Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 4 - Integrals - 4.3 The Fundamental Theorem of Calculus - 4.3 Exercises - Page 328: 38

Answer

$$\int_{-2}^{2} f(x) d x = \frac{28}{3}$$

Work Step by Step

Given $$\int_{-2}^{2} f(x) d x \quad \text { where } f(x)= \begin{cases}2 & \text { if }-2 \leqslant x \leqslant 0 \\ 4-x^{2} & \text { if } 0\lt x \leqslant 2\end{cases}$$ Since \begin{aligned} \int_{-2}^{2} f(x) d x &=\int_{-2}^{0} (2) d x +\int_{0}^{2} (4-x^2) d x \\ &=2x\bigg|_{-2}^0+ (4x-\frac{1}{3}x^3)\bigg|_0^2\\ &= (0)-(-4)+ [(8-\frac{8}{3}) - (0) ]\\ &= \frac{28}{3} \end{aligned}
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