(1) When you integrate with respect to $x$, you’re chopping the region into vertical slices. Look at the shaded region in your picture. First of all, $x$ ranges from $-5$ to $4$, so if you could do the calculation as a single integral, it would be $\int_{-5}^4\text{ something }dx$. But when $x$ is between $-5$ and $0$, vertical slices from from $y=x+2$ at the top down to $y=-\sqrt{x-4}$ at the bottom, while for $x$ between $0$ and $4$ they run from $\sqrt{x-4}$ at the top down to $-\sqrt{x-4}$ at the bottom. In other words, for $-5\le x\le 0$ the slice at $x$ has length
$$\begin{align*}\text{top}-\text{bottom}&=(x+2)-(-\sqrt{x-4})\\
&=x+2+\sqrt{x-4}\;,
\end{align*}$$
but for $0\le x\le 4$ it has length $$\begin{align*}\text{top}-\text{bottom}&=\sqrt{x-4}-(-\sqrt{x-4})\\
&=2\sqrt{x-4}\;.
\end{align*}$$
Since these are not the same, you have to break the calculation into two parts: the area is
$$\int_{-5}^0(x+2+\sqrt{4-x})\,dx+\int_0^4 2\sqrt{4-x}\,dx\;.$$
Notice that I can’t simply say that $y=\sqrt{4-x}$: $y$ is $\sqrt{4-x}$ for points on the upper branch of the parabola, but for points on the lower branch $y=-\sqrt{4-x}$.
(2) Again, look at the shaded region: when you cut a horizontal slice across it, the slice starts on the left at a point on the straight line and ends at a point on the right on the parabola. For a given $y$, the $x$-coordinate on the parabola is given by $x=4-y^2$, so that’s the $x$-coordinate at the righthand end of the slice; the $x$-coordinate on the straight line is given by $x=y-2$, so that’s the $x$-coordinate at the lefthand end of the slice. The length of the slice is therefore $$\text{right}-\text{left}=(4-y^2)-(y-2)=6-y-y^2\;,$$ and the infinitesimal bit of area contributed by it is its length times its width $dy$:
$$dA=(6-y-y^2)\,dy\;.$$ (For this one you do have the correct limits of integration.)
You need to find the intersections of the two parabola.
These occur when $9-y^2=y^2-9$, which is equivalent to $y^2-9=0$.
So you have two intersections, namely $(0,-3)$ and $(0,3)$.
Draw a picture.
The area is given by
$$
\int_{-3}^3(9-y^2-(y^2-9))dy=2\int_0^3(18-2y^2)dy.
$$
I guess you can finish the calculation.
Best Answer
Answer:
$$ -\int_{-1}^1 y^2-2 \,dy + \int_{-1}^1 e^y \,dy= e- \frac{1}{e} +\frac{10}{3}$$