Using integration, find the area of the region bounded by the triangle whose vertices are (2, - 2), (4, 5) and (6, 2).

Let assume that A(2, -2), B(4, 5) and C=(6, 2)
Now, we will draw the points on the axes and get the triangle ABC as shown in the figure:

Now, we will find the equation of line AB,
$\mathrm{y}-{\mathrm{y}}_1=\frac{{\mathrm{y}}_2-{\mathrm{y}}_1}{{\mathrm{x}}_2-{\mathrm{x}}_1}\left(\mathrm{x}-{\mathrm{x}}_1\right)$
Where, $\mathrm{A}(2,-2)=\mathrm{A}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)\text{ and }\mathrm{B}(4,5)=\mathrm{B}\left({\mathrm{x}}_2,{\mathrm{y}}_2\right)$
$\begin{array}{l}\therefore \mathrm{y}-(-2)=\frac{5-(-2)}{4-2}(\mathrm{x}-2)\\ \therefore \mathrm{y}+2=\frac{5+2}{2}(\mathrm{x}-2)\\ \therefore \mathrm{y}+2=\frac{7}{2}(\mathrm{x}-2)\\ \therefore \frac{2}{7}\mathrm{y}+2\times \frac{2}{7}=\mathrm{x}-2\\ \therefore \frac{2}{7}\mathrm{y}+\frac{4}{7}=\mathrm{x}-2\\ \therefore \mathrm{x}=\frac{2}{7}\mathrm{y}+\frac{4}{7}+2\\ \therefore \mathrm{x}=\frac{2}{7}\mathrm{y}+\frac{18}{7}\\ \therefore \mathrm{x}=\frac{2}{7}(\mathrm{y}+9)\end{array}$
Similarly, Equation of BC and AC are respectively $\mathrm{x}=-\frac{2}{3}(\mathrm{y}-11)\text{ and }\mathrm{x}=\mathrm{y}+4$
From the graph,
Area of Δ𝐴𝐵𝐶$={\int }_{-2}^2 (\mathrm{y}+4)\mathrm{dy}+\left(-\frac{2}{3}\right){\int }_2^5 (\mathrm{y}-11)\mathrm{dy}-{\int }_{-2}^5 \frac{2}{7}(\mathrm{y}+9)\mathrm{dy}$
$\begin{array}{l}=\frac{1}{2}{\left[(\mathrm{y}+4{)}^2\right]}_{-2}^2+\left(-\frac{2}{3}\right)\left(\frac{1}{2}\right){\left[(\mathrm{y}-11{)}^2\right]}_2^5-\frac{2}{7}\left(\frac{1}{2}\right){\left[(\mathrm{y}+9{)}^2\right]}_{-2}^5\\ =\frac{1}{2}\left((2+4{)}^2-(-2+4{)}^2\right)+\left(-\frac{2}{3}\right)\left(\frac{1}{2}\right)\left\{(5-11{)}^2-(2-11{)}^2\right\}-\\ \frac{2}{7}\left(\frac{1}{2}\right)\left\{(5+9{)}^2-(-2+9{)}^2\right\}\end{array}$
$\begin{array}{l}=\frac{1}{2}\left(6^2-2^2\right)-\frac{1}{3}\left((-6{)}^2-(-9{)}^2\right)-\frac{1}{7}\left({14}^2-7^2\right)\\ =\frac{1}{2}(36-4)-\frac{1}{3}(36-81)-\frac{1}{7}(196-49)\\ =\frac{1}{2}\left(32\right)-\frac{1}{3}(-45)-\frac{1}{7}\left(147\right)\\ =16+15-21=10\end{array}$
Therefore, area of the region bounded by the triangle is 10 units.