Calculus: Early Transcendentals 8th Edition

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

Chapter 7 - Section 7.6 - Integration Using Tables and Computer Algebra Systems - 7.6 Exercises - Page 513: 6

Answer

$$\int_{0}^{2}x^{2}\sqrt{4-x^{2}}dx=\pi$$

Work Step by Step

Table 31: $\int u^{2}\sqrt{a^{2}-u^{2}}du=\frac{u}{8}(2u^{2}-a^{2})\sqrt{a^{2}-u^{2}}+\frac{a^{4}}{8}sin^{-1}(\frac{u}{a})+C$ $$\int_{0}^{2}x^{2}\sqrt{4-x^{2}}dx=\left [\frac{x}{8}(2x^{2}-4)\sqrt{4-x^{2}}+\frac{16}{8}sin^{-1}(\frac{x}{2}) \right ]\tfrac{0}{2}$$ $$=2sin^{-1}1=\pi$$
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