Thomas' Calculus 13th Edition

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

Chapter 5: Integrals - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises 5.5 - Page 295: 56

Answer

$$y = 3 (x^{2} + 8)^{\frac{2}{3}} + C $$ $$ C=-12 $$

Work Step by Step

$$\frac{dy}{dx} = 4x (x^{2} + 8)^{\frac{-1}{3}} , y (0) = 0 $$ $\frac{dy}{dx} = 4x (x^{2} + 8)^{\frac{-1}{3}}$ $ dy = 4x (x^{2} + 8)^{\frac{-1}{3}} dx $ $u = x^{2} +8 $ $ du = (2x)dx $ , $ 2du= (4x)dx $ $dy = 4x (x^{2} + 8)^{\frac{-1}{3}} dx$ Replace u and du $dy = 2du (u)^{\frac{-1}{3}} $ $\int dy = \int 2du (u)^{\frac{-1}{3}}$ $ y= 2 \int (u)^\frac{-1}{3} $ $ 2 \int \frac{u^{\frac{-1}{3}+1}}{\frac{-1}{3}+1} $ $ 2 * \frac{u^{\frac{2}{3}}}{\frac{2}{3}} + C$ $ 2 * \frac{3u^{\frac{2}{3}}}{2} + C $ $ 3u^{\frac{2}{3}} + C $ Replace u $ y = 3(x^{2} +8)^{\frac{2}{3}} + C $ $$ y = 3(x^{2} +8)^{\frac{2}{3}} + C $$ $y(0)=0$ $ 0 = 3(0^{2} +8)^{\frac{2}{3}} + C $ $ 0 = 3(4) + C $ $ 0 = 12 + C $ $ C=-12 $ $$ C=-12 $$
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