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

Explanation:

Step1: Find intersection points

Set $\sqrt{x}+2=\frac{1}{2}x + 2$. Subtract 2 from both sides: $\sqrt{x}=\frac{1}{2}x$. Let $t = \sqrt{x}(t\geq0)$, then $t=\frac{1}{2}t^{2}$. Rearrange to $\frac{1}{2}t^{2}-t = 0$, factor out $t$: $t(\frac{1}{2}t - 1)=0$. 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 $x = 0$ and $x = 4$.

Step2: Find the length of the leg of the isosceles right - triangle

The length of the leg $L$ of the isosceles right - triangle at a given $x$ is $L=(\sqrt{x}+2)-(\frac{1}{2}x + 2)=\sqrt{x}-\frac{1}{2}x$.

Step3: Find the area of the cross - section

The area of an isosceles right - triangle with leg length $L$ is $A=\frac{1}{2}L^{2}$. Substitute $L=\sqrt{x}-\frac{1}{2}x$ into the formula: $A(x)=\frac{1}{2}(\sqrt{x}-\frac{1}{2}x)^{2}=\frac{1}{2}(x - x^{\frac{3}{2}}+\frac{1}{4}x^{2})$.

Step4: Find the volume using the integral

The volume $V$ of the solid with cross - sectional area $A(x)$ from $x = a$ to $x = b$ is given by $V=\int_{a}^{b}A(x)dx$. Here, $a = 0$, $b = 4$, and $A(x)=\frac{1}{2}(x - x^{\frac{3}{2}}+\frac{1}{4}x^{2})$. So $V=\frac{1}{2}\int_{0}^{4}(x - x^{\frac{3}{2}}+\frac{1}{4}x^{2})dx$. Integrate term - by - term: $\frac{1}{2}\left[\frac{1}{2}x^{2}-\frac{2}{5}x^{\frac{5}{2}}+\frac{1}{12}x^{3}\right]_{0}^{4}$. First, substitute $x = 4$: $\frac{1}{2}\left(\frac{1}{2}(4)^{2}-\frac{2}{5}(4)^{\frac{5}{2}}+\frac{1}{12}(4)^{3}\right)=\frac{1}{2}\left(8-\frac{2}{5}\times32+\frac{64}{12}\right)$. $=\frac{1}{2}\left(8-\frac{64}{5}+\frac{16}{3}\right)$. Find a common denominator of 15: $\frac{1}{2}\left(\frac{120 - 192+80}{15}\right)=\frac{1}{2}\times\frac{8}{15}=\frac{4}{15}\approx0.267$.

Answer:

$0.267$