logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
0 votes

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} $

Can you answer this question?
 
 

1 Answer

0 votes
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}$$+c$.
  • Equation of a line where two points are given is $\large\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$
Step 1:
Given :Vertices of the triangle $A(-1,1),B(0,5)$ and$C(3,2)$.
First let us plot the points on the graph to find the required region.
The shaded portion is the required region.
To find the area,we need the equations of $AB,BC$ and $CA$
Equation of the line AB is $A(x_1,y_1)=(-1,1)$ and $B(x_2,y_2)=(0,5)$
Equation of the line $AB=\large\frac{y-1}{5-1}=\frac{x+1}{0+1}$
$\Rightarrow (y-1)=4(x+1)\Rightarrow y=4x+5$-------(1)
Step 2:
Equation of the line BC is $B(x_1,y_1)=(0,5)$ and $C(x_2,y_2)=(3,2)$
Equation of the line $BC=\large\frac{y-5}{2-5}=\frac{x-1}{3-0}$
$\Rightarrow 3(y-5)=-3x\Rightarrow y=\large\frac{-3x+15}{3}$$=-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.
answered May 8, 2013 by sreemathi.v
 
Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...