6
Techniques of Integration
6.1
Integration by Parts
In section 5.2 we discussed solving integrals by making u-substitutions. It was explained
then that this is the integration analogue of the Chain Rule for differentiation. In this
section we discuss the integration analogue of the product rule.
Suppose u and v are functions of x. Then the product rule states
(uv)0 = u0 v + uv 0
We can integrate both sides of this
Z
Z
Z
0
0
(uv) dx = u v dx + uv 0 dx
Now the left hand side is just an antiderivative of (uv)0 , ie, uv. Thus
Z
Z
0
uv = u v dx + uv 0 dx
This can be rewritten
Z
Z
0
uv dx = uv −
u0 v dx
This is called the integration by parts formula. It is useful when we have to find the
integral of a product of two functions. In that case, we can let one of the functions be u
and the other one be v 0 and then use the formula.
Examples:
1. Evaluate
R
xe3x dx.
solution: This integral can’t be done with any kind of simple substitution, so we try
integration by parts. We have to choose a u and a v 0 such that xe3x = uv 0 and that
R 0
u v dx will be easier to solve. In this case, we let
u=x
v 0 = e3x
To use the formula we need u0 and v, so we solve for them by taking the derivative of
u and the antiderivative of v 0 :
u0 = 1
1
v = e3x
3
Now we can use the integration by parts formula:
Z
Z
0
uv dx = uv − u0 v dx
Z
Z
1
1 3x
3x
− (1) e3x dx
e
xe dx = x
3
3
1 3x 1 3x
=
xe − e + C
3
9
R
2. Evaluate x ln(x) dx.
solution: Once again we would like to use integration by parts, and it’s clear that one
of the parts should be x and the other should be ln(x). We shouldn’t make v 0 = ln(x)
because we don’t yet know how to take ln’s antiderivative. So we set
u = ln(x)
v0 = x
Which gives us
u0 =
1
x
v=
x2
2
So integration by parts gives us
Z
Z
0
uv dx = uv − u0 v dx
Z
Z
x2
1 x2
x ln(x) dx =
ln(x) −
·
dx
2
x 2
Z
x
1 2
x ln(x) −
dx
=
2
2
1 2
x2
=
x ln(x) −
+C
2
4
R
3. Evaluate ln(x) dx.
solution: As before we don’t know how to take ln’s antiderivative, but we do know how
to take its derivative. So if we take
u = ln(x)
v0 = 1
Then uv 0 = ln(x) and we can use integration by parts. The above gives us
1
u0 =
v=x
x
So we have
Z
Z
0
uv dx = uv − u0 v dx
Z
Z
1
ln(x) dx = x ln(x) −
(x) dx
x
Z
= x ln(x) − (1) dx
= x ln(x) − x + C
Notes:
• It is crucial to make the right choice. Making the wrong choice for u and v 0 will usually
give you a harder integral than the one you started with. For example, say you were
asked to solve
Z
xex dx
and chose
u = ex
v0 = x
u0 = ex
v=
giving you
x2
2
Then integration by parts would give you
Z
Z 2
x2 x
x x
x
xe = e −
e dx
2
2
| {z }
even harder
We ended up with a worse integral, so we know we made the wrong choice for u and
v0.
• Sometimes you have to do integration by parts twice. For example, say you were asked
to solve
Z
x2 ex dx
The reasonable thing to do would be to choose
u = x2
v 0 = ex
u0 = 2x
v = ex
giving you
Then integration by parts gives you
Z
Z
2 x
2 x
x e dx = x e − 2xex dx
The second integral we can now do, but it also requires parts. We take
u = 2x
v 0 = ex
u0 = 2
v = ex
giving us
So we have
Z
2 x
2 x
x
x e dx = x e − 2xe −
Z
x
2e dx = x2 ex − 2xex + 2ex + C
In general, you need to do n integration by parts to evaluate
R
xn ex dx.
• In the case of definite integrals, the integration by parts formula becomes
Z b
b Z b
0
u0 v dx
uv dx = uv −
a
a
a
Application: Present Value
Definition 6.1 The present value of a future payment is the amount that would have
to be deposited today to produce the future payment.
If, for instance, someone has promised to pay you $1000 in two years, the present value
of that payment is the amount he would have to pay you now to give you the same value
as $1000 two years from now. One would assume that the present value of this payment
would be less than $1000 because in two years inflation would have likely caused the $1000
to have less value than it does today.
Formulas:
If c(t) represents a continuous income function in dollars per year and the annual inflation
rate is r then we have
1. The actual income over t1 years is
Z
t1
c(t) dt
0
2. The present value of that income is
Z
t1
c(t)e−rt dt
0
This formula also applies if the money is invested at a rate of r% (and you ignore
inflation).
Examples:
1. You have won the lottery for $1,000,000. You will be paid $50,000 per year for 20 years.
Assuming an annual inflation rate of 6%, what is the present value of this income?
solution: First let’s calculate the actual value:
20
Z 20
50000 dt = 50000t = 1,000,000
0
0
as expected. Now we calculate the present value:
20
Z 20
50000
−0.06t
−0.06t 50000e
dt =
e
≈ $582,338
−0.06
0
0
So this would be the value today of that money paid out over 20 years.
2. Suppose c(t) = 30000 + 500t, r = 7% and t1 = 6. Find the present value.
solution: We evaluate
Z
Z 6
−0.07t
(30000 + 500t)e
dt =
0
6
30000e
−0.07t
0
Z
dt +
6
500te−0.07t dt
0
For the second integral we need parts. We leave it as an exercise to evaluate it.
6.5
Improper Integrals
Consider y = e−x , and the area under it and above [1, 7].
R7
We know how to calculate this already - it’s just 1 e−x dx. But suppose that we wanted
to know the area under y = e−x but above [1, ∞). Since e−x approaches the x-axis very
quickly it’s possible that the total area is finite.
R∞
R 100
What we want is 1 e−x dx. But what does this mean? We can calculate 1 e−x dx,
R 1000000 −x
R ∞ −x
e
dx
and
so
on.
Each
provides
a
better
approximation
to
e dx. So our
1
1
R ∞ −x
intuition should be that 1 e dx is the limit of this.
Definition 6.2 Suppose f (x) is continuous on [a, ∞) then we define
Z ∞
Z b
f (x) dx = lim
f (x) dx
a
a→∞
a
and call this the improper integral of f from a to ∞.
So an improper integral is an ordinary integral followed by an ordinary limit.
Examples:
1. Evaluate
R∞
1
e−x dx.
solution:
Z
∞
e
1
−x
Z
b
e−x dx
1
b 
= lim −e−x 
b→∞
1
−b
= lim −e + e−1
dx =
lim
b→∞
b→∞
=
1
e
So the area under the curve y = e−x and above [1, ∞) is 1e .
R∞
2. Calculate 2 √1x dx
solution: The area in question looks like this:
The function y = √1x approaches the x-axis as x goes to infinity, so it is possible that
there will only be a finite area.
Z b
Z ∞
1
1
√ dx = lim
√ dx
b→∞
x
x
2
2
b 
1/2 x 
= lim 
b→∞
1/2 2
h √
√ i
= lim 2 b − 2 2
b→∞
= ∞
Thus the area under the curve is infinite. In cases where the limit doesn’t exist (ie is
infinite) we say the improper integral diverges.
R0
3. Calculate −∞ ex dx.
solution: Here our limit is negative infinity, but we deal with it in the same way
Z
0
x
Z
−∞
0
ex dx
b 
0
= lim ex 
b→−∞
b
= lim 1 − eb
e dx =
lim
b→−∞
b→−∞
= 1
So the total area is 1.
There is another type of improper integral, though it is not as obvious to spot. Suppose
we want to find
Z 2
1
dx.
2
0 x
This represents the following area
Here we are once again dealing with a possibly infinite area, and so we need to use a limit.
Z 2
Z 2
1
1
dx = lim+
dx
2
2
b→0
0 x
b x

2 
1 = lim+ − 
b→0
x
b
1 1
= lim+ − +
b→0
2 b
= ∞
Here the limit diverges to infinity, so the area is infinite. Notice that we take the limit
through values greater than 0.
This type of improper integral has to be done whenever we are integrating a function f and
one of the endpoints of the integral is not in the domain of f .
4. Calculate
R4
0
√1
x
dx.
solution: Here the right endpoint is not in the domain of the function being integrated,
so this integral is improper.
Z 4
Z 4
1
1
√ dx = lim
√ dx
b→0+ b
x
x
0

4 
√ = lim+ 2 x 
b→0
b
h √
√ i
= lim+ 2 4 − 2 b
b→0
= 4.
Application: Present Value
The present value of a payment given out in perpetuity can be calculated using improper
integrals.
Example: Suppose an alumnus of a university wishes to make a gift to the university of a
scholarship fund for business majors. The find will give out $18,000 per year every year.
Assuming an interest rate of 10% (compounded continuously), what is the present value of
the endowment?
solution: Since the endowment should be paying out money forever, we get that the present
value is
Z ∞
Z b
−0.10t
c(t)e
dt = lim
18000e−0.10t dt
b→∞
0
0
b 
18000 −0.10t 

e
= lim
b→∞ −0.10
0
= lim 180000 − 180000e−0.10b
b→∞
= $180,000
So the present value is $180,000.