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