Answer
$\frac{20}{3}$
Work Step by Step
$\int_{-1}^{1} (x^2-2x+3)dx=[\frac{x^3}{3}-x^2+3x]_{-1}^{1}=[\frac{1^3}{3}-1^2+3\cdot1]-[\frac{(-1)^3}{3}-(-1)^2+3\cdot(-1)]=[\frac{1}{3}-1+3]-[\frac{-1}{3}-1-3]=\frac{7}{3}-(-\frac{13}{3})=\frac{20}{3}$
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
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 320: 2
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