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EXAMPLE 5.20 CHAPTER 5 Indeterminate Forms
Calculate the area above the x-axisand under the curve
1
y = 4/3 , 0 <x ≤ 1/2.
x · ln x
SOLUTION
According to the preceding discussion, this area is equal to the value of the
improper integral
1/2 1 1/2 1
dx = lim dx.
0 x · ln 4/3 x (→0 + ( x · ln 4/3 x
For clarity we let ϕ(x) = ln x, ϕ (x) = 1/x. Then the (indefinite) integral
becomes
ϕ (x) 3
dx =− .
ϕ 4/3 (x) ϕ 1/3 (x)
Thus
1/2
1/2 1 3
lim dx = lim −
(→0 + ( x · ln 4/3 x (→0 + ln 1/3
x (
−3 −3
= lim − .
(→0 + [− ln 2] 1/3 [ln (] 1/3
1/3
Now as ( → 0 then ln ( →−∞ hence 1/[ln (] → 0. We conclude that our
improper integral converges and the area under the curve and above the x-axis
1/3
equals 3/[ln 2] .
5.4 More on Improper Integrals
5.4.1 INTRODUCTION
Suppose that we want to calculate the integral of a continuous function f(x) over
an unbounded interval of the form [A, +∞) or (−∞,B]. The theory of the integral
that we learned earlier does not cover this situation, and some new concepts are
needed. We treat improper integrals on infinite intervals in this section, and give
some applications at the end.