Question

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

Answer 1

Let we have the vertices of $△ABC$ as $A(-1,1),B(0,5)$ and $C(3,2)$.

$\therefore$ Equation of $AB$ is $y-1=\left(\frac{5-1}{0+1}\right)(x+1)$

$\Rightarrow y-1=4x+4$

$\Rightarrow y=4x+5....\left(i\right)$

And equation of $BC$ is $y-5=\left(\frac{2-5}{3-0}\right)(x-0)$

$\Rightarrow y-5=\frac{-3}{3}\left(x\right)$

$\Rightarrow y=5-x....\left(ii\right)$

Similarly, equation of $AC$ is $y-1=\left(\frac{2-1}{3+1}\right)(x+1)$

$\Rightarrow y-1=\frac{1}{4}(x+1)$

$\Rightarrow 4y=x+5....\left(iii\right)$

$\therefore$ Area of shaded region $={\int }_{-1}^0(y_1-y_2)dx+{\int }_0^3(y_1-y_2)dx$

$={\int }_{-1}^0[4x+5-\frac{x+5}{4}]dx+{\int }_0^3[5-x-\frac{x+5}{4}]dx$

$={[\frac{4x^2}{2}+5x-\frac{x^2}{8}-\frac{5x}{4}]}_{-1}^0+{[5x-\frac{x^2}{2}-\frac{x^2}{8}-\frac{5x}{4}]}_0^3$

$=[0-(4.\frac{1}{2}+5(-1)-\frac{1}{8}+\frac{5}{\$})]+[(15-\frac{9}{2}-\frac{9}{8}-\frac{15}{4})-0]$

$=[-2+5+\frac{1}{8}-\frac{5}{4}+15-\frac{9}{2}-\frac{9}{8}-\frac{15}{4}]$

$=18+\left(\frac{1-10-36-9-30}{8}\right)$

$=18+(-\frac{84}{8})=18-\frac{21}{2}=\frac{15}{2}squnits$.