Using integration,find the area of the traiangular region whose sides have the equations y= 2x+1,y= 3x+1\: and\: x= 4.

 

 

 

 
 
 
 
 

Answers (1)

(A)     The equation of side of triangle are
           y= 2x+1-\left ( i \right )
           y= 3x+1-\left ( ii \right )
           and  x= 4-\left ( iii \right )


The equation y= 2x+1 meets x and y axes at\left ( \frac{-1}{2},0 \right )  and \left ( 0,1 \right ). by joining there two points we obtain the graph of x+2y= 2. similarly,graph of other equation are drawn.
solving equation (i),(ii) and (iii) in pairs,we obtain the co-ordinates of vertices of \triangle ABC are 
A\left ( 0,1 \right ),B\left ( 4,13 \right ) and\: C\left ( 4,9 \right )
Then area of \triangle ABC= Area\: \left ( 0hBAO \right )-Area\: \left ( 0hCAO \right )
\Rightarrow \int_{0}^{4}\left ( 3x+1 \right )dx-\int_{0}^{4}\left ( 2x+1 \right )dx\Rightarrow \int_{0}^{4}\left ( 3x+1-2x-1 \right )dx
\Rightarrow \int_{0}^{4}xdx\Rightarrow \left [ \frac{x^{2}}{2} \right ]\Rightarrow \frac{1}{2}\times 4\times 4= 8square\: units

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