Chapter 5 - Section 5.3 - The Fundamental Theorem of Calculus - 5.3 Exercises - Page 400: 44

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

We can evaluate the integral: $\int_{-2}^{2}f(x)~dx$ $=\int_{-2}^{0}f(x)~dx + \int_{0}^{2}f(x)~dx$ $=\int_{-2}^{0}2~dx + \int_{0}^{2}(4-x^2)~dx$ $=~2x~\vert_{-2}^{0} + (4x-\frac{x^3}{3})~\vert_{0}^{2}$ $=[2(0)-2(-2)] + [(4(2)-\frac{2^3}{3})-(4(0)-\frac{0^3}{3})]$ $=(4) + (\frac{16}{3})$ $= \frac{28}{3}$