Answer
7
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
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
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
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