be a function defined piecewise by the rule
⎧
√
⎪
⎨
x
f (x) = ⎪
⎩
if 0 ≤ x ≤ 1
1
x
if 1 < x ≤ 2
– By Property 5 of definite integrals, we have
0
2
f (x)dx =
0
1
√
xdx +
1
2
1
dx
x
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Finding the Area Under a Curve
• Example 4: Find the area of the region R under the
graph of f (x) = e(1/2)x from x = −1 to x = 1 (Figure
6.25)
– Its area is given by
A=
1
−1
e(1/2)x dx
– Let u = 12 x, so that du = 12 dx. When
x = −1, u = − 12 , and when x = 1, u = 12 . Then
A=2
1/2
−1/2
1/2
eu du = 2eu |−1/2 = 2(e1/2 − e−1/2 )
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• Example: A straight line y = ax. Find the area from
x = 0 to x = b. This is a triangle with base b and
height ab.
– High-school math:
(b)·(ab)
2
=
ab2
2
– By calculus:
b
x b2
axdx = a = a ·
2
2
0
2 b
0
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Average Value of a Function
• Average value of a set of n numbers is
y1 +y2 +···+yn
n
• Suppose f is continuous on [a, b]. Choose points
x1 , x2 , · · · , xn in the first, second, · · ·, and nth
subintervals of equal length (b − a)/n.
• Then, the average value
f (x1 ) + f (x2 ) + · · · + f (xn )
n
is an approximation of the average of all the values of
f (x) on [a, b].
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b−a
1
1
1
=⇒
f (x1 ) · + f (x2 ) · + · · · + f (xn ) ·
b−a
n
n
n
1
b−a
b−a
b−a
=
f (x1 ) ·
+ f (x2 ) ·
+ · · · + f (xn ) ·
b−a
n
n
n
1
=
[f (x1 )Δx + f (x2 )Δx + · · · + f (xn )Δx]
b−a
• Average value of f over [a, b]
f (x1 ) + · · · + f (xn )
lim
n→∞
n
1
1 b
=
f (x)dx
lim [f (x1 )Δx + · · · + f (xn )Δx] =
n→∞
b−a
b−a a
Calculus Chapter 6
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Applications
• Example 6: Interest rates on auto loans for used cars
over a certain 6-month period are approximated by
1 3 7 2
r(t) = − t + t − 3t + 12
12
8
where t is measured in months and r(t) is the annual
percentage rate.
– Average rate over the 6-month period
1 3 7 2
1 6
− t + t − 3t + 12 dt
6−0 0
12
8
6
1
1 4
7 3 3 2
=
− t + t − t + 12t = 9
6
48
24
2
0
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or 9% per year.
• Example 7: Amount of a certain drug in a patient’s
body t days after it has been administered is
C(t) = 5e−0.2t units. Average amount for the first 4
days is
1
4−0
0
4
5e−0.2t dt ≈ 3.44
or approximately 3.44 units.
• Geometric interpretation of the average value of f
over [a, b] (Figure 6.26)
– Suppose f (x) ≥ 0 so that the definite integral
b
a f (x)dx gives the area under the graph of f
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from a to b.
– Consider a constant function g(x) = k such that
the areas under each of the two funtions f and g
are the same.
k(b − a) =
b
a
1 b
f (x)dx ⇒ k =
f (x)dx
b−a a
so that k is the average value of f over [a, b].
Calculus Chapter 6
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6.6 Area Between Two Curves
• Let f and g be continuous function such that
f (x) ≥ g(x) on [a, b]. Then, the area of the region
bounded above by y = f (x) and below by y = g(x)
on [a, b] is
b
a
[f (x) − g(x)]dx
• Example 1: Find the area of the region bounded by
the x-axis, the graph of y = −x2 + 4x − 8, and the
lines x = −1 and x = 4.
– The region R is shown in Figure 6.31.
– The maximum point of y = −x2 + 4x − 8 is
Calculus Chapter 6
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achieved at x = 2 with a value of −4 (Chapter 4).
– The area
b
a
[f (x) − g(x)]dx =
=
=
4
−1
4
−1
[0 − (−x2 + 4x − 8)]dx
(x2 − 4x + 8)dx
4
1 3
2
x − 2x + 8x
3
−1
2
= 31
3
• Example 3: Find the area of the region R that is
completely enclosed by the graphs of the functions
f (x) = 2x − 1 and g(x) = x2 − 4.
– The region R is shown in Figure 6.33.
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– Points of intersection of the two curves
2x − 1 = x2 − 4 ⇒ x = −1, 3
– Verify f (x) ≥ g(x) for x ∈ [−1, 3]. Define h(x) ≡
f (x) − g(x) = (2x − 1) − (x2 − 4) = −x2 + 2x + 3
⇒ h(x) = (−x + 3)(x + 1) ≥ 0,
– The area
b
a
3
∀x ∈ [−1, 3]
2
[f (x)−g(x)]dx =
[(2x−1)−(x −4)]dx = 10
3
−1
2
• Example 4: Find the area of the region R bounded
by the graphs of the functions f (x) = x2 − 2x − 1
and g(x) = −ex − 1 and the vertical lines x = −1 and
Calculus Chapter 6
x = 1.
– The region R is shown in Figure 6.34.
– Define h(x) ≡ f (x) − g(x) =
(x2 − 2x − 1) − (−ex − 1) = x2 − 2x + ex . Want to
prove h(x) > 0 over [−1, 1] (could be skipped)
∗ Then h (x) = 2x − 2 + ex , h (x) = 2 + ex
∗ Since h (x) > 0 for all x ∈ [−1, 1], we know
that h (x) is an increasing function. Also,
h (−1) = −4 + e−1 < 0 and h (1) = 2 − 2 + e > 0
∗ Using the method of bisection in Section 2.5,
we get h (0.315) ≈ 0.
∗ Because h(x) is continuous on [−1, 1], the
function h(x) has absolute minimum and
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Calculus Chapter 6
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maximum by Theorem 3 in Section 4.4.
∗ Because h(−1) = 3 + e−1 > 3 and
h(1) = −1 + e > 1, we have h(0.315) = 0.839 as
the absolute minimum.
– The desired area is
1
2
1
[(x − 2x − 1) − (−e − 1)]dx = + e −
3
e
−1
2
x
• Example 5: Find the area of the region bounded by
the function f (x) = x3 , the x-axis, and the lines
x = −1 and x = 1.
– The region R is shown in Figure 6.35.
– Method 1: The area
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R1 +R2 =
0
3
−1
(0−x )dx+
1
0
1 1
1
(x −0)dx = + =
4 4
2
3
– Method 2: By making use of symmetry, we could
have obtain the same result by computing
−2
0
3
−1 x dx =
1
,
2
or
2
1
0
x3 dx =
1
2
0
– Method 3: Consider − −1 x3 dx. Change of
variable u = −x. Then du = −dx.
−
0
−1
3
x dx =
1
0
3
(−u) du =
0
1
u3 du
• Example 6: Find the area of the region completely
enclosed by the graphs of the functions
Calculus Chapter 6
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f (x) = x3 − 3x + 3 and g(x) = x + 3
– The region R is shown in Figure 6.36.
– Points of intersection
x3 − 3x + 3 = x + 3 ⇒ x = 0, −2, 2
– Can prove f (x) ≥ g(x) over [−2, 0] and
g(x) ≥ f (x) over [0, 2] as before
– Area
0
−2
[f (x) − g(x)]dx +
0
2
[g(x) − f (x)]dx = 8
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Application
• Example 7: If the Energy Conservation Bill were
implemented in 1995, the country’s oil consumption
for the next 5 years would be expected to grow in
accordance with the model R(t) = 20e0.05t where t is
measured in years and R(t) in millions of barrels per
year. Without the government-imposed conservation
measures, however, the expected rate of growth of oil
consumption would be given by R1 (t) = 20e0.08t
– How much oil would have been saved from 1995
to 2000?
0
5
[R1 (t) − R(t)]dt =
0
5
[20e0.08t − 20e0.05t ]dt ≈ 9.3
Calculus Chapter 6
6.7 Applications of the Definite Integral to Business and
Economics
• Consumers’s surplus:
– Suppose p = D(x) is the demand function.
– Suppose a fixed unit market price has been
established for the commodity and corresponding
to this unit price the quantity demanded is x
units (Figure 6.40).
– Then, those consumers who would be willing to
pay a unit price higher than p for the commodity
would in effect experience a savings.
– This difference between what the consumers
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Calculus Chapter 6
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would be willing to pay for x units of the
commodity and what they actually pay for them
is called the consumers’ surplus.
– The consumers’ surplus is given by
CS =
0
x
[D(x) − p]dx =
0
x
D(x)dx − px
• Producers’ surplus:
– Suppose p = S(x) is the supply function.
– Suppose a fixed market price p has been
established for the commodity and, corresponding
to this unit price, a quantity of x units will be
made available in the market by the supplier
(Figure 6.42).
Calculus Chapter 6
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– Then, the suppliers who would be willing to make
the commodity available at a lower price stand a
gain from the fact that the market price is set as
such.
– This difference between what the suppliers
actually receive and what they would be willing
to receive is called the producers’ surplus.
– The producers’ surplus is given by (Figure 6.42)
PS =
0
x
[p − S(x)]dx = px −
0
x
S(x)dx
• Example 1: Demand function for a certain make of
10-speed bicycle is p = D(x) = −0.001x2 + 250. The
supply function is p = S(x) = 0.0006x2 + 0.02x + 100.
Calculus Chapter 6
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Determine the consumers’ surplus and the producers’
surplus if the market price of a bicycle is set at the
equilibrium price. (Figure 6.43)
– Equilibrium price: Point of intersection
0.0006x2 +0.02x+100 = −0.001x2 +250 ⇒ x = −312.5, 300
– If x = 300, then p = 160
– Consumers’ surplus
0
300
(−0.001x2 + 250)dx − (160)(300) = 18, 000
– Producers’ surplus
(160)(300)−
0
300
(0.0006x2 +0.02x+100)dx = 11, 700
Calculus Chapter 6
Future and Present Value of an Income Stream
• Suppose
– R(t) = Rate of income generated at any time t
(Dollars per year)
– r = Interest rate compounded continuously
– T = Term (In years)
• Divide the time interavl [0, T ] into n subintervals of
equal length Δt = T /n and denote the right end
points of these intervals by t1 , t2 , · · · , tn = T
• If R is a continuous function on [0, T ], then the
income generated over the time interval [0, t1 ] is
approximately
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Calculus Chapter 6
R(t1 )Δt
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(Constant rate of income · Length of time )
• The future value of this amount, T years from now,
calculated as if it were earned at time t1 , is
[R(t1 )Δt]er(T −t1 )
(Equation 10 in Section 5.3)
• Therefore, the sum of the future values of the income
stream generated over the time interval [0, T ] is
approximately
R(t1 )er(T −t1 ) Δt + R(t2 )er(T −t2 ) Δt + · · · + R(tn )er(T −tn ) Δt
= erT [R(t1 )e−rt1 Δt + R(t2 )e−rt2 Δt + · · · + R(tn )e−rtn Δt]
Calculus Chapter 6
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dollars. But this sum is just the Riemann sum of the
function erT R(t)e−rt over [0, T ] with representative
points t1 , t2 , · · · , tn .
• The accumulated, or total, future value after T years
of an income stream of R(t) dollars per year, earning
interest at the rate of r per year compounded
continuously, is given by
rT
A=e
0
T
R(t)e−rt dt
• Example 2: Crystal Car Wash recently bought an
automatic car-washing machine that is expected to
generate $40, 000 in revenue per year, t years from
now, for the next 5 years. If the income is reinvested
Calculus Chapter 6
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in a business earning interest at the rate of 12% per
year compounded continuously, find the total
accumalted value of this income stream at the end of
5 years.
– We have R(t) = 40, 000, r = 0.12, T = 5, so the
required value is
0.12(5)
e
0
5
40, 000e−0.12t dt ≈ 274, 039.60
• The present value (PV) of an income stream of R(t)
dollars per year over a term of T years, earning
interest at the rate of r per year compounded
continuously, is the principal P that will yield the
same accumulated value as the income stream itself
Calculus Chapter 6
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when P is invested today for a period of T years at
the same rate of interest. In other words,
rT
Pe
rT
=e
0
T
−rt
R(t)e
dt ⇒ P V =
0
T
R(t)e−rt dt
• Example 3: Two alternative plans for renovating and
improving the theater. Plan A calls for an immediate
cash outlay of $250, 000, whereas plan B requires an
immediate cash outlay of $180, 000. Adopting plan A
would result in a net income stream generated at the
rate of f (t) = 630, 000 dolalrs per year, whereas
adopting plan B would result in a net income stream
generated at the rate of g(t) = 580, 000 dolalrs per
year for the next 3 years. If the prevailing interest
Calculus Chapter 6
rate for the next 5 years is 10% per year, which plan
will generate a higher net income by the end of 3
years?
– Plan A data: initial outlay $250, 000,
R(t) = 630, 000, r = 0.1, T = 3. Present value of
the net income
3
0
630, 000e−0.1t dt − 250, 000 ≈ 1, 382, 845
– Plan B data: initial outlay $180, 000,
R(t) = 580, 000, r = 0.1, T = 3. Present value of
the net income
0
3
580, 000e−0.1t dt − 180, 000 ≈ 1, 323, 254
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Calculus Chapter 6
Amount and Present Value of an Annuity
• An annuity is a sequence of payments made at
regular time intervals. The time period in which
these payments are made is called the term of the
annuity.
• Examples of annuities are regular deposits to a
savings account, monthly home mortgage payments,
and monthly insurance payments.
• The amount of an annuity is the sum of the
payments plus the interest earned.
• Let
– P = Size of each payment in the annuity
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Calculus Chapter 6
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– r = Interest rate compounded continuously
– T = Term of the annuity (in years)
– m = Number of payments per year
• The payments into the annuity constitute a constant
income stream of R(t) = mP dollars per year.
• The amount of an annuity is
A = erT
0
T
R(t)e−rt dt = erT
0
T
mP e−rt dt
mP rT
=
(e − 1)
r
• Example 4: On January 1, 1990, Marcus Chapman
Calculus Chapter 6
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deposited $2000 into an Individual Retirement
Account (IRA) paying interest at the rate of 10% per
year compounded continuously. Assuming that he
deposits $2000 annualy into the account, how much
will he have in his IRA at the beginning of the year
2006?
– We have P = 2000, r = 0.1, T = 16, m = 1, then
2000 1.6
(e − 1) ≈ 79, 060.65
A=
0.1
• The present value of an annuity is given by
mP
PV =
(1 − e−rT )
r
• Example 5: Tomas Perez, the proprietor of a
Calculus Chapter 6
hardware store, wants to establish a fund from which
he will withdraw $1000 per month for the next 10
years. If the fund earns interest at the rate of 9% per
year compounded continuously, how much money
does he need to establish the fund?
– We have P = 1000, r = 0.09, T = 10, m = 12, then
12, 000
(1 − e−(0.09)(10) ) ≈ 79, 124.05
PV =
0.09
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Calculus Chapter 6
6.8 Volumes of Solids of Revolution
• Suppose the plane region under the curve defined by
a nonnegative continuous function y = f (x) between
x = a and x = b is revolved about the x-axis (Figure
6.48).
• The volume of the solid of revolution is approximated
by the sum of the volumes of n disks (Figure 6.51).
• The volume of a disk (cylinder) is approximately
equal to π[f (pi )]2 Δx. Then the volume of the solid of
revolution is approximated by
π[f (p1 )]2 Δx + π[f (p2 )]2 Δx + · · · + π[f (pn )]2 Δx
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Calculus Chapter 6
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• Letting n approach infinity, the volume V is
V =π
b
[f (x)]2 dx
a
• Example 1: Find the volume of the solid of
revolution obtained by revolving the region under the
curve y = f (x) = e−x from x = 0 to x = 1 about the
x-axis (Figure 6.52).
1
1
π
−x 2
−2x
e dx = (1 − e−2 )
π (e ) dx = π
2
0
0
• Example 3: y = ex and y = x2 + 12 from x = 0 to
x = 1 about the x-axis (Figure 6.55).
π
0
1
1
1
1
x 2
2
2
2x
4
2
[e ] − [x + ] dx = π
e −x −x −
dx
2
4
0
Calculus Chapter 6
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• Approximate volume of a football:
x2
– Ellipse a2 +
semi-ellipse.
y2
b2
=1⇒y =b 1−
x2
a2
for the upper
– Revolve about the x-axis
2
2
x
4πb
a
2
b (1 − 2 )dx =
π
a
3
−a
a
– Regulation size: f (x) = −0.0944x2 + 3.4, where
−5.5 ≤ x ≤ 5.5 inches.
V =π
5.5
2
f
(x)
dx ≈ 232 cubic inches
−5.5
• If a = b, then the volume of a ball is
4πa3
3