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Evaluate −3 + e x d x . State whether the improper integral converges or diverges.

e 3 , converges

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Integrating a discontinuous integrand

Now let’s examine integrals of functions containing an infinite discontinuity in the interval over which the integration occurs. Consider an integral of the form a b f ( x ) d x , where f ( x ) is continuous over [ a , b ) and discontinuous at b . Since the function f ( x ) is continuous over [ a , t ] for all values of t satisfying a < t < b , the integral a t f ( x ) d x is defined for all such values of t . Thus, it makes sense to consider the values of a t f ( x ) d x as t approaches b for a < t < b . That is, we define a b f ( x ) d x = lim t b a t f ( x ) d x , provided this limit exists. [link] illustrates a t f ( x ) d x as areas of regions for values of t approaching b .

This figure has three graphs. All the graphs have the same curve, which is f(x). The curve is non-negative, only in the first quadrant, and increasing. Under all three curves is a shaded region bounded by a on the x-axis an t on the x-axis. There is also a vertical asymptote at x = b. The region in the first curve is small, and progressively gets wider under the second and third graph as t gets further from a, and closer to b on the x-axis.
As t approaches b from the left, the value of the area from a to t approaches the area from a to b .

We use a similar approach to define a b f ( x ) d x , where f ( x ) is continuous over ( a , b ] and discontinuous at a . We now proceed with a formal definition.

Definition

  1. Let f ( x ) be continuous over [ a , b ) . Then,
    a b f ( x ) d x = lim t b a t f ( x ) d x .
  2. Let f ( x ) be continuous over ( a , b ] . Then,
    a b f ( x ) d x = lim t a + t b f ( x ) d x .

    In each case, if the limit exists, then the improper integral is said to converge. If the limit does not exist, then the improper integral is said to diverge.
  3. If f ( x ) is continuous over [ a , b ] except at a point c in ( a , b ) , then
    a b f ( x ) d x = a c f ( x ) d x + c b f ( x ) d x ,

    provided both a c f ( x ) d x and c b f ( x ) d x converge. If either of these integrals diverges, then a b f ( x ) d x diverges.

The following examples demonstrate the application of this definition.

Integrating a discontinuous integrand

Evaluate 0 4 1 4 x d x , if possible. State whether the integral converges or diverges.

The function f ( x ) = 1 4 x is continuous over [ 0 , 4 ) and discontinuous at 4. Using [link] from the definition, rewrite 0 4 1 4 x d x as a limit:

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

The improper integral converges.

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Integrating a discontinuous integrand

Evaluate 0 2 x ln x d x . State whether the integral converges or diverges.

Since f ( x ) = x ln x is continuous over ( 0 , 2 ] and is discontinuous at zero, we can rewrite the integral in limit form using [link] :

0 2 x ln x d x = lim t 0 + t 2 x ln x d x Rewrite as a limit. = lim t 0 + ( 1 2 x 2 ln x 1 4 x 2 ) | t 2 Evaluate x ln x d x using integration by parts with u = ln x and d v = x . = lim t 0 + ( 2 ln 2 1 1 2 t 2 ln t + 1 4 t 2 ) . Evaluate the antiderivative. = 2 ln 2 1 . Evaluate the limit. lim t 0 + t 2 ln t is indeterminate. To evaluate it, rewrite as a quotient and apply L’Hôpital’s rule.

The improper integral converges.

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Integrating a discontinuous integrand

Evaluate −1 1 1 x 3 d x . State whether the improper integral converges or diverges.

Since f ( x ) = 1 / x 3 is discontinuous at zero, using [link] , we can write

−1 1 1 x 3 d x = −1 0 1 x 3 d x + 0 1 1 x 3 d x .

If either of the two integrals diverges, then the original integral diverges. Begin with −1 0 1 x 3 d x :

−1 0 1 x 3 d x = lim t 0 −1 t 1 x 3 d x Rewrite as a limit. = lim t 0 ( 1 2 x 2 ) | −1 t Find the antiderivative. = lim t 0 ( 1 2 t 2 + 1 2 ) Evaluate the antiderivative. = + . Evaluate the limit.

Therefore, −1 0 1 x 3 d x diverges. Since −1 0 1 x 3 d x diverges, −1 1 1 x 3 d x diverges.

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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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