Calculus: Early Transcendentals (2nd Edition)

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

Chapter 5 - Integration - 5.4 Working with Integrals - 5.4 Exercises: 11

Answer

$\int_{-2}^2 (x^9 - 3x^5 + 2x^2 -10) dx = \frac{-88}{3}$

Work Step by Step

$\int_{-2}^2 (x^9 - 3x^5 + 2x^2 -10) dx = \int_{-2}^2 x^9 dx - \int_{-2}^2 3x^5 dx + \int_{-2}^2 2x^2 dx - \int_{-2}^2 10 dx$ Because $f(x) = x^9$ and $f(x) = -3x^5$ are odd functions and $f(x)=2x^2$ is an even function: $= 0 + 0 + 2\int_0^2 2x^2 dx - \int_{-2}^2 10 dx = 4\int_0^2 x^2 dx - \int_{-2}^2 10dx = 4(\frac{1}{3}(2)^3) - (10(2) - 10(-2)) = \frac{32}{3} - 40 = \frac{-88}{3}$
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