## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 5: Integrals - Section 5.4 - The Fundamental Theorem of Calculus - Exercises 5.4 - Page 286: 2

20/3

#### Work Step by Step

Integrate with respect to x, and then plug in the limits: $=\frac{1}{3}x^3-x^2+3x$ with limits from -1 to 1 $=\frac{1}{3}(1)^3-1+3-\frac{1}{3}(-1)^3+(-1)^2-3(-1)=1/3+2+1/3+1+3=20/3$

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