Calculus: Early Transcendentals 8th Edition

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

Chapter 5 - Section 5.4 - Indefinite Integrals and the Net Change Theorem - 5.4 Exercises - Page 409: 40


$$\int^{10}_{-10}\frac{2e^x}{\sinh x+\cosh x}dx=40$$

Work Step by Step

$$A=\int^{10}_{-10}\frac{2e^x}{\sinh x+\cosh x}dx$$ $$A=\int^{10}_{-10}Bdx$$ *Consider B: $$B=\frac{2e^x}{\sinh x+\cosh x}$$ We know that $$\sinh x=\frac{e^x-e^{-x}}{2}$$ $$\cosh x=\frac{e^x+e^{-x}}{2}$$ Therefore, $$B=\frac{2e^x}{\frac{e^x-e^{-x}}{2}+\frac{e^x+e^{-x}}{2}}$$ $$B=\frac{2e^x}{\frac{2e^x}{2}}$$ $$B=\frac{2e^x}{e^x}$$ $$B=2$$ That means $$A=\int^{10}_{-10}2dx$$ According to Table 1, we have $$\int kdx=kx+C$$ Therefore, $$A=2x\Bigg]^{10}_{-10}$$ $$A=2[10-(-10)]$$ $$A=40$$
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.