Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Review - Exercises: 11

Answer

$\int_{1}^2(8x^3+3x^2)dx=37$

Work Step by Step

$\int_{1}^2(8x^3+3x^2)dx$ $=[\frac{8x^4}{4}+\frac{3x^3}{3}]^2_{1}$ $=[2x^4+x^3]^2_{1}$ Sub in the bounds and subtract the lower bound (1) from the upper bound (2). $=[2(2)^4+(2)^3]-[2(1)^4+(1)^3]$ $=40-3$ $=37$
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