The area of the region (x,y):|x-y| y 4x is

Question

The area of the region ${(x,y):|x-y|\le y\le 4\sqrt{x}}$ is

Accepted Answer

The condition $|x-y|\le y$ (with $y\ge0$) reduces to $0 \le x \le 2y$, i.e. $y \ge x/2$; combined with $y \le 4\sqrt x$, the curves $y=x/2$ and $y=4\sqrt x$ meet where $x/2 = 4\sqrt x \Rightarrow x = 64$.

$$A = \int_0^{64}\left(4\sqrt{x}-\frac{x}{2}\right)dx = \left[\frac{8x^{3/2}}{3}-\frac{x^2}{4}\right]_0^{64} = \frac{8(512)}{3}-1024 = \frac{4096}{3}-\frac{3072}{3} = \frac{1024}{3}$$

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