Answer
$$
y=4-x^{2} , \quad [0,3].
$$
The total area is given by
$$
\begin{aligned} A &=\int_{0}^{2}\left(4-x^{2}\right) d x+\left|\int_{2}^{3}\left(4-x^{2}\right) d x\right| \\
&=\frac{23}{3} \end{aligned}
$$
Work Step by Step
$$
y=4-x^{2} , \quad [0,4].
$$
The Figure shows the required region. Part of the region is below the x-axis. The definite integral over that interval will have a negative value. To find the area, integrate the negative and positive portions separately and take the absolute value of the second result before combining the two results to get the total area.
The total area is given by
$$
\begin{aligned} A &=\int_{0}^{2}\left(4-x^{2}\right) d x+\left|\int_{2}^{3}\left(4-x^{2}\right) d x\right| \\
&=\left(4 x-\frac{x^{3}}{3}\right)|_{0}^{2}+\left.\left|\left(4 x-\frac{x^{3}}{3}\right)\right|_{2}^{3}\right| \\
&=\left[\left(8-\frac{8}{3}\right)-0\right]
+\left|(12-9)-\left(8-\frac{8}{3}\right)\right| \\
&=\frac{16}{3}+\left|3-\frac{16}{3}\right| \\
&=\frac{16}{3}+\frac{7}{3}\\
&=\frac{23}{3} \end{aligned}
$$
