Integral Calculus
Daniel Rakotonirina
February 9, 2017
1
Integration by Parts
It is a useful method to evaluate integrals of products of functions. Given two differentiable functions u and v,
the product rule states:
d
(u(x) · v(x)) = u0 (x) · v(x) + u(x) · v 0 (x)
dx
By integrating both sides, we can write this rule in terms of an indefinite integral:
Z
Z
Z
d
0
(u(x) · v(x)) = u (x) · v(x)dx + u(x) · v 0 (x)dx
u(x) · v(x) =
dx
Rearranging the expression, we have:
Z
Z
u(x) · v 0 (x)dx = u(x) · v(x) − v(x) · u0 (x)dx
| {z }
| {z }
dv
du
Here, the goal is to suppress the independent variable x. We can write it as:
Indefinite
Z
Z
u dv = uv −
v du
Definite
Z
a
b
b Z
u(x) v 0 (x)dx = u(x) v(x) −
b
v(x) u0 (x)dx
a
a
OR
b
Z
Z
b
u dv =uv −
a
v du
a
Example 1.1. Integration by parts. Evaluate
R
xex dx
Functions in original integral
u=x
dv = ex dx
Functions in new integral
du = dx
v = ex
Z
x
x
x e| {z
dx} = |{z}
x |{z}
e −
|{z}
u
dv
u
v
1
Z
ex |{z}
dx
|{z}
v
du
We have:
Z
x ex dx = xex −
Z
ex dx = xex − ex + c = (x − 1)ex + c
When is the ”Integration by parts" the best choice to evaluate an integral?
1. The integrand is a product of functions (u · dv)
2. u is easy to differentiate and v is easy to integrate
The integrand is a product of functions
Many books mention the ILATE rule of thumb here. These are supposed to be memory devices to help you
choose your u and dv in an integration by parts question. We have:
• I = inverse trigonometric
• L = logarithmic
• A = algebraic
• T = trigonometric
• E = exponential
ILATE rule is supposed to suggest the order in which we choose the u.
Example 1.2.
Z
−1
−1
x tan
|{z}
| {z x} dx =⇒ u = tan x
A
dv = xdx
I
Z
x2 |{z}
ln x dx =⇒ u = ln x
|{z}
A
dv = x2 dx
L
Example 1.3. Integration by parts. Evaluate the following integral:
Z e
x ln xdx
1
Here, we have an algebraic and a logarithmic functions. He have:
Functions in original integral
Functions in new integral
u = ln x
1
du = dx
x
dv = xdx
x2
v=
2
e Z e 2
e Z e
x2 x 1
x2 x
−
dx
=
ln
x
dx
−
2
2
x
2
1
1
1 2
1
1
e
Z e
e2
12
1 x2 e2
1
x ln xdx = ln(e)
− ln(1)
−
=
ln(e)
− (e2 − 1)
2
2
2 2 1
2
4
1
Z e
e2
1
e2 + 1
x ln xdx =
+ =
4
4
4
1
Z
e
x ln xdx = ln x
Related Exercises sec. 7.2 7–22
2
Example 1.4. Repeated use of integration by parts. Evaluate the following integral:
Z
x2 ex dx
From the ILATE rule, we can choose u = x2 and dv = ex dx Hence, we have:
Functions in original integral
u = x2
dv = ex dx
Functions in new integral
du = 2xdx
v = ex
Z
x2 ex dx = x2 ex −
Z
ex 2x dx = x2 ex − 2
Z
Here we have to apply again the Integration by Parts on
Z
xex dx
xex dx. Therefore, we choose again u and dv.
Functions in original integral
u=x
dv = ex dx
Functions in new integral
du = dx
v = ex
Finally, we have:
Z
Z
So for
Z
x
x
x e dx = x e − 2 xe − e dx = x2 ex − 2[(x − 1)ex ] + c
2 x
2
x
xn ex dx, where n is a positive integer, we apply the IBP n−times. Hence,
Z
xn ex dx = xn ex − n
3
Z
xn−1 ex dx