## University Calculus: Early Transcendentals (3rd Edition)

The area of the shaded region is $16/3$.
Here we see that the shaded region is bounded by the curve $y=x\sqrt{4-x^2}$ and the line $y=0$. However, it can be divided into two parts: - The first part is where $x\in(-2,0)$. Here the given region is bounded above by $y=0$ and below by $y=x\sqrt{4-x^2}$. So the area of this region would be $$A_1=\int^{0}_{-2}(0-x\sqrt{4-x^2})dx=-\int^{0}_{-2}x\sqrt{4-x^2}dx$$ - The second part is where $x\in(0,2)$. Here the given region is bounded above by $y=x\sqrt{4-x^2}$ and below by $y=0$. So the area of this region would be $$A_2=\int^{2}_{0}(x\sqrt{4-x^2}-0)dx=\int^{2}_{0}x\sqrt{4-x^2}dx$$ Therefore, the area of the whole shaded region is $$A=A_1+A_2=-\int^{0}_{-2}x\sqrt{4-x^2}dx+\int^{2}_{0}x\sqrt{4-x^2}dx$$ We set $u=4-x^2$, which means $$du=-2xdx$$ $$xdx=-\frac{1}{2}du$$ - For $x=-2$, we have $u=4-4=0$ - For $x=0$, we have $u=4-0=4$ - For $x=2$, we have $u=4-4=0$ Therefore, $$A=-\Big(-\frac{1}{2}\int^4_0\sqrt udu\Big)-\frac{1}{2}\int^0_4\sqrt udu$$ $$A=\frac{1}{2}\int^4_0\sqrt udu+\frac{1}{2}\int^4_0\sqrt udu$$ $$A=\int^4_0\sqrt udu$$ $$A=\frac{2u^{3/2}}{3}\Big]^4_0=\frac{2}{3}(4^{3/2})=\frac{16}{3}$$