Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 8 - Techniques of Integration - Chapter Review Exercises - Page 461: 80


The integral converges and its value is $\frac{1}{4}e^{36}$.

Work Step by Step

We have $$ \int_{-\infty}^{9} e^{4x}dx=\frac{1}{4}e^{4x}|_{-\infty}^9\\ =\frac{1}{4}(e^{-\infty}-e^{36})=\frac{1}{4}(0-e^{36})=\frac{1}{4}e^{36}. $$ Hence, the integral converges and its value is $\frac{1}{4}e^{36}$.
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