Thomas' Calculus 13th Edition

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

Chapter 5: Integrals - Section 5.4 - The Fundamental Theorem of Calculus - Exercises 5.4 - Page 286: 1



Work Step by Step

First, factor the inside of the integral: $\int x^2-3x$ with limits from 0 to 2 Then, integrate and simplify: $=1/3x^3-3/2x$ with limits from 0 to 2 $=1/3(2)^3-3/2(2)-0=8/3-3=-1/3$
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