Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - 5.4 The Fundamental Theorem of Calculus, Part I - Exercises - Page 257: 2



Work Step by Step

$f(x)=2x-x^2$ $Area=\int_0^2(2x-x^2)dx$ $A=[x^2-\frac{x^3}{3}]_0^2$ $A=4-\frac{8}{3}$ $A=\frac{4}{3}$
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