Let's observe picture bellow.Total area $A$ is represented as sum of blue and yellow area:
$A=A_1+A_2$ , where $A_1=\int\limits_{-1}^{0} \sqrt{1+x}\, dx$ ,and $A_2=\int\limits_{0}^{1} (1-x) \, dx$
So we have that:
$A=\frac{2}{3}+\frac{1}{2}=\frac{7}{6}$
(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.)
Best Answer
Notice that $3 \sqrt{4 - x}$ is only defined for $x \le 4$, where it's values are non-negative. Similarly, $3 \sqrt{x - 2}$ is only defined for $x \ge 2$, where it's values are non-negative.
As you've correctly observed, the two curves intersect at $x = 3$, where $y = 3$ as well. The two curves, together with the $x$-axis (which is yet another curve, namely $y = 0$) form the boundary of a region.
Sketch the picture. (You ought to know what the graph of $y = \sqrt{x}$ looks like. Hint: it's a parabola. Now, your curves are obtained by some transformations.)
Now, the area is calculated as: $$ \begin{align} A &= \int_2^4 \min\{ 3 \sqrt{x - 2}, 3 \sqrt{4 - x} \}\;dx \\ &= \int_2^3 3 \sqrt{x - 2}\;dx + \int_3^4 3 \sqrt{4 - x}\;dx. \end{align} $$