Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 4 - Integrals - Review - Exercises - Page 349: 11



Work Step by Step

Let I= $\int^{2}_{1}(8x^{3}+3x^{2})dx$ $\int (8x^{3}+3x^{2})dx=8\int x^{3}dx+3\int x^{2}dx$ $=8\times\frac{x^{4}}{4}+3\times\frac{x^{3}}{3}+C$ $=2x^{4}+x^{3}+ C= F(x)$ Therefore, by the second fundamental theorem of calculus, we get I= F(2)-F(1)= $(2\times2^{4}+2^{3})-(2\times1^{4}+1^{3})=40-3=37$
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