94. Let's calculate the area of the figure bounded by the lines

$$
y=x^{2}-5 x+4 \text{ and } y=2 x-2 .
$$
Solution. For calculating the area of a curvilinear trapezoid, two formulas are known:

$$
\begin{aligned}
S & =\int_{a}^{b} f(x) d x \\
S & =-\int_{a}^{b} f(x) d x
\end{aligned}
$$

Formula (1) is used when the function $f(x)$ is non-negative, and formula (2) is used when the function is negative.

In our example, we need to determine the area of the shaded figure (Fig. $43, a$). We perform a parallel translation by the vector $\bar{r}(0 ;-10)$, obtaining an equal figure (Fig. $43, \sigma$). The equations of the lines will become: $y=2 x-12$ and $y=x^{2}-5 x-6$. We have $S_{B C D}=$
![](https://cdn.mathpix.com/cropped/2024_05_21_b1a79769c6c63d19f37fg-57.jpg?height=828&width=1370&top_left_y=1505&top_left_x=343)

Fig. 43

56
$=S_{A B C D}-S_{A B D}$. But by formula (2) $S_{A B C D}=-\int_{1}^{6}\left(x^{2}-5 x-6\right) d x=45 \frac{5}{6}$

$$
S_{A B D}=-\int_{1}^{6}(2 x-12) d x=25
$$

Thus,

$$
S_{B C D}=45 \frac{5}{6}-25=20 \frac{5}{6}
$$