Page 156 - Calculus Demystified

Indeterminate Forms
CHAPTER 5
The second integral is evaluated similarly: 143
0 1 0
lim dx = lim Tan −1
x
M 1 + x 2 M
M→−∞ M→−∞

= lim Tan −1 0 − Tan −1 M
M→−∞
π
= .
2
Since each of the integrals on the half line is convergent, we conclude that the
original improper integral over the entire real line is convergent and that its
value is
π π
+ = π.
2 2
∞
You Try It: Discuss (1 + x) −1 dx.
1
5.4.3 SOME APPLICATIONS

Now we may use improper integrals over infinite intervals to calculate area.
EXAMPLE 5.24
4
Calculate the area under the curve y = 1/[x · (ln x) ] and above the x-axis,
2 ≤ x< ∞.
SOLUTION
The area is given by the improper integral
N
+∞ 1 1
dx = lim dx.
x · (ln x) 4 x · (ln x) 4
2 N→+∞ 2
For clarity, we let ϕ(x) = ln x, ϕ (x) = 1/x. Thus the (indefinite) integral

becomes

ϕ (x) 1/3
dx =− .
4
3
ϕ (x) ϕ (x)
Thus
N 1 1/3 N
lim dx = lim −
3
N→+∞ 2 x · (ln x) 4 N→+∞ ln x 2

1/3 1/3
= lim − 3 − 3
N→+∞ ln N ln 2
1/3
= 3 .
ln 2
3
Thus the area under the curve and above the x-axis is 1/(3ln 2).