Answer
$$\int_{0}^{x^{3}} y e^{-y / x} d y=x^{2}\left(1-e^{-x^{2}}-x^{2} e^{-x^{2}}\right)$$
Work Step by Step
Given $$\int_{0}^{x^{3}} y e^{-y / x} d y$$
by using integration by partition, let
$$u=y, d u=d y, \ \ \ \ \ \ \ d v=e^{-y / x} d y, v=-x e^{-y / x}$$
so, we get
\begin{align} \int_{0}^{x^{3}} y e^{-y / x} d y=&\left[-x y e^{-y / x}\right]_{0}^{x^{3}}+x \int_{0}^{x^{3}} e^{-y / x} d y\\
&=-x^{4} e^{-x^{2}}-\left[x^{2} e^{-y / x}\right]_{0}^{3}\\
&=x^{2}\left(1-e^{-x^{2}}-x^{2} e^{-x^{2}}\right)
\end{align}
