let the region r be the area enclosed by the function f(x)=√x - 2 and g(x)=1/2x - 2. if the region r is the…

Explanation:

Step1: Find intersection points

Set $\sqrt{x}-2=\frac{1}{2}x - 2$. Then $\sqrt{x}=\frac{1}{2}x$. Let $t = \sqrt{x}(t\geq0)$, so $t=\frac{1}{2}t^{2}$. Rearranging gives $\frac{1}{2}t^{2}-t = 0$, factoring out $t$ we get $t(\frac{1}{2}t - 1)=0$. So $t = 0$ or $t = 2$. When $t = 0,x = 0$; when $t = 2,x = 4$. The intersection - points of $y=\sqrt{x}-2$ and $y=\frac{1}{2}x - 2$ are $(0, - 2)$ and $(4,0)$.

Step2: Determine the diameter of the semi - circle

The diameter $d$ of each semi - circle perpendicular to the $x$ - axis is $d=(\sqrt{x}-2)-(\frac{1}{2}x - 2)=\sqrt{x}-\frac{1}{2}x$.

Step3: Find the radius of the semi - circle

The radius $r$ of the semi - circle is $r=\frac{1}{2}(\sqrt{x}-\frac{1}{2}x)$.

Step4: Find the area of the semi - circle

The area of a semi - circle is $A=\frac{1}{2}\pi r^{2}=\frac{1}{2}\pi(\frac{1}{2}(\sqrt{x}-\frac{1}{2}x))^{2}=\frac{\pi}{8}(x - x^{\frac{3}{2}}+\frac{1}{4}x^{2})$.

Step5: Calculate the volume using the integral

The volume $V$ of the solid with cross - sectional area $A(x)$ from $x = 0$ to $x = 4$ is given by the integral $V=\int_{0}^{4}A(x)dx=\int_{0}^{4}\frac{\pi}{8}(x - x^{\frac{3}{2}}+\frac{1}{4}x^{2})dx$. [ \begin{align*} V&=\frac{\pi}{8}\int_{0}^{4}(x - x^{\frac{3}{2}}+\frac{1}{4}x^{2})dx\ &=\frac{\pi}{8}\left[\frac{1}{2}x^{2}-\frac{2}{5}x^{\frac{5}{2}}+\frac{1}{12}x^{3}\right]_{0}^{4}\ &=\frac{\pi}{8}\left(\frac{1}{2}(4)^{2}-\frac{2}{5}(4)^{\frac{5}{2}}+\frac{1}{12}(4)^{3}\right)\ &=\frac{\pi}{8}\left(8-\frac{2}{5}\times32+\frac{64}{12}\right)\ &=\frac{\pi}{8}\left(8-\frac{64}{5}+\frac{16}{3}\right)\ &=\frac{\pi}{8}\left(\frac{120 - 192+80}{15}\right)\ &=\frac{\pi}{8}\times\frac{8}{15}\ &=\frac{\pi}{15}\approx0.209 \end{align*} ]

Answer:

$0.209$