Answer
\[
=\frac{1-e^{-16}}{8}
\]
Work Step by Step
\[
\int_{0}^{1} \int_{4 x}^{4} e^{-y^{2}} d y d x=\iint_{D} e^{-y^{2}} d A
\]
$D$ has been interpreted as a type I region (using Vertical cross-sections).
We can also Interpret $D$ as a type II region (using horizontal cross-sections).
So:
\[
\begin{array}{l}
\iint_{D} e^{-y^{2}} d A=\int_{0}^{4} \int_{0}^{y / 4} e^{-y^{2}} d x d y \\
\quad=\int_{0}^{4}\left[x e^{-y^{2}}\right]_{x=0}^{x=y / 4} d y \\
\quad=\int_{0}^{4} \frac{y}{4} e^{-y^{2}} d y \\
=-\frac{1}{8} \int_{0}^{4}-2 y e^{-y^{2}} d y
\end{array}
\]
Replace $du=-2 y d y$ and $u=-y^{2}$
\[
=-\frac{1}{8} \int_{0}^{-16} e^{u} d u
\]
\[
=-\frac{1}{8}\left[e^{u}\right]_{0}^{-16}
\]
\[
=-\left[e^{16}-e^{0}\right]\frac{1}{8}
\]
\[
=\frac{1-e^{-16}}{8}
\]
