BEE1020 — Basic Mathematical Economics Dieter Balkenborg
Week 16, Lecture Thursday 20.02.03
Department of Economics
The Indefinite Integral
University of Exeter
1
Introduction
The integral is for instance needed for calculating consumer surplus, to find the cumulative
distribution associated with a density or to solve differential equations. In this handout
we discuss the indefinite integral and the most basic techniques of integration. We follow
closely the textbook by Hoffmann and Bradley, Chapter 5, Sections 1, 2 and 4.
2
The Indefinite Integral
Integration is the inverse operation of differentiation. 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 differentiating. However, two antiderivative can differ 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
α
x , α 6= −1
ex
F (x)
C
x+C
ln |x| + C
1
xα+1 + C
α+1
ex + c
x|
1
For positive x one has ln |x| = ln x and hence d|ln
dx = x . For negative xone has ln |x| = ln (−x) and
hence
d ln |x|
1
1
=
× (−1) =
dx
(−x)
x
by the chain rule.
1
1
Notation and terminology: If F (x) is an antiderivative of f (x) one writes
Z
f (x) dx = F (x) + C
Z
e.g.
2xdx = x2 + C
R
and calls f (x) dx the indefinite integral because the result is definite only up to a
constant. f (x) is called the integrand.
3
Rules of Integration
These rules are just inverted rules
findings —
Z
0dx =
Z
kdx =
Z
x−1 dx =
Z
xα dx =
Z
ex dx =
of differentiation. We have — just restating the above
C
kx + C
ln |x| + C
1
xα+1 + C
α+1
for α 6= −1
ex + C
Moreover,
Z
Z
Z
(f (x) + g (x)) dx =
f (x) dx + g (x) dx
(rule for sums)
Z
Z
αf (x) dx = α f (x) dx
(rule for multiplicative constants)
Example 1
Z µ
√
3
5ex + x2 +
Z
Z
Z
2
x
5 e dx + x 3 dx + 5
¶
Z ³
´
2
5
dx =
5ex + x 3 + 5x−1 dx =
x
3 5
x−1 dx = 5ex + C1 + x 3 + C2 + 5 ln |x| + C3
5
3 5
= 5ex + x 3 + 5 ln |x| + C
5
where C = C1 + C2 + C3 .
2
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: Differentiating 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
× (−2) and hence
Similarly ln |3 − 2x| has the derivative 3−2x
Z
1
1
dx = − ln |3 − 2x| + 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
antiderivative of g (u), i.e., du
= g (u). Let u = u (x). By the chain rule the function
y = G (u (x)) has the derivative
dy du
du
dy
=
= 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
Z
xex dx
= 5 or du = 5dx or dx = 15 du. Substituting 5x + 2 by u and dx by
Set u = 5x + 2, so du
dx
1
dx we obtain
5
Z
Z
Z
1
1
1
1
5x+2
u
eu du = eu + C = e5x+2 + C
e
dx = e × du =
5
5
5
5
Example 3
2
1
du. Therefore
Set u = x2 . Then du = 2xdx or dx = 2x
Z
Z
x2
xe dx = xeu dx =?
3
Example 4
2
2
Z
xex dx
Z
x2 ex dx
2
1
Try u = ex . Then du = 2xex dx = 2xudx. Hence dx = 2xu
du
Z
Z
Z
1
1
1 2
1
x2
du =
du = u + C = ex + C
xe dx = xu
2xu
2
2
2
Example 5
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 different method.
0
3.1.1
Partial Integration
Partial integration is the inverse of the product rule
(uv)0 = u0 v + uv 0
for functions u (x), v (x). We can rewrite this as
u0 v = (uv)0 − uv0
or
Z
0
u vdx = uv −
since uv is the antiderivative of (uv)0 .
Z
uv 0 dx
Remark 1 To apply the formula the integrand must be a product of two factors, one of
which we know the integral (this factor will serve as u0 ) and one which we can differentiate. If we can integrate both factors there are two ways in which we can apply partial
integration, once as above with u0 being the first and v the second factor or as
Z
Z
0
uv dx = uv − u0 vdx
with u being the first and v being the second factor. Sometimes one must try both
versions.
4
Example 6
Z
x2 ex dx
Set v = x2 and u0 = ex . Then v0 = 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 v 0 = 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 v0 = ex and therefore by partial integration
Z
Z
1 3 x 1
2 x
x3 ex dx
x e dx = x e −
3
3
This does not simplify matters.
5