the region bounded by the graph of f(x) = 1 / (1 + x^2) and the x - axis between x = 0 and x = 2 is revolved…

Explanation:

Step1: Recall volume - of - revolution formula

The formula for the volume $V$ of the solid generated by revolving the curve $y = f(x)$ about the $x$-axis from $x=a$ to $x = b$ is $V=\pi\int_{a}^{b}[f(x)]^{2}dx$. Here, $a = 0$, $b = 2$ and $f(x)=\frac{1}{1 + x^{2}}$. So, $V=\pi\int_{0}^{2}\left(\frac{1}{1 + x^{2}}\right)^{2}dx$.

Step2: Use the reduction formula for $\int\frac{1}{(1 + x^{2})^{n}}dx$

For $n = 2$, $\int\frac{1}{(1 + x^{2})^{2}}dx=\frac{x}{2(1 + x^{2})}+\frac{1}{2}\arctan(x)+C$.

Step3: Evaluate the definite integral

$V=\pi\left[\frac{x}{2(1 + x^{2})}+\frac{1}{2}\arctan(x)\right]_{0}^{2}$. First, substitute $x = 2$: $\frac{2}{2(1 + 2^{2})}+\frac{1}{2}\arctan(2)=\frac{1}{5}+\frac{1}{2}\arctan(2)$. Then substitute $x = 0$: $\frac{0}{2(1+0^{2})}+\frac{1}{2}\arctan(0)=0$. So, $V=\pi\left(\frac{1}{5}+\frac{1}{2}\arctan(2)\right)$.

Step4: Calculate the numerical value

$V=\pi\left(\frac{1}{5}+\frac{1}{2}\arctan(2)\right)\approx\pi\left(0.2 + \frac{1}{2}\times1.107149\right)\approx\pi(0.2+0.553575)\approx2.3618$.

Answer:

$2.3618$