## Calculus: Early Transcendentals (2nd Edition)

$85.3333$
First, we can write the problem as an equation in order to solve it. Since we're trying to find the area between $f$ and the x-axis, that implies the use of an integral. The ends of our interval (-4 and 4) will be the limits. $\int\limits_{ -4}^4 16-x^2 dx$ Apply the sum rule $\int\limits_{ -4}^4 16dx$ - $\int\limits_{ -4}^4x^2dx$ $\int\limits_{ -4}^4 16dx=16x|_{-4}^{4}$ (fundamental thereom of calculus) $=16(4)-16(-4)=128$ $\int\limits_{ -4}^4x^2dx=\frac{1}{3}x^3|_{-4}^4$ $=\frac{1}{3}(4)^3-\frac{1}{3}(-4)^3=\frac{128}{3}\approx42.6667$ $128-42.6667=85.3333$