The first curve, $x + y^2 - 4 = 0$ represents a parabola that opens to the left, while the second curve $x + y = 2$ represents a line. We must first find where these curves intersect each other. This can be done in a variety of ways. One way would be to take the second equation $x + y = 2$ and isolate for either variable, say $x = 2 - y$. Then substitute this into the first equation (the parabola) to get:

(1)
\begin{align} x + y^2 - 4 &= 0 \\ (2 - y) + y^2 - 4 &= 0 \\ y^2 - y - 2 &= 0 \\ (y - 2)(y + 1) &= 0 \end{align}

Thus we get that $y = 2$ and $y = -1$. These are the $y$-coordinates of the points for which the parabola and line intersect each other. These $y$-coordinates can be plugged back into either equation to get the corresponding $x$-coordinates. If we plug $y = 2$ into the equation $x + y = 2$ we get that $x = 0$ and so $(0, 2)$ is a point of intersection. If we plug $y = -1$ into the equation $x + y = 2$ we get that $x = 3$ and so $(3, -1)$ is a point of intersection.

Now since one of our curves contains a $y^2$ term and can easily be rewritten with $x$ isolated, that is $x + y^2 - 4 = 0 \Leftrightarrow x = 4 - y^2$, it will be best to integrate with respect to $y$. A quick sketch of both curves shows us that our "top" or "rightmost" curve is the parabola $x = 4 - y^2$, and the "bottom" or "leftmost" curve is the straight line $x = 2 - y$.

Furthermore, since we are dealing with the AREA between curves, we must also check to see if any portion of our area goes to the left of the $y$-axis. In this case it does not.

Since we will set up an integral with respect to $y$, our lower limit of integration is $-1$ and our upper limit of integration is $2$ (which are the corresponding $y$-values of the points of intersection we found above).

(2)
\begin{align} \quad A &= \int_{-1}^{2} [4 - y^2] - [2 - y] \: dy \\ \quad A &= \int_{-1}^{2} 2 + y - y^2 \: dy \\ \quad A &= \left [ 2y + \frac{y^2}{2} - \frac{y^3}{3} \right]_{-1}^2 \\ \quad A &= \left [ 4 + 2 - \frac{8}{3} \right ] - \left [ -2 + \frac{1}{2} + \frac{1}{3} \right ] = \frac{9}{2} \end{align}