Answer
$$
\int_{0}^{1} y e^{-2 y^{2}} d y =\frac{1}{4}\left(1-e^{-2}\right)
$$
Work Step by Step
$$
\int_{0}^{1} y e^{-2 y^{2}} d y
$$
$$\text { Let } u=-2 y^{2} . \text { Then } d u=-4 y d y \\
y : 0 \rightarrow 1 . \text { Then } u : 0 \rightarrow -2
$$
substituting in the given integral we have :
$$
\begin{aligned}
\int_{0}^{1} y e^{-2 y^{2}} d y &
=\int_{0}^{-2} e^{u}\left(-\frac{1}{4} d u\right) \\
&=-\frac{1}{4}\left[e^{u}\right]_{0}^{-2} \\
&=-\frac{1}{4}\left(e^{-2}-1\right) \\
&=\frac{1}{4}\left(1-e^{-2}\right)
\end{aligned}
$$
