Calculate $\iint\limits_D {e^{y^2+1}}\,dA$ where $D$ is a triangle with $(0,0),(-2,4)$ and $(8,4)$. Calculate $\iint\limits_D {e^{y^2+1}}\,dA$ where $D$ is a triangle with $(0,0),(-2,4)$ and $(8,4)$.
So far I have visualized an upside down triangle where there is a point on the origin, and the longest side is directly opposite of that point.
Based on our text, I'm thinking to go with Type II approach:  $\iint\limits {e^{y^2+1}} dxdy$. Is this correct? 
From my rough (and hopefully in the right direction) calculations that my $y$ bounds go from $0$ to $4$, and that the lower $x$ bound is $-y/2$. Am I on the right track? How shall I proceed to evaluate?
 A: The task at hand is to evaluate 
$${\large{\iint\limits_D {e^{y^2+1}}\,dA}}$$
where $D$ is region in the $xy$-plane bounded by the triangle with vertices $(0,0),(-2,4),(8,4)$.

Trying to integrate $dy\,dx$ would be problematic since the inner integral would be missing a much needed helping factor.

Hence we should try the reverse order $dx\,dy$ with the hope that it will work out more naturally (it does).

Thus, noting that


*

*The left boundary of $D$ is the line whose equation is $y=-2x\;$or equivalently, $x=-{\large{\frac{y}{2}}}$.$\\[6pt]$

*The right boundary of $D$ is the line whose equation is $y={\large{\frac{x}{2}}}\;$or equivalently, $x=2y$.


the region $D$ can be epressed as the set of points $(x,y)$ in the $xy$-plane satisfying the constraints
$$
\left\lbrace
\begin{align*}
-\frac{y}{2}\le\;&x\le 2y\\[4pt]
0\le\;&y\le 4\\[4pt]
\end{align*}
\right.
$$
hence
\begin{align*}
{\large{\iint\limits_D {e^{y^2+1}}\,dA}}
&={\large{\int_0^4\int_{-\frac{y}{2}}^{2y} e^{y^2+1}\,dx\,dy}}\\[4pt]
&={\large{\int_0^4e^{y^2+1}\left(\int_{-\frac{y}{2}}^{2y} \,dx\right)\,dy}}\\[4pt]
&={\large{\int_0^4e^{y^2+1}\left({\small{\frac{5}{2}}}y\right)\,dy}}\\[4pt]
&=\frac{5}{4}{\large{\int_0^4e^{y^2+1}(2y)\,dy}}\\[4pt]
&=\frac{5}{4}{\large{\int_1^{17}e^u \,du}}\\[4pt]
&=\frac{5}{4}\left({\large{e^{17}-e}}\right)\\[4pt]
\end{align*}
which agrees with the answer you found, as noted in your comment.
