Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 5 - Integration - 5.2 Definite Integrals - 5.2 Exercises - Page 358: 8


It can be rewritten as $$\int_1^6(2x^3-4x)dx=\int_1^62x^3dx-\int_1^64xdx$$.

Work Step by Step

In the problem, it is asked only to rewrite the integral as the difference of two integrals, not to evaluate it. By the difference rule given in table 5.4 we have $$\int_1^6(2x^3-4x)dx=\int_1^62x^3dx-\int_1^64xdx.$$
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