Answer
$\frac{64}{3}$
Work Step by Step
Step 1. For $x\leq -2$ or $x\geq 2$, we have $y=|x^2-4|=x^2-4$ and the function intersects with $y=\frac{x^2}{2}+4$ at $(-4,12)$ and $(4,12)$
Step 2. For $-2\leq x\leq 2$, we have $y=|x^2-4|=4-x^2$ and the function intersects with $y=\frac{x^2}{2}+4$ at $(0,4)$ as shown in the figure.
Step 3. The enclosed area between the two curves can be calculated as
$A=\int_{-4}^{-2}(\frac{x^2}{2}+4-x^2+4)dx + \int_{-2}^{2}(\frac{x^2}{2}+4+x^2-4)dx + \int_{2}^{4}(\frac{x^2}{2}+4-x^2+4)dx
=\int_{-4}^{-2}(8-\frac{x^2}{2})dx + \int_{-2}^{2}(\frac{3x^2}{2})dx + \int_{2}^{4}(8-\frac{x^2}{2})dx
=(8x-\frac{1}{6}x^3)|_{-4}^{-2}+(\frac{1}{2}x^3)|_{-2}^{2}+(8x-\frac{1}{6}x^3)|_{2}^{4}
=(8(-2)-\frac{1}{6}(-2)^3)-(8(-4)-\frac{1}{6}(-4)^3)+(\frac{1}{2}(2)^3)-(\frac{1}{2}(-2)^3)+(8(4)-\frac{1}{6}(4)^3)-(8(2)-\frac{1}{6}(2)^3)=\frac{64}{3}$ (combine fractions with the same denominator).
