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In conclusion, although the area of the region between the x -axis and the graph of f ( x ) = 1 / x over the interval [ 1 , + ) is infinite, the volume of the solid generated by revolving this region about the x -axis is finite. The solid generated is known as Gabriel’s Horn .

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Chapter opener: traffic accidents in a city

This is a picture of a city street with a traffic signal. The picture has very busy lanes of traffic in both directions.
(credit: modification of work by David McKelvey, Flickr)

In the chapter opener, we stated the following problem: Suppose that at a busy intersection, traffic accidents occur at an average rate of one every three months. After residents complained, changes were made to the traffic lights at the intersection. It has now been eight months since the changes were made and there have been no accidents. Were the changes effective or is the 8-month interval without an accident a result of chance?

Probability theory tells us that if the average time between events is k , the probability that X , the time between events, is between a and b is given by

P ( a x b ) = a b f ( x ) d x where f ( x ) = { 0 if x < 0 k e k x if x 0 .

Thus, if accidents are occurring at a rate of one every 3 months, then the probability that X , the time between accidents, is between a and b is given by

P ( a x b ) = a b f ( x ) d x where f ( x ) = { 0 if x < 0 3 e −3 x if x 0 .

To answer the question, we must compute P ( X 8 ) = 8 + 3 e −3 x d x and decide whether it is likely that 8 months could have passed without an accident if there had been no improvement in the traffic situation.

We need to calculate the probability as an improper integral:

P ( X 8 ) = 8 + 3 e −3 x d x = lim t + 8 t 3 e −3 x d x = lim t + e −3 x | 8 t = lim t + ( e −3 t + e −24 ) 3.8 × 10 −11 .

The value 3.8 × 10 −11 represents the probability of no accidents in 8 months under the initial conditions. Since this value is very, very small, it is reasonable to conclude the changes were effective.

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Evaluating an improper integral over an infinite interval

Evaluate 0 1 x 2 + 4 d x . State whether the improper integral converges or diverges.

Begin by rewriting 0 1 x 2 + 4 d x as a limit using [link] from the definition. Thus,

0 1 x 2 + 4 d x = lim t t 0 1 x 2 + 4 d x Rewrite as a limit. = lim t tan −1 x 2 | t 0 Find the antiderivative. = lim t ( tan −1 0 tan −1 t 2 ) Evaluate the antiderivative. = π 2 . Evaluate the limit and simplify.

The improper integral converges to π 2 .

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Evaluating an improper integral on ( , + )

Evaluate + x e x d x . State whether the improper integral converges or diverges.

Start by splitting up the integral:

+ x e x d x = 0 x e x d x + 0 + x e x d x .

If either 0 x e x d x or 0 + x e x d x diverges, then + x e x d x diverges. Compute each integral separately. For the first integral,

0 x e x d x = lim t t 0 x e x d x Rewrite as a limit. = lim t ( x e x e x ) | t 0 Use integration by parts to find the antiderivative. (Here u = x and d v = e x . ) = lim t ( −1 t e t + e t ) Evaluate the antiderivative. = −1 . Evaluate the limit. Note: lim t t e t is indeterminate of the form 0 · . Thus, lim t t e t = lim t t e t = lim t −1 e t = lim t e t = 0 by L’Hôpital’s Rule.

The first improper integral converges. For the second integral,

0 + x e x d x = lim t + 0 t x e x d x Rewrite as a limit. = lim t + ( x e x e x ) | 0 t Find the antiderivative. = lim t + ( t e t e t + 1 ) Evaluate the antiderivative. = lim t + ( ( t 1 ) e t + 1 ) Rewrite. ( t e t e t is indeterminate.) = + . Evaluate the limit.

Thus, 0 + x e x d x diverges. Since this integral diverges, + x e x d x diverges as well.

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Practice Key Terms 1

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Source:  OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2
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