University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises - Page 321: 59



Work Step by Step

We can calculate the area, by integrating the function from 0 to 2. $\int_{0}^{2} (x^3-3x^2+2x)dx =[\frac{x^4}{4}-x^3+x^2]_{0}^{2}=[\frac{2^4}{4}-2^3+2^2]-[\frac{0^4}{4}-0^3+0^2]=[4-8+4]-0=0$
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