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# Improper Integrals Help (page 2)

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By McGraw-Hill Professional
Updated on Aug 31, 2011

#### Example 4

Evaluate the improper integral

#### Solution 4

The integrand is unbounded as x tends to −1. Therefore we evaluate separately the two improper integrals

The first of these has the value

The second integral has the value

We conclude that the original integral converges and

You Try It : Evaluate the improper integral .

It is dangerous to try to save work by not dividing the integral at the singularity. The next example illustrates what can go wrong.

#### Example 5

Evaluate the improper integral

#### Solution 5

What we should do is divide this problem into the two integrals

Suppose that instead we try to save work and just antidifferentiate:

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

diverges.

#### Example 6

Analyze the integral

#### Solution 6

The key idea is that we can only handle one singularity at a time. This integrand is singular at both endpoints 0 and 1. Therefore we divide the domain of integration somewhere in the middle—at 1/2 say (it does not really matter where we divide)—and treat the two singularities separately.

First we treat the integral

Since the integrand has a singularity at 0, we consider

This is a tricky integral to evaluate directly. But notice that

when 0 < ∈ ≤ x ≤ 1/2. Thus

We evaluate the integral: it equals ln(1/2) − ln . Finally,

The first of our integrals therefore diverges.

But the full integral

converges if and only if each of the component integrals

and

converges. Since the first integral diverges, we conclude that the original integral diverges as well.

You Try It : Calculate as an improper integral.

## An Application to Area

Suppose that f is a non-negative, continuous function on the interval ( a, b ] which is unbounded as xa + . Look at Fig. 5.5. Let us consider the area under the graph of f and above the x -axis over the interval ( a, b ]. The area of the part of the region over the interval [ a + ∈, b ], ∈ > 0, is

Fig. 5.5

Therefore it is natural to consider the area of the entire region, over the interval ( a, b ], to be

This is just the improper integral

#### Example

Calculate the area above the x -axis and under the curve

#### Solution

According to the preceding discussion, this area is equal to the value of the improper integral

For clarity we let φ( x ) = ln x , φ′( x ) = 1/ x . Then the (indefinite) integral becomes

Thus

Now as ∈ → 0 then ln ∈ → −∞ hence 1/[ln ∈] 1/3 → 0. We conclude that our improper integral converges and the area under the curve and above the x -axis equals 3/[ln 2] 1/3 .

Find practice problems and solutions for these concepts at: Indeterminate Forms Practice Test.

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