BEE1004 { Basic Mathematics for Economists
Dieter Balkenborg
BEE1005 { Introduction to Mathematical Economics
Department of Economics
Week 16, Lecture 1.2, Notes: The Inde¯nite Integral 11/02/2002
University of Exeter
1
Introduction
The integral is for instance needed for calculating consumer surplus, to ¯nd the cumulative
distribution associated with a density or to solve di®erential equations. In this handout
we discuss the inde¯nite integral and the most basic techniques of integration. We follow
closely the textbook by Ho®mann and Bradley, Chapter 5, Sections 1, 2 and 4.
2
The Inde¯nite Integral
Integration is the inverse operation of di®erentiation. Consider for instance the function
f (x) = 2x:
A function which has f (x) as derivative is
F (x) = x2 :
We call F (x) the antiderivative of a function f (x) when the derivative of F (x) is f (x):
dF
= f (x).
dx
The function F (x) = x2 is not the only antiderivative of f (x) = 2x. Also the functions
G (x) = x2 + 10 and H (x) = x2 ¡ 5 are antiderivatives. This is so because additive
constants vanish when di®erentiating. However, two antiderivative can di®er only by an
additive constant.
Whereas an antiderivative is determined only up to a constant, the following question
has a unique answer: Find the antiderivative F (x) of f (x) = 2x with F (5) = 2. Solution:
An antiderivative of f (x) must be of the form F (x) = x2 + C with a suitable constant
C. To have F (5) = 2 we must have
F (5) = 52 + C = 25 + C = 2
C = 2 ¡ 25 = ¡23
Hence F (x) = x2 ¡ 23 solves the problem.
Here are some important antiderivatives:1
f (x)
0
1
x¡1
x® , ® 6= ¡1
ex
F (x)
C
x+C
ln jxj + C
1
x®+1 + C
®+1
ex + c
Notation and terminology: If F (x) is an antiderivative of f (x) one writes
Z
f (x) dx = F (x) + C
Z
1
e.g.
xdx = x2 + C
2
R
and calls f (x) dx the inde¯nite integral because the result is de¯nite only up to a
constant.
3
Rules of Integration
These rules are just inverted rules
¯ndings {
Z
0dx =
Z
kdx =
Z
x¡1 dx =
Z
x® dx =
Z
ex dx =
of di®erentiation. We have { just restating the above
C
kx + C
ln jxj + C
1
x®+1 + C
®+1
for ® 6= ¡1
ex + C
Moreover,
Z
Z
Z
(f (x) + g (x)) dx =
f (x) dx + g (x) dx
(rule for sums)
Z
Z
®f (x) dx = ® f (x) dx
(rule for multiplicative constants)
1
For positive x one has ln jxj = ln x and hence
hence
djln xj
dx
= x1 . For negative xone has ln jxj = ln (¡x) and
1
1
d ln jxj
£ (¡1) =
=
dx
(¡x)
x
by the chain rule.
2
Example 1
Z µ
p
3
5e + x2 +
Z
Z
Z
2
x
5 e dx + x 3 dx + 5
x
5
x
¶
Z ³
´
2
¡1
x
3
dx =
5e + x + 5x
dx =
3 5
x¡1 dx = 5ex + C1 + x 3 + C2 + 5 ln jxj + C3
5
5
3
= 5ex + x 3 + 5 ln jxj + C
5
where C = C1 + C2 + C3 .
3.1
Integration by substitution
1
With the above rules we cannot yet integrate simple functions like e5x+2 or 3¡2x
. However,
the solutions can easily be guessed using the chain rule: Di®erentiating a function y =
ae5x+2 gives 5ae5x+2 according to the chain rule: Set u = 5x + 2. Then du
= 5, y = aeu ,
dx
dy
= aeu and
dx
dy
dy du
=
= aeu £ 5 = 5ae5x+2 :
dx
du dx
Hence
Z
1
e5x+2 dx = e5x+2 + C
5
1
Similarly ln j3 ¡ 2xj has the derivative 3¡2x
£ (¡2) and hence
Z
1
1
dx = ¡ ln j3 ¡ 2xj + C:
3 ¡ 2x
2
More generally, integration by substitution
Z
Z
du
g (u (x)) dx = g (u) du
dx
is the inversion of the chain rule. This can be seen as follows: Suppose y = G (u) is an
dy
= g (u). Let u = u (x). By the chain rule the function
antiderivative of g (u), i.e., du
y = G (u (x)) has the derivative
dy
dy du
du
=
= g (u (x))
dx
du dx
dx
.
so G (u (x)) is an antiderivative of g (u (x)) du
dx
Example 2
Z
e5x+2 dx
Set u = 5x + 2, so du
= 5 or du = 5dx or dx = 15 du. Substituting 5x + 2 by u and dx by
dx
1
dx we obtain
5
Z
Z
Z
1
1
1
1
5x+2
u
dx = e £ du =
e
eu du = eu + C = e5x+2 + C
5
5
5
5
3
Example 3
Z
2
xex dx
1
Set u = x2 . Then du = 2xdx or dx = 2x
du. Therefore
Z
Z
x2
xe dx = xeu dx =?
Example 4
Z
2
2
xex dx
2
1
Try u = ex . Then du = 2xex dx = 2xudx. Hence dx = 2xu
du
Z
Z
Z
1
1
1 2
1
x2
xe dx = xu
du =
du = u + C = ex + C
2xu
2
2
2
Example 5
Z
x2 ex dx
Try u = ex . Then du = ex dx = udx, x = ln u and hence
µ ¶
Z
Z
Z
Z
Z
1
2 x
2
2
2
du = x du = (ln u)2 du
x e dx = x udx = x u
u
which does not really help.
Method: Substitute some sub-expression of x in the integral by u. Then du =
u (x) dx yields dx = u01(x) du. Replace dx by u01(x) du in the integral. Hopefully u0 (x) is a
factor in the integral and hence cancels against u01(x) so that a simple expression in u only
remains. Otherwise try a di®erent method.
0
3.1.1
Partial Integration
Partial integration is the inverse of the product rule
(uv)0 = u0 v + uv0
for functions u (x), v (x). We can rewrite this as
u0 v = (uv)0 ¡ uv 0
or
Z
0
u vdx = uv ¡
since uv is the antiderivative of (uv)0 .
4
Z
uv 0 dx
Example 6
Z
x2 ex dx
Set v = x2 and u0 = ex . Then v 0 = 2x and u = ex . Therefore
Z
Z
2 x
2 x
x e dx = x e ¡ 2 xex dx:
To solve the last integral set v = x and u0 = ex. Then v0 = 1 and u = ex . Therefore
Z
Z
x
x
xe dx = xe ¡ 1 £ ex dx = xex ¡ ex + C
and overall
Z
x2 ex dx = x2 ex ¡ 2xex + 2ex ¡ C
Example 7
Z
x2 ex dx
Try u0 = x2 , v = ex . Then u = 13 x3 and v 0 = ex and therefore by partial integration
Z
Z
1 3 x 1
2 x
x e dx = x e ¡
x3 exdx
3
3
This does not simplify matters.
5