Page 150 - Calculus Demystified

Indeterminate Forms
CHAPTER 5
It is dangerous to try to save work by not dividing the integral at the singularity. 137
The next example illustrates what can go wrong.
EXAMPLE 5.18
Evaluate the improper integral
2

−4
x dx.
−2
SOLUTION
What we should do is divide this problem into the two integrals

0 2

−4 −4
x dx and x dx. (∗)
−2 0
Suppose that instead we try to save work and just antidifferentiate:
2 1 2 1

x −4 dx =− x −3 =− .
−2 3 −2 12
A glance at Fig. 5.4 shows that something is wrong. The function x −4 is
positive hence its integral should be positive too. However, since we used
an incorrect method, we got a negative answer.

Fig. 5.4

In fact each of the integrals in line (∗) diverges, so by definition the improper
integral
2

x −4 dx
−2
diverges.