Let R be the region \(x,y): |x^2-4|≤ y≤ 5\. The area of R is ______.

The curve $y=|x^2-4|$ equals $4-x^2$ for $|x|\le2$ and $x^2-4$ for $|x|\ge2$. It meets the line $y=5$ where $x^2-4=5$, i.e. $x=\pm3$ (the branch $4-x^2$ never reaches $5$). So $R$ lies between $x=-3$ and $x=3$ with upper boundary $y=5$. By symmetry about the $y$-axis, area $=2\left[\int_0^2\left(5-(4-x^2)\right)dx+\int_2^3\left(5-(x^2-4)\right)dx\right]=2\left[\int_0^2(1+x^2)\,dx+\int_2^3(9-x^2)\,dx\right]$. The first integral is $2+\frac83=\frac{14}3$; the second is $9-\frac{27-8}{3}=\frac{8}3$. Hence the area is $2\left(\frac{14}3+\frac83\right)=\frac{44}3\approx14.667$, i.e. $14.67$. Computing the area as a rectangle minus the region under the curve gives $30-\frac{46}3=\frac{44}3$ as a check.