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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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Using the method of integration find the area of the region bounded by lines: \(2x + y = 4, 3x - 2y = 6\) and \(x - 3y + 5 = 0\)

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  • Suppose three lines intersect at three different points,the enclosed area will be the area bounded by the three lines at their points intersection.
  • The limits can be obtained by solving these equations finding the points of intersection.
The graph for the three lines 2x+y=4,3x-2y=6 and x-3y+5=0 can be sketched and drawn as shown in th fig.
The required area is the region bounded by the lines in the area of the $\bigtriangleup ABC.$
To find the limits,let us find the pints of intersection of the lines.
Let 2x+y=4------(1)
$\;\;\;\;$3x-2y=6------(2)
$\;\;\;\;$x-3y=-5------(3)
on solving (1) and (2) we get
$(\times 3)\;$2x+ y=4
$(\times 2)\;$3x-2y=6
---------------------
$\;$6x+3y=12
-6x+3y=12
--------------------
$\;\;\;\;\;$7y= 0
$\;\;\;\;\;\;$y= 0
$\;\;\;\;\;\;$x= 2
Hence the point of intersection for line (1) and (2) is (2,0)
on solving (2) and (3) we get
$\;\;\;\;\;\;$3x-2y= 6
$(\times 3)\;$ x-3y =-5
-------------------------
$\;\;\;\;\;$3x-2y= 6
$\;\;\;\;\;$3x-9y=-15
-----------------------
$\;\;\;\;\;\;\;$7y=21
$\;\;\;\;\;\;\;$y= 3.Hence x=4
Hence the point of intersection between emu(2) and equ(3) is (4,3)
on solving equ(3) and (1) we get
$(\times 2)$x-3y=-5
$\;\;\;\;\;$2x+ y=4
----------------------
$\;\;\;\;\;$2x-6y=-10
$\;\;\;\;\;$2x+ y= 4
----------------------
$\;\;\;\;\;$-7y=-14
$\;\;\;\;\;$y=2.Hence x=1
Hence the point of intersection between emu(3) and emu(1) is (1,2).
Hence the required area A is
A=(Area enclosed between the line AC and x-axis)+(Area enclosed between the line AB and x-axis)+(Area enclosed between the line BC and x-axis)
$A=\int_1^4\bigg(\frac{x+5}{3}\bigg)dx+\int_2^1(4-2x)dx+\int_4^2\bigg(\frac{3x-6}{2}\bigg)dx$
$A=\int_1^4\bigg(\frac{x+5}{3}\bigg)dx-\int_1^2(4-2x)dx-\int_2^4\bigg(\frac{3x-6}{2}\bigg)dx$
on integrating we get,
$A=\frac{1}{3}\begin{bmatrix}\frac{x^2}{2}+5x\end{bmatrix}_1^4-\begin{bmatrix}4x-x^2\end{bmatrix}_1^2-\frac{1}{3}\begin{bmatrix}\frac{3x^2}{2}-6x\end{bmatrix}_2^4$
$A=\frac{1}{3}[8+20- {\frac{1}{2}}-5]_1^4-[8-4-4+1]_1^2- {\frac{1}{2}}[24-24-6+12]_2^4$
$A=(\frac{1}{3}\times \frac{45}{2}-(1)-\frac{1}{2}(6)$
$A=\frac{15}{2}-1-3$
$\;\;\;=\frac{7}{2}sq.units$
Hence the required area is $\frac{7}{2}sq. units.$
answered Dec 21, 2013 by yamini.v
 

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