Using integration,find the area of the triang | vedclass

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(B) The equations of the sides of the triangle are $y=2x+1$,$y=3x+1$,and $x=4$.On solving these equations,we obtain the vertices of the triangle as $A

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$8 	ext{ sq. units}$

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(B) The equations of the sides of the triangle are $y=2x+1$,$y=3x+1$,and $x=4$.On solving these equations,we obtain the vertices of the triangle as $A(0, 1)$,$B(4, 13)$,and $C(4, 9)$.It can be observed that the area of the triangular region is the area between the two lines $y=3x+1$ and $y=2x+1$ from $x=0$ to $x=4$.$	ext{Area} = \int_{0}^{4} [(3x+1) - (2x+1)] \, dx$$= \int_{0}^{4} (3x + 1 - 2x - 1) \, dx$$= \int_{0}^{4} x \, dx$$= \left[ \frac{x^2}{2} ight]_{0}^{4}$$= \frac{16}{2} - 0$$= 8 	ext{ sq. units}$

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

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