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: 2

Answer

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

Work Step by Step

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

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.