1287. Calculate the area of the figure bounded by the parabola $y=-x^{2}+$ $+6 x-5$ and the coordinate axes.
Solution. If the figure is located on opposite sides of the $O x$ axis (Fig. $82, a$), then the area $S$ should be calculated using the formula

$$
S=\int_{a}^{c} f(x) d x+\left|\int_{c}^{b} f(x) d x\right|
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_1b9a1c1720bae4af5bc5g-248.jpg?height=345&width=471&top_left_y=1424&top_left_x=58)

Fig. 81
![](https://cdn.mathpix.com/cropped/2024_05_22_1b9a1c1720bae4af5bc5g-248.jpg?height=898&width=592&top_left_y=878&top_left_x=537)

Fig. 82

From Fig. 82, b, it is clear that the area of the given figure will be equal to

$$
\begin{aligned}
& S=\left|\int_{0}^{1}\left(-x^{2}+6 x-5\right) d x\right|+\int_{1}^{5}\left(-x^{2}+6 x-5\right) d x= \\
& =\left|\left(-\frac{x^{3}}{3}+3 x^{2}-5 x\right)\right|_{0}^{1}+\left.\left(-\frac{x^{3}}{3}+3 x^{2}-5 x\right)\right|_{1} ^{5}=13
\end{aligned}
$$