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

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

Answers (1)

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}

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