## Thomas' Calculus 13th Edition

$\dfrac{32}{3}$
Consider $f(x)=\int_{-2}^{2} (4-y^2) dy$ Use formula such as follows: $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+C$ This implies that $[4y-\dfrac{y^{2+1}}{2+1}]_{-2}^{2}=4(2+2)-\dfrac{1}{3}(8+8)$ Thus, $f(x)=\int_{-2}^{2} (4-y^2) dy=\dfrac{32}{3}$