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: 5

Answer

Integrate with respect to x. $$area = e - \frac{1}{e} + \frac{4}{3}$$
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Work Step by Step

$y = e^x, y = x^2 - 1, x = -1, x=1$ Before solving for the area, the aspects of the curves need to be identified. $\underline{y = e^x}$: y-intercept: $e^0 = y$ $y = 1$ $(0, 1)$ $\underline{x^2 - 1}$: x-intercepts: $x^2 - 1 = 0$ $x = \pm1$ $(-1,0),(1,0)$ y-intercept: $y = 0^2 -1$ $y = -1 \rightarrow (0,-1)$ vertex (when $y\prime$ = 0): $y\prime = 2x$ $2x = 0$ $x = 0$ $(0,-1)$ With both graphs' aspects calculated, the graphs can be drawn. To find the area, integrate with respect to $x$. Based on graphs, $f(x) = e^x$ and $g(x) = x^2 - 1$ Limits of integration: $a = -1$ and $b = 1$ $$\int_a^b f(x) - g(x) dx$$ $$\int^1_{-1} e^x - (x^2 - 1) dx$$ Solve the definite integral and simplify completely. $$= \int^1_{-1} (e^x - x^2 + 1)dx$$ $$= (e^x - \frac{x^3}{3} + 1)|^1_{-1}$$ $$= [e^1 - \frac{1^3}{3} + 1] - [e^{-1} - \frac{(-1)^3}{3} + (-1)]$$ $$= e - \frac{1}{3} + 1 - \frac{1}{e} - \frac{1}{3} + 1$$ $$area = e - \frac{1}{e} + \frac{4}{3}$$
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