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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.

$\begin{array}{1 1}15\;sq.units. \\ 30\;sq.units. \\ \frac{15}{2}\;sq.units. \\ \frac{15}{4}\;sq.units\end{array}$

Toolbox:
• The area enclosed by the curve $y=f(x)$,the $x$-axis and the ordinates $x=a$ and $x=b$ is given by $\int_a^b ydx.$
• $\int x^n dx=\large\frac{x^{n+1}}{n+1}$$+c. • Equation of a line where two points are given is \large\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1} Step 1: Given :Vertices of the triangle A(-1,1),B(0,5) andC(3,2). First let us plot the points on the graph to find the required region. The shaded portion is the required region. To find the area,we need the equations of AB,BC and CA Equation of the line AB is A(x_1,y_1)=(-1,1) and B(x_2,y_2)=(0,5) Equation of the line AB=\large\frac{y-1}{5-1}=\frac{x+1}{0+1} \Rightarrow (y-1)=4(x+1)\Rightarrow y=4x+5-------(1) Step 2: Equation of the line BC is B(x_1,y_1)=(0,5) and C(x_2,y_2)=(3,2) Equation of the line BC=\large\frac{y-5}{2-5}=\frac{x-1}{3-0} \Rightarrow 3(y-5)=-3x\Rightarrow y=\large\frac{-3x+15}{3}$$=-x+5$-------(2)
Equation of the line CA is $C(x_1,y_1)=(3,2)$ and $A(x_2,y_2)=(-1,1)$
Equation of the line $CA=\large\frac{y-2}{1-2}=\frac{x-3}{-1-3}$
$\Rightarrow (y-2)(-4)=(-1)(x-3)$
$-4y+8=-x+3$
$y=\large\frac{-x-5}{-4}$
$y=\large\frac{x+5}{4}$-----(3)
Step 3:
The required area $A=\int_{-1}^0y_1dx+\int_0^3y_2dx-\int_{-1}^3y_3dx.$
Where $y_1=4x+5,y_2=-x+5$ and $y_3=\large\frac{x+5}{4}$
$\Rightarrow A=\int_{-1}^0(4x+5)dx+\int_0^3(-x+5)dx+\int_{-1}^3\large\big(\frac{x+5}{4}\big)$$dx. On integrating we get, \;\;=\begin{bmatrix}\large\frac{4x^2}{2}\normalsize+5x\end{bmatrix}_{-1}^0+\begin{bmatrix}\large\frac{-x^2}{2}\normalsize+5x\end{bmatrix}_{0}^3-\large\frac{1}{4}\begin{bmatrix}\large\frac{x^2}{2}\normalsize+5x\end{bmatrix}_{-1}^3 \;\;=\begin{bmatrix}2x^2+5x\end{bmatrix}_{-1}^0+\begin{bmatrix}\large\frac{-x^2}{2}\normalsize+5x\end{bmatrix}_{0}^3-\large\frac{1}{4}\begin{bmatrix}\large\frac{x^2}{2}\normalsize+5x\end{bmatrix}_{-1}^3 On applying limits we get, \;\;=\large\frac{4}{2}$$[(0-1)]+5[(0+1)]+[\large\frac{-1}{2}$$(9-0)+5(3-0)]+\large\frac{1}{4}[\frac{1}{2}\normalsize (9-1)+5(3+1)] \;\;=3+\large\frac{21}{2}$$-6=\large\frac{15}{2}$sq.units.
Hence the required area is $\large\frac{15}{2}$sq.units.