Answer
$$1.87691$$
Work Step by Step
Given $$y=e^{ -x^2} ; \quad\left[0,1\right] ; \quad x \text { -axis; } \quad T_{8}$$
Since
$$ V= \pi \int_{0}^{1} [f(x)]^2dx= \pi \int_{0}^{1}e^{ -2x^2}dx$$
Now, we will evaluate the integral using $T_8$, since $\Delta x=\dfrac{b-a}{n}=\dfrac{1}{8}$
\begin{align*}
T_{n}&=\dfrac{1}{2}\left[f(x_0)+2f(x_1)+2f(x_2)+.....+2f(x_{n-1})+f(x_n)\right]\Delta x\\
T_{8}&=\dfrac{1}{2}\left[f(x_0)+2f(x_1)+2f(x_2)+\cdots +2f(x_{7})+f(x_8)\right]\Delta x\\
&=\dfrac{1}{16}\left[f(0)+2f(1/8)+2f(2/8)+2f(3/8)+2f(4/8)+2f(5/8)+2f(6/8)+2f(7/8)+f(1)\right] \\
&\approx 0.59746
\end{align*}
Hence, $V\approx \pi *0.59746 \approx 1.87691$
