University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.3 - The Definite Integral - Exercises - Page 309: 49



Work Step by Step

$\int^{2}_{0}(3x^{2}+x-5)dx $$=3\int^{2}_{0}x^{2}dx+\int^{2}_{0}xdx-5\int^{2}_{0}dx$ $=(3\times\frac{2^{3}}{3})+(\frac{2^{2}}{2})-(5\times2)$ $= 8+2-10= 0$
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