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: 49



Work Step by Step

$\int^{2}_{0}(3x^2+x-5)$dx =3$\int^{2}_{0}x^2dx+\int^{2}_{0}xdx-\int^{2}_{0}5dx$ =3$[\frac{2^3}{3}-\frac{0^3}{3}]+[\frac{2^2}{2}-\frac{0^2}{2}]$ =5[2-0]=(8+2)-10 =0
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