Answer
The improper integral converges to $\dfrac{1}{18}$.
Work Step by Step
We will solve the integral $I=\int x^2 e^{-6x^3} \ dx$ by using u-substitution method.
Let us consider that $u=-6x^3 \implies dx=\dfrac{-1}{18x^2}du$
So, we have, $I=\int x^2 e^u (\dfrac{-1}{18x^2}) du=-\dfrac{1}{18}\int e^u du=-\dfrac{1}{18}e^u+C=-\dfrac{1}{18}e^{-6x^3}+C$
Now, $\lim\limits_{a \to \infty} [-\dfrac{1}{18}e^{-6x^3}]_0^a=\dfrac{-1}{18} \lim\limits_{a \to + \infty} [e^{-6a^3}-e^0]$
or, $=\dfrac{-1}{18} [e^{-6(\infty)^3}-1]$
or, $=\dfrac{-1}{18} (0-1)$
or, $=\dfrac{1}{18}$
Therefore, the given improper integral converges to $\dfrac{1}{18}$.
