Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - Chapter Review Exercises - Page 279: 90



Work Step by Step

By making use of (FTC II), we have $$ G'(x)=\frac{d}{d x} \int_{0}^{x^3} \sqrt{t+1} dt =\left(\frac{d}{d u} \int_{0}^{u}\sqrt{t+1} dt\right)\frac{du}{dx}=\sqrt{u+1} ( 3 x^2)\\ =3x^2 \sqrt{x^3+1}. $$ Hence, $$G'(x)=3(2^2) \sqrt{8+1}=36.$$
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