Final Review
Riemann Sums
If we want to estimate the area of a region,
we break it up into little rectangles; we then
add up the areas of the individual rectangles. This is the main idea of Riemann
sums. Let’s say we wanted to approximate
the area under the graph of a function f (x)
on an interval from x = a to x = b. First,
we partition the interval [a, b] by choosing
sample points
x0 = a < x1 < x2 < · · · < xn−1 < xn = b.
The width of the ith rectangle is ∆xi =
xi − xi−1 . The height is f (ci ), where ci is
chosen so that xi−1 ≤ ci ≤ xi . Adding
up the areas of the rectangles gives you an
approximate answer
Area ≈
n
X
Problem 4. For the functions f1 (x) =
5x2 + 7x + 1, f2 (x) = 3x2 − 5x + 15,
f3 (x) = 10x2 − x + 1, find closed formulas
for the approximate area under their graphs
from x = 0 to x = 4 using n rectangles,
right-endpoint approximations (i.e. calculate Rn for each function).
By taking finer partitions of the interval
[a, b], we get better approximations. We
take a limit as the widths ∆xi go to 0. In
other words, we use more rectangles, but
make them skinnier and skinnier.
Problem 5. In problem ?? above, calculate the limit lim Rn to determine the area
n→∞
under the graph of each function, f1 , f2 , f3 ,
for the bounds x = 0 to x = 4.
Definite Integrals
f (ci )∆xi .
i=1
Problem 1. Estimate the area under the
graph of f (x) = 3x + 1 from x = 1 to x =
2 using 5 rectangles. Use the left-endpoint
and right-endpoint Riemann sums.
Problem 2. Estimate the area under the
graph of f (x) = 3x + 1 from x = 1 to x = 2
using n rectangles. Use the left-endpoints
and right-endpoint Riemann sums.
Problem 3. Estimate the area under the
graphs of s(x) = sin(x) and c(x) = cos(x)
from x = 0 to x = π using 6 rectangles.
Use left-endpoint and right-endpoint Riemann sums.
Next, we introduce the integral sign to
mean the following:
Z
b
f (x)dx = lim
n→∞
a
n
X
f (ci )∆xi .
i=1
As long as f (x) has at most finitely many
discontinuities on the interval [a, b], the
limit above will not depend on the chosen
partition of [a, b]. With this interpretation
of a definite integral as “signed” area, we
can answer the following questions.
Problem 6. ZWhat is the value of the defi5√
25 − x2 dx?
nite integral
0
Problem 7. ZWhat is the value of the defi2
nite integral
|x − 1|dx?
−1
Problem 8. What
Z 3 is the value of the defbxcdx, where bxc is the
inite integral
0
greatest integer less than or equal to x?
The Fundamental Theorem of
Calculus
Next, we were introduced to a very important theorem. It’s so important that it has
two parts and goes by the grandiose title
“The Fundamental Theorem of Calculus”.
This theorem unites the area problem with
differential calculus. The first part says
Theorem 9 (FTC1). Let f be a continuous
function. Define the function g(x) by the
rule
Z x
g(x) =
f (t)dt,
a
where a is a constant. Then g(x) is differentiable and g 0 (x) = f (x).
In some sense, FTC1 says that if you integrate a function f , and then differentiate
the result, you get the original function f
back.
Problem 10. Differentiate the following
functions:
Z x
2
• f (x) =
et dt,
5
Z
• f (x) =
x2
sin(t)dt,
0
Z
ex
• f (x) =
sin(5t)dx.
x7
Problem 11. Compute the limit
Rx t
e dt
lim 1
.
x→1 x − 1
The second part of the Fundamental Theorem of Calculus says
Theorem 12 (FTC2). Let f be a continuous function on an interval [a, b] and let F
be ANY antiderivative of f , then
Z b
F (b) − F (a) =
f (x)dx.
a
The FTC2 makes computing areas much
easier than using the Riemann sum definition, provided we know an antiderivative.
Problem 13. Use the FTC2 to compute
the area under the graphs of the functions
listed in problem ?? above, for the bounds
x = 0 to x = 4. Check that you get the
same answer as before.
Problem 14. What is the area under the
1
graph of the function f (x) =
from
1 + x2
x = 0 to x = 1?
Also, we introduce the notation of indefinite integrals
Z
f (x)dx
R
to mean f (x)dx is any antiderivative of
f . For example
Z
x5 dx =
where C is a constant.
x6
+ C,
6
Integration Techniques
Once we saw that antiderivatives were
important for computing areas, we next
sought ways to calculate antiderivatives.
One of the first techniques we encountered
was the substitution rule. This is a chain
rule for antiderivatives.
Problem 15. Make an appropriate usubstitution to compute the following:
Z
• (3x + 2)50 dx,
the integrand has something of the form
√
√
√
a2 + x2 , x2 − a2 , or a2 − x2 , this gives
us a clue that a trig substitution may be
appropriate.
Problem 18. Integrate the following (the
first one is the most difficult):
Z 5 r
1
1
•
1 + 4 dx,
x
1 x
Z 1√
•
x2 + 1dx,
0
Z
Z
•
5x
√
6
•
(x4 + 2x2 + 5)3/2
x7 + 1dx,
√
Z p
1+ x
√
•
dx.
x
Some substitutions are less obvious than
others.
Problem 16. Integrate using any method
you see fit. (Hint: The substitution u =
tan(x/2) will always transform the following integrals into rational functions in u,
but you may find other techniques easier for
some of these):
Z
dx
,
•
1 − sin x
Z π/2
dx
•
,
4 sin x + 3 cos x
0
Z
3 sin x − cos x
•
dx.
1 + cos x
Problem 17. Use the substitution u =
R
sec x + tan x to integrate sec xdx.
Another technique for integration is called
trigonometric substitution. Basically, when
xdx
.
Another technique for integration is called
integration by parts. This is a product rule
for antiderivatives.
Z f (x)g (x) + f (x)g(x) dx = f (x)g(x) + C
0
0
In other words,
Z
Z
0
f (x)g (x)dx = f (x)g(x)− f 0 (x)g(x)dx.
Problem 19. Integrate the following:
Z
•
xex dx,
Z
•
Z
•
Z
•
Z
•
ln(x4 + 3x3 )dx,
tan−1 (x)dx,
ex sin(x)dx,
x2 cos(x)dx.
Finally, we saw one other useful integration technique called partial fractions. This
is essentially algebra rather than calculus.
The idea is to simplify an integrand of the
(x)
form fg(x)
into something more manageable,
where f (x) and g(x) are polynomials. For
instance,
1
1
1
1
=
−
.
x2 − 1
2 x−1 x+1
We first check whether or not deg(f ) <
deg(g). If not, then we need to use polynomial division to write
f (x)
r(x)
= q(x) +
,
g(x)
g(x)
where q(x), r(x) are polynomials with
0 ≤ deg(r) < deg(g).
Problem 20. Use polynomial division to
simplify the following rational functions
(i.e. find the quotient q(x) and the remainder r(x), as described above).
•
x4
,
x4 − 1
•
x5 + 2x
,
x2
•
x2 − 2x + 1
.
x−1
3
•
•
x −x +x−1
.
x2 + 1
10
,
2
x − 7x + 10
x3
2x
,
x3 − x2 + 4x − 4
•
1
.
(x − 1)2 (x + 2)
Problem 22. Integrate the following:
Z
x4
dx,
•
x4 − 1
Z 5
x + 2x
•
dx,
x2
Z 2
x − 2x + 1
dx.
•
x−1
Z 3
x − x2 + x − 1
•
dx,
x2 + 1
Z
10
dx,
•
2
x − 7x + 10
Z
x−9
•
dx,
3
x − 6x2 + 11x − 6
Z
2x
•
dx,
x3 − x2 + 4x − 4
Z
1
dx.
•
(x − 1)2 (x + 2)
Applications of Integration
2
Problem 21. Write the partial fraction decomposition for the following:
•
•
x−9
,
− 6x2 + 11x − 6
One of the first applications of integration
that we saw was called the net change theorem. The FTC2 can be rewritten as follows
The Net Change Theorem
Let f (x) be a differentiable function
defined on an interval [a, b], then
Rb
f (b) − f (a) = a f 0 (x)dx.
With this theorem, we can answer the following questions:
Problem 23. A particular chickadee flies
2
with velocity v(t) = t27t+1 miles per hour,
where t is the number of hours after
12:00pm on March 3, 1745. How far did
the chickadee travel between the hours of
3:00pm and 4:30pm on March 3, 1745?
Problem 24. Let p(t) represent the worldwide wombat population (in millions) t
months after January 1, 1822. Suppose the
100e10t
population growth rate at time t is (1+e
10t )2
wombats
per month. If p(0) = 5, what is
ln(10)
p 10 ? What value does p(t) approach
as t goes to infinity?
Problem 25. A train is initially at rest
and accelerates at a rate of ct2 meters per
second, where t is the number of seconds
after the train starts moving. Suppose the
train travels 6 meters in the first 2 seconds.
What is c?
Problem 26. A ball is rolled across the
floor with an initially velocity of 15 meters
per second. Due to friction, it slows down
at a rate of 3 meters per second squared.
When does the ball stop? How far has it
travelled?
favg =
1
b−a
Rb
a
f (x)dx
Problem 27. What is the average value of
1
the function f (x) = 1+x
2 on the interval
[−1, 1]?
Problem 28. Professor Johnson plots the
attendance rate in the calculus course each
day and notices the function
t
1 2
s(t) = 110 cosh
+
t − 40t + 60
10
3
seems to model the data very nicely. Here,
t represents the number of weeks since the
quarter started and cosh(x) = 12 (ex + e−x )
is the hyperbolic cosine function. What is
the average attendance rate over the 10week quarter?
Problem 29. Over the course of a year,
a particular company has an average profit
of 100 dollars per day. On the first day of
the year, the company had just opened and
only made 3 dollars profit. Also, the function P (t) which gives the profit (in dollars)
on the t-th day after the start of the year
has the form P (t) = mt + b. What is P (t)
explicitly?
Work
Average Values of Functions
We’ve seen how integration can be used to
calculate average values of functions. The
formula for the average value of a function
f (x) between x = a and x = b is
We’ve also seen how integration can be used
to calculate the amount of work done in
moving on object from point A to point B.
The formula for work is
Work =Force·Distance
Problem 30. Suppose of force of 5 Newtons is used to move an object 2 meters.
What is the work done?
The answer to the above question is 10
Newton meters (or 10 Joules). You simply multiply the force with the distance.
Now suppose the force varies as the object
moves.
Problem 31. Suppose an object is moving
along the x-axis, and a force of F (x) = 3x2
Newtons acts on the object when it is at position x along the axis. How much work is
done to move the object from x = 0 meters
to x = 5 meters?
To answer the above question, we partition
the interval [0, 5] into little pieces. When
the object is at a point x∗i , the force acting
on it is F (x∗i ) and the distance it moves is
∆xi . Thus, we add up everything to get an
approximation
X
Work done ≈
F (x∗i )∆xi .
i
By partitioning the interval into finer
pieces, we get better approximations. To
calculate the work done to move an object
from x = a to x = b, we compute the integral
Z b
Work done =
F (x)dx.
a
Sometimes, the force may be constant, but
the object may need to be split up into tiny
pieces.
Problem 32. Suppose an aquarium is full
of water and is in the shape of a cube with
all sides measuring 2 meters. How much
work is required to lift all of the water to
the top of the aquarium, thereby emptying
it. The density of water is ρ = 1000kg/m3
and the acceleration due to gravity is g =
9.8m/s2 .
Problem 33. A rope, 50 ft long, weights
0.5 lb/ft and hangs over the edge of a building 200 ft high. How much work is done in
pulling the rope to the top of the building?
Volumes
We also saw that integration can be used
to calculate volumes. When we calculate
the volume of an object, the idea is to split
the object up by taking cross-sections. We
get an approximate volume by adding up
the volumes of the individual cross-sections,
or slabs. When we slice the object thinner
and thinner, we get better approximations
to the actual volume,
X
Volume ≈
A(x∗i )∆xi .
Here, the object is split up by partitioning
the x-axis into little pieces, and A(x∗i ) represents the cross-sectional area at the sample point x∗i . Partitioning the x-axis into
finer pieces gives us the formula
Z b
Volume =
A(x)dx.
a
Problem 34. Find the volume of the solid
whose base is the region bounded by the xaxis and the arch of the curve y = sin(x)
for x = 0 to x = π and for which each
cross section perpendicular to the x-axis is
a square whose base lies in the region.
For volumes of revolution, we saw two ways
to compute the volume, the cylindrical shell
method and the washer (or disk) method.
We use the formulas
Cylindrical Shell Method
Rb
Volume = 2π a d(x)h(x)dx.
Washer Method
Rb
Volume = π a (R(x)2 − r(x)2 ) dx.
In the above formulas, d(x) is the distance
from the rectangle to the axis of revolution,
h(x) is the height of the rectangle at the
sample point x, R(x) is the outer radius,
and r(x) is the inner radius.
Remark: There are also formulas for integration along the y-axis. Depending on the
geometry of the problem, it may be better
to use those formulas instead.
Problem 35. The graph of f (x) = x1 from
x = 1 to x = 6 is spun around the x-axis
to make the shape of a horn. Compute its
volume.
Problem 36. In problem ??, you computed
the volume of a horn. What happens to
the volume when the horn gets longer and
longer? In other words, compute the volume from x = 1 to x = t and then see
what happens as t goes to infinity? Does the
volume converge, and if so, to what value?
( Remark: This example is called Gabriel’s
horn.)
Problem 37. Find the volume of the solid
obtained by rotating the region bounded by
the given curves about the specified line.
Sketch the region, the solid, and a typical
disk or washer.
• y = 2 − 12 x, y = 0, x = 1, x = 2, about
the x-axis,
√
• x = 2 y, x = 0 , y = 9, about the
y-axis,
• y 2 = x, x = 2y, about the y-axis.
• x = 1 + y 2 , x = 0, y = 0, y = 1, about
the x-axis,
Improper Integrals
Sometimes the bounds of integration may
be infinite. In this case, we call the integral improper. For example, the so-called
Gamma function Γ(x) is defined by
Z ∞
tx−1 e−t dt.
Γ(x) =
0
We didn’t spend time in class talking about
the gamma function; it is provided here
only as an example of how some functions
in mathematics are defined via improper integrals. When the upper bound is infinity,
what we really mean is
Z
Γ(x) = lim
b→∞
b
tx−1 e−t dt.
0
In other words, we compute a definite integral over finite bounds, and then take a
limit as the bound goes to infinity.
Problem 38. What is Γ(1)? (i.e. ComR∞
pute 0 e−t dt.)
Problem 39. Let n be a positive integer.
Use integration by parts to show Γ(n + 1) =
nΓ(n).
Remark: An interesting result that follows from problems (??) and (??) is
Γ(n + 1) = n! Here, n! is pronounced
“n factorial” and is the product of the
numbers 1, 2, ..., n. Thus, the Gamma
function generalizes the factorial function
to non-integer values.
R∞
An improper integral a f (x)dx might converge to a finite value, or it may diverge.
For example, we have
( p+1
Z ∞
− ap+1
if p < −1,
xp dx =
∞
if p ≥ −1.
a
When p < −1, the integral converges to
p+1
− ap+1 , and when p ≥ −1 the integral diverges. This is an important example because it gives us a nice basis for comparison.
Recall that the comparison test is a useful tool to determine whether an integral
R∞
f (x)dx converges or diverges. Oftena
times, it is useful to compare the integrand
f (x) with xp .
Problem 40. Use the comparison test to
determine whether the following integrals
converge or diverge
R∞
2 (x)
• 1 1+sin
dx,
x3
R∞
7
• 5000000 xx7 +1
dx,
−1
R∞
2
• 6 e−x dx,
R∞
dx.
• 2 sin(x)+2
x
Arc Length
If we want to compute the length of a curve
y = f (x) from x = a to x = b, we showed
how the curve can be split up into tiny
pieces approximated by line segments.
Problem 41. Approximate the length of
the curve given by the graph of f (x) = 1/x
between x = 1 and x = 5 by splitting
up the curve into 4 line segments at the
points (1, 1), (2, 1/2), (3, 1/3), (4, 1/4) and
(5, 1/5).
To get the actual length, we partition the
interval [a, b] into finer pieces and the formula becomes
Arc Length Formula
Rbp
L = a 1 + (f 0 (x))2 dx
Problem 42. Find the length of the following curves:
• y = 1 + 6x3/2 , 0 ≤ x ≤ 1,
• y = ln(sec x), 0 ≤ x ≤ π/4,
Rx√
• y = 1 t3 − 1dt, 1 ≤ x ≤ 4.
Surface Area
We can use integration to compute the area
of a surface of revolution. The idea we use is
to approximate the surface by bands. The
bands can be considered as a portion of a
circular cone. If we want to compute the
surface area of an object obtain by rotating
a curve y = f (x) about the x-axis from
x = a to x = b, we use the formula
Surface Area = 2π
Rb
a
f (x)ds,
where ds is the arc length
s
2
dy
dx.
ds = 1 +
dx
For an object obtained by rotating a curve
x = f (y) about the y-axis from y = c to
y = d, we use the formula
Rd
Surface Area = 2π c f (y)ds,
where ds is the arc length
s
2
dx
ds = 1 +
dy.
dy
Problem 43. Find the area of the surface
obtained by rotating the following curves
about the x-axis:
√
• y = 1 + 4x, 1 ≤ x ≤ 5,
• y = sin(πx), 0 ≤ x ≤ 1.
Problem 44. Find the area of the surface
obtained by rotating the following curves
about the y-axis:
• y = x1/3 , 1 ≤ y ≤ 2,
p
• x = 1 − y 2 , 0 ≤ y ≤ 1/2.
Problem 45. Rotate the curve y = 1/x
for 1 ≤ x ≤ 5 around the x-axis to make a
surface in the shape of a horn. Compute its
surface area. (Hint: see Problem ??)
Problem 46. Now rotate the same curve
y = 1/x from Problem ?? around the xaxis, but change the upper bound to t and
let t go to infinity. This will build a horn
that is infinitely long, called Gabriel’s horn.
Does its surface area converge to a specific
value, or does it diverge?