Answer
$2 \pi$
Work Step by Step
We need to use shell mode:
$Volume =\int_{0}^{\infty} 2 \pi x e^{-x} dx =2 \pi \lim\limits_{a \to \infty} \int_0^a xe^{-x} dx$
Now, $2 \pi \lim\limits_{a \to \infty} \int_0^a xe^{-x} dx=2 \pi \lim\limits_{a \to \infty} [ (-x e^{-x})_0^a +\int_0^a e^{-x} dx $
or, $=-a e^{-a} +[-e^{-x}]_0^a$
or, $= 2 \pi \lim\limits_{a \to \infty} [-a e^{-a} +(-e^{-a} +e^{-0})]$
or, $= 2 \pi \lim\limits_{a \to \infty} [-a e^{-a}]-2\pi \times (0)+2 \pi$
Now, apply L'Hospital's rule.
$V=2 \pi \lim\limits_{a \to \infty} [\dfrac{-a}{e^{a}}]+2 \pi=2 \pi \lim\limits_{a \to \infty} [\dfrac{-1}{e^{a}}]+2 \pi= 2\pi$
