Answer
$$
\begin{split}
\int_{0}^{\infty} x^{2} e^{-x^{2}} d x &=\lim _{t \rightarrow \infty} \int_{0}^{t} x^{2} e^{-x^{2}} d x \\
& \quad\quad\quad\left[\text { use integration by parts with }
\right] \\
&\quad\quad\quad \left[\begin{array}{c}{u=x, \quad\quad dv= x e^{-x^{2}} dx } \\ {d u= dx, \quad\quad v= \frac{1}{-2}e^{-x^{2}} }\end{array}\right] , \text { then }\\
&=\lim _{t \rightarrow \infty}\left[\frac{1}{-2}xe^{-x^{2}} \right]_0^{t}+\frac{1}{2}\int_{0}^{\infty} e^{-x^{2}} d t \\
&=\lim _{t \rightarrow \infty}\left[\frac{t}{-2e^{t^{2}}} \right] +\frac{1}{2}\int_{0}^{\infty} e^{-x^{2}} d t \\
\\
& \quad\quad\quad\left[\text {The fraction part is } \frac{\infty}{\infty} \text { so we can use L'Hospital's } \right] \\
&=\lim _{t \rightarrow \infty}\left[\frac{1}{-4te^{t^{2}}} \right] +\frac{1}{2}\int_{0}^{\infty} e^{-x^{2}} d t \\
&=0 +\frac{1}{2} \int_{0}^{\infty} e^{-x^{2}} d t\\
&=\frac{1}{2} \int_{0}^{\infty} e^{-x^{2}} d t
\end{split}
$$
Therefore,
$$
\int_{0}^{\infty} x^{2} e^{-x^{2}} d x = \frac{1}{2} \int_{0}^{\infty} e^{-x^{2}} d t.
$$
Work Step by Step
$$
\begin{split}
\int_{0}^{\infty} x^{2} e^{-x^{2}} d x &=\lim _{t \rightarrow \infty} \int_{0}^{t} x^{2} e^{-x^{2}} d x \\
& \quad\quad\quad\left[\text { use integration by parts with }
\right] \\
&\quad\quad\quad \left[\begin{array}{c}{u=x, \quad\quad dv= x e^{-x^{2}} dx } \\ {d u= dx, \quad\quad v= \frac{1}{-2}e^{-x^{2}} }\end{array}\right] , \text { then }\\
&=\lim _{t \rightarrow \infty}\left[\frac{1}{-2}xe^{-x^{2}} \right]_0^{t}+\frac{1}{2}\int_{0}^{\infty} e^{-x^{2}} d t \\
&=\lim _{t \rightarrow \infty}\left[\frac{t}{-2e^{t^{2}}} \right] +\frac{1}{2}\int_{0}^{\infty} e^{-x^{2}} d t \\
\\
& \quad\quad\quad\left[\text {The fraction part is } \frac{\infty}{\infty} \text { so we can use L'Hospital's } \right] \\
&=\lim _{t \rightarrow \infty}\left[\frac{1}{-4te^{t^{2}}} \right] +\frac{1}{2}\int_{0}^{\infty} e^{-x^{2}} d t \\
&=0 +\frac{1}{2} \int_{0}^{\infty} e^{-x^{2}} d t\\
&=\frac{1}{2} \int_{0}^{\infty} e^{-x^{2}} d t
\end{split}
$$
Therefore,
$$
\int_{0}^{\infty} x^{2} e^{-x^{2}} d x = \frac{1}{2} \int_{0}^{\infty} e^{-x^{2}} d t.
$$
