Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.3 - The Fundamental Theorem of Calculus - 5.3 Exercises: 48

Answer

$= \frac{4}{3} $
1512920635

Work Step by Step

$\int^{2}_0 2x-x^{2} dx$ $= 2\frac{x^{2}}{2} - \frac{x^{3}}{3} |^{2}_0$ $= x^{2} - \frac{x^{3}}{3} |^{2}_0$ $= [2^{2} - \frac{2^{3}}{3}] - [0^{2} - \frac{0^{3}}{3}]$ $= [4 - \frac{8}{3}] - 0$ $= [4 - \frac{8}{3}] $ $= \frac{4}{3} $
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