Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Review - Concept Check - Page 537: 7


The statement is False, so $$ \int_{-\infty}^{\infty} f(x) d x \ne \lim _{t \rightarrow \infty} \int_{-t}^{t}f( x) d x $$

Work Step by Step

The statement is false, so $$ \int_{-\infty}^{\infty} f(x) d x \ne \lim _{t \rightarrow \infty} \int_{-t}^{t}f( x) d x $$ Now, we give an example that disproves the statement. Let $f(x)=x$ $$ \int_{-\infty}^{\infty} x d x \ne \lim _{t \rightarrow \infty} \int_{-t}^{t} xd x $$ (a) The given integral $$ \int_{-\infty}^{\infty} x d x $$ is an improper integral, hence $$ \int_{-\infty}^{\infty} x d x =\int_{-\infty}^{0} x d x+\int_{0}^{\infty} x d x $$ and $$ \begin{split} \int_{0}^{\infty} x d x & = \lim _{t \rightarrow \infty} \int_{0}^{t} x d x\\ & =\lim _{t \rightarrow \infty}\left[\frac{1}{2} x^{2}\right]_{0}^{t} \\ & =\lim _{t \rightarrow \infty}\left[\frac{1}{2} t^{2}-0\right] \\ & = \infty, \end{split} $$ so the integral $$ \int_{0}^{\infty} x d x $$ is divergent, and hence, $$ \int_{-\infty}^{\infty} x d x $$ is divergent. (b) $$ \begin{split} \int_{-t}^{t} x d x & = \left[\frac{1}{2} x^{2}\right]_{-t}^{t} \\ & =\left[\frac{1}{2} t^{2}-\frac{1}{2} t^{2}\right] \\ & = 0, \end{split} $$ so $$ \lim _{t \rightarrow \infty} \int_{-t}^{t} x d x=0 $$ Therefore, from (a) and (b) we get $$ \lim _{t \rightarrow \infty} \int_{-t}^{t} x d x \ne \int_{-\infty}^{\infty} x d x $$
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