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 - Page 434: 19


$$A = \frac{2}{\pi}+\frac{2}{3}$$

Work Step by Step

From the graph, we can see that $\cos (\pi x) \geq 4x^{2}-1$ on the interval $\big[-\frac{1}{2},\frac{1}{2}\big]$; therefore, the area between the curves is $$ A= \displaystyle\int_{-\frac{1}{2}}^{\frac{1}{2}} \cos (\pi x) - (4x^{2}-1)dx$$ $$= \displaystyle\int_{-\frac{1}{2}}^{\frac{1}{2}} \cos(\pi x)-4x^{2}+1 dx$$ $$=\left[\frac{\sin (\pi x)}{\pi}-\frac{4x^{3}}{3}+x \right]^{\frac{1}{2}}_{-\frac{1}{2}}$$ $$=2\bigg[\frac{1}{\pi}-\frac{1}{6}+\frac{1}{2}\bigg]$$ $$=\frac{2}{\pi}+\frac{2}{3}$$
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