Cambridge International AS and A Level Mathematics Pure Mathematics 1

P1^

6

Improper

(^) integrals
207
At first sight, (^1112)
x 1
x
x
∞ ∞
∫ =−




d doesn’t look like a particularly daunting integral.
However, the upper limit is infinity, which is not a number; so when you get an
answer of 1 −^1
∞
, you cannot work it out. Instead, you should start by looking at
the case where you are finding the finite area between 1 and b (as you did in the
activity). You can then say what happens to the value of 1

1

−

b

as b approaches (or

tends to) infinity. This process of taking ever larger values of b, is called taking a

limit. In this case you are finding the value of    1 −b^1 in the limit as b tends to ∞.

You can write this formally as: 

As b → ∞ then (^112)
x
x
b
∫ d^ becomes^ blim
b
x
x
→∞∫
1
1 2 d^ =^ bli→∞m()−+b
(^11) = 1.
●?^ What^ is^ the^ value^ of^

1

ax^2 x

∞

∫ d?^

What can you say about (^012)

x
x
∞
∫ d?
Integrals where one of the limits is infinity are called improper integrals.
There is a second type of improper integral, which is when the expression you
want to integrate is not defined over the whole region between the two limits. In
the example that follows the expression is 1
x
and it is not defined when x = 0.
EXAMPLE 6.17 Evaluate 091
x
∫ dx.
SOLUTION
The diagram shows the graph of y
x

=

1

.

11

11

1

(^11)
(^1) x^2 x x 1
b
b
b b
∫ =−




=−()−−()
=−()+
d
y
Oa 9 x
y =^1 x
Figure 6.24