Calculus: Early Transcendentals 8th Edition

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

Chapter 7 - Section 7.3 - Trigonometric Substitution - 7.3 Exercises - Page 491: 11

Answer

$$\displaystyle{\int_0^{\frac{1}{2}}x\sqrt{1-4x^2}dx=\frac{1}{12}}\\$$

Work Step by Step

$\displaystyle{I=\int_0^{\frac{1}{2}}x\sqrt{1-4x^2}dx}\\ \displaystyle{I=\int_0^{\frac{1}{2}}x\sqrt{1-{\left(2x\right)}^2}dx}\\$ $\displaystyle \left[\begin{array}{ll} 2x=\sin\theta & 4 x^2=\sin^2\theta \\ & \\ \frac{dx}{d\theta}=\frac{1}{2}\cos\theta & dx=\frac{1}{2}\cos\theta\ d\theta \end{array}\right]$ Integration by substitution $\displaystyle{I=\int_0^{\frac{\pi}{2}}\frac{1}{2}\sin\theta\sqrt{1-\sin^2\theta}\times\frac{1}{2}\cos\theta\ d\theta}\\ \displaystyle{I=\int_0^{\frac{\pi}{2}}\frac{1}{2}\sin\theta\cos\theta\times\frac{1}{2}\cos\theta\ d\theta}\\ \displaystyle{I=\int_0^{\frac{\pi}{2}}\frac{1}{4}\sin\theta\cos^2\theta\ d\theta}\\ \displaystyle{I=\frac{1}{4}\int_0^{\frac{\pi}{2}}\frac{1}{3}\times3\sin\theta\cos^2\theta\ d\theta}\\ \displaystyle{I=\frac{1}{12}\int_0^{\frac{\pi}{2}}3\sin\theta\cos^2\theta\ d\theta}\\ \displaystyle{I=\frac{1}{12}\left[-\cos^3\theta\right]_0^{\frac{\pi}{2}}}\\ \displaystyle{I=-\frac{1}{12}\cos^3\left(\frac{\pi}{2}\right)\ +\frac{1}{12}\cos^3\left(0\right)}\\ \displaystyle{I=0 +\frac{1}{12}}\\ \displaystyle{I=\frac{1}{12}}\\ $
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