Thomas' Calculus 13th Edition

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

Chapter 5: Integrals - Section 5.3 - The Definite Integral - Exercises 5.3 - Page 275: 47



Work Step by Step

$\int^{2}_{1}3u^2 du$ =3$\int^{2}_{1}u^2du$ =$3[\int^{2}_{0}u^2du-\int^{1}_{0}u^2du$ =3($[\frac{2^3}{2}-\frac{0^2}{3}] - [\frac{1^3}{3}-\frac{0^3}{3}])$ =$3[\frac{2^3}{3}-\frac{1^3}{3}]$ =3($\frac{7}{3})$ =7
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