Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 4 - Integration - 4.4 Exercises - Page 288: 25

Answer

$\displaystyle \frac{64}{3}$

Work Step by Step

The indefinite integral is over the interval [0,4], where f(x) has a zero, x=3. $f(x)=\left\{\begin{array}{lll} x^{2}-9, & when & 4 \geq x \geq 3\\ -(x^{2}-9) & when & 0 \leq x < 3 \end{array}\right.$ Apply Th.4.6 from section 4-3, Additive Interval Property $\displaystyle \int_{0}^{4}f(x)dx=\int_{0}^{3}f(x)dx+\int_{3}^{4}f(x)dx$ $\displaystyle \int_{0}^{4}|x^{2}-9|dx=\int_{0}^{3}(9-x^{2})dx+\int_{3}^{4}(x^{2}-9)dx$ Use the table "Basic Integration Rules", p.246 $=[9x-\displaystyle \frac{x^{3}}{3}]_{0}^{3}+[\frac{x^{3}}{3}-9x]_{3}^{4}$ $=(27-9)+(\displaystyle \frac{64}{3}-36)-(9-27)$ $=\displaystyle \frac{64}{3}$ Check: using desmos.com (see image below)
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