Questions: Find the volume of the solid obtained by rotating the region under the graph of the function (f(x)=frac2x+1) about the x-axis over the interval ([0,5]).

$Find the volume of the solid obtained by rotating the region under the graph of the function (f(x)=frac2x+1) about the x-axis over the interval ([0,5]).$

Transcript text: (1 point) Find the volume of the solid obtained by rotating the region under the graph of the function $f(x)=\frac{2}{x+1}$ about the $x$-axis over the interval $[0,5]$.

Solution

Solution Steps

To find the volume of the solid obtained by rotating the region under the graph of the function $ f(x) = \frac{2}{x+1} $ about the x-axis over the interval $[0, 5]$, we can use the disk method. The volume $ V $ is given by the integral of $\pi [f(x)]^2$ from 0 to 5.

Step 1: Set Up the Integral for Volume Calculation

To find the volume of the solid obtained by rotating the region under the graph of $ f(x) = \frac{2}{x+1} $ about the x-axis over the interval $[0, 5]$, we use the disk method. The formula for the volume $ V $ is given by:

\[ V = \pi \int_{0}^{5} \left( \frac{2}{x+1} \right)^2 \, dx \]

Step 2: Evaluate the Integral

We evaluate the integral:

\[ \int_{0}^{5} \left( \frac{2}{x+1} \right)^2 \, dx = \int_{0}^{5} \frac{4}{(x+1)^2} \, dx \]

The antiderivative of $\frac{4}{(x+1)^2}$ is $-\frac{4}{x+1}$. Therefore, we have:

\[ \left[ -\frac{4}{x+1} \right]_{0}^{5} = \left( -\frac{4}{6} \right) - \left( -\frac{4}{1} \right) = -\frac{2}{3} + 4 = \frac{10}{3} \]

Step 3: Calculate the Volume

Substitute the evaluated integral back into the volume formula:

\[ V = \pi \times \frac{10}{3} = \frac{10\pi}{3} \]