The area (in sq. units) of the region bounded | vedclass

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(C) The curves $y=3x+1$ and $y=4x+1$ intersect at the point where $3x+1 = 4x+1$,which gives $x=0$. Given the boundary $x=3$,the region is bounded betw

Vedclass · Accepted Answer

$\frac{9}{2}$

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(C) The curves $y=3x+1$ and $y=4x+1$ intersect at the point where $3x+1 = 4x+1$,which gives $x=0$. Given the boundary $x=3$,the region is bounded between $x=0$ and $x=3$. In this interval,$4x+1 \geq 3x+1$. Therefore,the required area is given by the integral: $	ext{Area} = \int_{0}^{3} [(4x+1) - (3x+1)] \, dx$ $= \int_{0}^{3} x \, dx$ $= \left[ \frac{x^2}{2} ight]_{0}^{3}$ $= \frac{3^2}{2} - \frac{0^2}{2} = \frac{9}{2} 	ext{ sq. units}$.

The area (in sq. units) of the region bounded by the curves $y=3x+1$,$y=4x+1$ and the line $x=3$ is:

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