Example 3. Determine the area bounded by the parabola $y=x^{2}-5 x+6$ and the coordinate axes (and adjacent to both axes).
Solution. Let's find the limits of integration. Since the figure is bounded on the left by the $O y$ axis, the lower limit of integration $a=0$. The curve $y=x^{2}-5 x+6$ intersects the $O x$ axis at two points: $x_{1}=2, x_{2}=3$ (Fig. 5.7), so the upper limit $b=2$. Thus,

$$
S=\int_{0}^{2}\left(x^{2}-5 x+6\right) d x=\int_{0}^{2} x^{2} d x-5 \int_{0}^{2} x d x+6 \int_{0}^{2} d x=
$$

$$
\begin{gathered}
=\left.\frac{x^{3}}{3}\right|_{0} ^{2}-\left.5 \frac{x^{2}}{2}\right|_{0} ^{2}+\left.6 x\right|_{0} ^{2}=\frac{2^{3}}{3}-5 \cdot \frac{2^{2}}{2}+6 \cdot 2= \\
=\frac{8}{3}-10+12=\frac{14}{3}=4 \frac{2}{3}
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_7ef2114a321cf223ba99g-210.jpg?height=431&width=394&top_left_y=311&top_left_x=39)

Fig. 5.7

![](https://cdn.mathpix.com/cropped/2024_05_22_7ef2114a321cf223ba99g-210.jpg?height=396&width=380&top_left_y=318&top_left_x=727)

Fig. 5.8