326. $y=-x^{2}+5$ and $y=x+3$.
Solution. Let's find the abscissas of the points of intersection of the parabola $y = -x^{2}+5$ and the line $y=x+3$. For this, we solve the system $\left\{\begin{array}{l}y=-x^{2}+5, \\ y=x+3,\end{array}\right.$ from which $x_{1}=-2, x_{2}=1$.

Let's find the area $S_{1}$ of the figure bounded by the parabola $y=-x^{2}+5$, the lines $x=-2, x=1$ and $y=0$ (Fig. 173). We get

$$
S_{1}=\int_{-2}^{1}\left(-x^{2}+5\right) d x=\left.\left(-\frac{x^{3}}{3}+5 x\right)\right|_{-2} ^{1}=12 \text { (sq. units) }
$$

Let's find the area $S_{2}$ of the figure bounded by the lines $y=x+3$, $x=-2, x=1$ and $y=0$:

$$
S_{2}=\int_{-2}^{1}(x+3) d x=\left.\left(\frac{x^{2}}{2}+3 x\right)\right|_{-2} ^{1}=7.5 \text { (sq. units) }.
$$

The area of the desired figure is $S=S_{1}-S_{2}=12-7.5=4.5$ (sq. units). 327. $y=x^{2}$ and $y=2 x$.