Divide the region into horizontal slices. At each value of $y$ between $-1$ and $1$, the slice at $y$ extends from $y^2-2$ on the left to $e^y$ on the right, so its length is $\text{right}-\text{left}=e^y-(y^2-2)=e^y-y^2+2\;,$ and its area is $dA=(e^y-y^2+2)\,dy\;,$ its length times its (vertical) width $dy$. Thus, you should be calculating $\int_{-1}^1(e^y-y^2+2)\,dy=\left[e^y-\frac{y^3}3+2y\right]_{-1}^1\;.$ Can you finish it from there?
Added: You asked how I knew to subtract the parabola from the exponential. Think about calculating the area of a horizontal strip of width $dy$ taken at some particular value of $y$: I need to multiply the width $dy$ by the length of the strip. The strip runs between two $x$-coordinates, $y^2-2$ and $e^y$, so its length is the larger of these numbers minus the smaller.
Digression: How long is a line segment that runs from $x=-1$ to $x=3$? The segment runs from $-1$ up to $0$, for a length of $1$, and then from $0$ to $3$, for an additional length of $3$, making a total length of $4$. It’s not coincidence that $3-(-1)=4$: the length is always the bigger $x$-coordinate minus the smaller one. In this example you can see that subtracting $-1$ has the effect of adding to $3$ the length of the part of the segment that’s to the left of $0$, which is exactly what’s needed to make it come out right.
Back to the problem: I know that $y^2-2, because between $y=-1$ and $y=1$ the parabola lies entirely to the left of the exponential, so it always gives the smaller $x$-coordinate. (For that matter, I also know it because $y^2-2<0$ when $-1\le y\le 1$, and $e^y$ is always positive.) Thus, I know that the length of a skinny rectangle at height $y$ must be $e^y-(y^2-2)$, and its area must be $\big(e^y-(y^2-2)\big)dy$.
Note that if the integration had been with respect to $x$, so that I had vertical strips of width $dx$, I’d subtract the bottom $y$-coordinate from the top one. It’s exactly the same idea: I’m getting the length of a strip by subtracting the small coordinate from the large one.