Calculus: Early Transcendentals 8th Edition

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

Chapter 6 - Section 6.1 - Areas Between Curves - 6.1 Exercises: 3

Answer

$$e - \frac{1}{e} + \frac{10}{3}$$

Work Step by Step

Area = $\int^d_c f(y) - g(y) dy$ In this question, $f(y) = e^y$ and $g(y) = y^2 - 2$ The limits of integration will equal the given lines $y = -1$ and $y = 1$ $$\int^d_c f(y) - g(y) dy$$ $$\int^1_{-1} (e^y)-(y^2 - 2) dy$$ Solve the produced definite integral and simplify until final answer. $$= \int^1_{-1} (e^y - y^2 + 2)dy$$ $$= (e^y - \frac{y^3}{3} + 2y)|^1_{-1}$$ $$= [e^1 - \frac{1^3}{3} + 2(1)] - [e^{-1} - \frac{(-1)^3}{3} + 2(-1)]$$ $$= e - \frac{1}{3} + 2 - \frac{1}{e} - \frac{1}{3} + 2$$ $$= e - \frac{1}{e} + \frac{10}{3}$$
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