Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

Chapter 5 - Section 5.4 - Indefinite Integrals and the Net Change Theorem - 5.4 Exercises - Page 409: 49

Answer

$\frac{4}{3}$

Work Step by Step

We can evaluate the integral to find the area of the region: $\int_{0}^{2}(2y-y^2)~dy$ $=(y^2-\frac{y^3}{3})~\vert_{0}^{2}$ $=(2^2-\frac{2^3}{3})-(0^2-\frac{0^3}{3})$ $=(4-\frac{8}{3})-(0)$ $= \frac{4}{3}$

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