#### Find the area of region bounded by the triangle whose vertices are (–1, 1), (0, 5) and (3, 2), using integration.

The vertices of the triangle are given as (-1, 1), (0, 5) and (3, 2)

The three vertices are given alphabets as P, Q and R respectively.

$\\\Rightarrow \text{Equation of } \mathrm{PQ } \: \: is \: \: y-1=\frac{5-1}{0+1}(x+1)\\$

$\\\therefore y=4 x+5 \Rightarrow Equation of Q R is y-5=\frac{2-5}{3-0}(x-0) \therefore y=-x+5 \Rightarrow Equation of \mathrm{RP} is y-2=\frac{1-2}{-1-3}(x-3) \therefore 4 y=x+5 \\$

Area of the region bounded by the curve $y=f(x)$, the x -axis and the ordinates x=a and x=b, where f(x) is a continuous function defined on [a, b], is given by $A=\int_{a}^{b} f(x) d x or \int_{a}^{b} y d x$

Required area

$=\int_{-1}^{0}(4 x+5) d x+\int_{0}^{3}(-x+5) d x-\int_{-1}^{3}\left(\frac{x}{4}+\frac{5}{4}\right) d x$

$\\=\left[2 x^{2}+5 x\right]_{-1}^{0}+\left[-\frac{x^{2}}{2}+5 x\right]_{0}^{3}-\left[\frac{x^{2}}{8}+\frac{5 x}{4}\right]_{-1}^{3}\\=((0+0)-(2-5))+\left(\left(-\frac{9}{2}+15\right)-(0+0)\right)-\left(\left(\frac{9}{8}+\frac{15}{4}\right)-\left(\frac{1}{8}-\frac{5}{4}\right)\right)\\=3+\frac{21}{2}-\frac{39}{8}-\frac{9}{8} \\$

$=\frac{15}{2}\text{sq.units}$