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

Explanation:

Step1: Find intersection points

Set $\sqrt{x}=\frac{1}{2}x$. Square both sides: $x = \frac{1}{4}x^{2}$. Rearrange to $x^{2}-4x = 0$, factor to $x(x - 4)=0$. So $x = 0$ and $x = 4$ are the intersection - points.

Step2: Determine base of cross - section

The base of each cross - section perpendicular to the $x$ - axis is $b=\sqrt{x}-\frac{1}{2}x$.

Step3: Determine height of cross - section

The height of each cross - section is $h=\frac{1}{2}(\sqrt{x}-\frac{1}{2}x)$.

Step4: Find area of cross - section

The area of a rectangle $A = b\times h$. Substitute $b$ and $h$: $A(x)=\frac{1}{2}(\sqrt{x}-\frac{1}{2}x)^{2}=\frac{1}{2}(x - x^{\frac{3}{2}}+\frac{1}{4}x^{2})$.

Step5: Calculate volume using integral

The volume $V=\int_{a}^{b}A(x)dx$, where $a = 0$, $b = 4$. So $V=\int_{0}^{4}\frac{1}{2}(x - x^{\frac{3}{2}}+\frac{1}{4}x^{2})dx$. [ \begin{align*} V&=\frac{1}{2}\left[\frac{1}{2}x^{2}-\frac{2}{5}x^{\frac{5}{2}}+\frac{1}{12}x^{3}\right]_{0}^{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)\ &=\frac{1}{2}\left(\frac{120 - 192+80}{15}\right)\ &=\frac{1}{2}\times\frac{8}{15}\ &=\frac{4}{15}\approx0.267 \end{align*} ]

Answer:

$0.267$