Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.3 - Exponential Functions - Exercises 7.3 - Page 390: 44

Answer

$$ e $$

Work Step by Step

$$\eqalign{ & \int_{\pi /4}^{\pi /2} {\left( {1 + {e^{\cot \theta }}} \right){{\csc }^2}\theta } d\theta \cr & {\text{Use substitution}}{\text{. Let }}u = \cot \theta,{\text{ so that }}du = - {\csc ^2}\theta d\theta \cr & {\text{The new limits on }}u{\text{ are found as follows}} \cr & \,\,\,\,\,\,{\text{If }}\theta = \pi /2,{\text{ }}u = \cot \left( {\pi /2} \right) = 0 \cr & \,\,\,\,\,\,{\text{If }}\theta = \pi /4,{\text{ }}u = \cot \left( {\pi /4} \right) = 1 \cr & {\text{Then}} \cr & \int_{\pi /4}^{\pi /2} {\left( {1 + {e^{\cot \theta }}} \right){{\csc }^2}\theta } d\theta = \int_1^0 {\left( {1 + {e^u}} \right)} \left( { - du} \right) \cr & = \int_0^1 {\left( {1 + {e^u}} \right)} du \cr & {\text{Integrate}} \cr & \int_0^1 {\left( {1 + {e^u}} \right)} du = \left( {u + {e^u}} \right)_0^1 \cr & {\text{Use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr & = \left( {1 + {e^1}} \right) - \left( {0 + {e^0}} \right) \cr & {\text{Simplifying}} \cr & = 1 + {e^1} - 0 - 1 \cr & = e \cr} $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.