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Q : 5         Using integration find the area of the triangular region whose sides have the equations  \small y=2x+1,y=3x+1  and  \small x=4.
 

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The equations of sides of the triangle are y=2x+1, y =3x+1,\ and\ x=4.

ON solving these equations, we will get the vertices of the triangle as A(0,1),B(4,13),\ and\ C(4,9)

Thus it can be seen that,

Area (\triangle ACB) = Area (OLBAO) -Area (OLCAO)

= \int_0^4 (3x+1)dx -\int_0^4(2x+1)dx

= \left [ \frac{3x^2}{2}+x \right ]_0^4 - \left [ \frac{2x^2}{2}+x \right ]_0^4

=(24+4) - (16+4) = 28-20 =8units.

Posted by

Divya Prakash Singh

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