Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 4 - Review - Exercises - Page 367: 65

Answer

$\frac{8}{3}x^{\frac{3}{2}}-2x^{3}+3x+C$

Work Step by Step

$\int (4\sqrt x-6x^{2}+3)dx$$= 4\int x^{1/2}dx-6\int x^{2}dx+3\int dx$ $=4\times\frac{2}{3}x^{3/2}-6\times\frac{x^{3}}{3}+3\times x+C$ $=\frac{8}{3}x^{\frac{3}{2}}-2x^{3}+3x+C$
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