Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 7 - Integration - 7.1 Antiderivatives - 7.1 Exercises - Page 366: 10

Answer

$\frac{5x^3}{3}-3x^2+3x+C$

Work Step by Step

$\int(5x^2-6x+3)dx$ $=\int 5x^2dx-\int 6xdx+\int 3dx$ $=5\int x^2dx-6\int x^1 dx+3\int x^0dx$ $=5*\frac{x^3}{3}-6*\frac{x^2}{2}+3*\frac{x^1}{1}+C$ $=\frac{5x^3}{3}-3x^2+3x+C$
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