Example 2.48. Calculate the area of the figure bounded by the parabola $y=-x^{2}+3 x-2$ and the coordinate axes.
Solution. Since the figure $y=-x^{2}+3 x-2$ is located in the regions with $y \geqslant 0$ and $y<0$ (Fig. 2.3), the area $S$ should be calculated separately for the part $y \geqslant 0$ and the part $y<0$, and then the absolute values of the obtained integrals should be added:

$$
\begin{gathered}
S=\left|\int_{0}^{1}\left(-x^{2}+3 x-2\right) d x\right|+\left|\int_{1}^{2}\left(-x^{2}+3 x-2\right) d x\right|= \\
\left.=\left.\left|\left(-\frac{x^{3}}{3}+\frac{3 x^{2}}{2}-2 x\right)\right|_{0}^{1}|+|\left(-\frac{x^{3}}{3}+\frac{3 x^{2}}{2}-2 x\right)\right|_{1} ^{2} \right\rvert\,= \\
=\left|-\frac{5}{6}\right|+\left|\frac{1}{6}\right|=1
\end{gathered}
$$