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: 2

Answer

$\frac{20}{3}$

Work Step by Step

$\int_{-1}^{1} (x^2-2x+3)dx=[\frac{x^3}{3}-x^2+3x]_{-1}^{1}=[\frac{1^3}{3}-1^2+3\cdot1]-[\frac{(-1)^3}{3}-(-1)^2+3\cdot(-1)]=[\frac{1}{3}-1+3]-[\frac{-1}{3}-1-3]=\frac{7}{3}-(-\frac{13}{3})=\frac{20}{3}$
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