Thomas' Calculus 13th Edition

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

Chapter 5: Integrals - Practice Exercises - Page 307: 20



Work Step by Step

Consider $f(x)=\int_{-2}^{2} (4-y^2) dy$ Use formula such as follows: $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+C$ This implies that $[4y-\dfrac{y^{2+1}}{2+1}]_{-2}^{2}=4(2+2)-\dfrac{1}{3}(8+8)$ Thus, $f(x)=\int_{-2}^{2} (4-y^2) dy=\dfrac{32}{3}$
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