Understanding Engineering Mathematics

Not so obvious is the fact that ifa<b<cthen
∫b

a

f(x)dx+

∫c

b

f(x)dx=F(b)−F(a)+F(c)−F(b)=F(c)−F(a)

=

∫c

a

f(x)dx

Remember to change limits if you make a substitution –andto ensure that the substitution
is consistent with the values of the limits.
When integrating a definite integral by parts you can substitute the limits in as you go
along: ∫
b

a

u

dv

dx

dx=[uv]ba−

∫b

a

v

du

dx

dx

When putting limits on integrals we must be careful to avoid, in the integration interval,
any points where the integral or integrand would not exist. For example

∫ 1

0

dx

2 x− 1

is animproper integralbecause the integrand is not defined atx=^12 , which is within
the range of integration. Another type of ‘improper integral’ occurs when one or both
of the limits involves infinity. We will see this specifically in Chapter 17, in the Laplace
transform, but for now we simply state the definition:
∫∞

0

f(x)dx=lim

a→∞

∫a

0

f(x)dx

That is, we first integrate using a finite limitaand then we letatend to infinity in the
result.

Solution to review question 9.1.12

A. (i) The indefinite integral of f(x)=x^2 +1isF(x)=

x^3

3

+x,

ignoring the arbitrary constant. So

∫ 1

0

(x^2 + 1 )dx=[F(x)]^10 =F( 1 )−F( 0 )=

[

x^3

3

+x

] 1

0

=

1

3

+ 1 − 0 =

4

3

(ii) We can fit the limits in as we go along in integration by parts:

∫ 1

0

xexdx=[xex]^10 −

∫ 1

0

exdx

=e− 0 −[ex]^10

=e−(e−e^0 )=1sincee^0 = 1