The area of the region bounded by the lines y | vedclass

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(D) The region is bounded by $y=3x+1$ (upper line),$y=2x+1$ (lower line),and $x=4$. The lines intersect at $x=0$,where $y=1$. Thus,the vertices of the

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$8$ sq. unit

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(D) The region is bounded by $y=3x+1$ (upper line),$y=2x+1$ (lower line),and $x=4$. The lines intersect at $x=0$,where $y=1$. Thus,the vertices of the triangle are $(0, 1)$,$(4, 9)$,and $(4, 13)$.The area $A$ can be calculated using integration:$A = \int_{0}^{4} [(3x+1) - (2x+1)] \, dx$$A = \int_{0}^{4} x \, dx$$A = \left[ \frac{x^2}{2} ight]_{0}^{4}$$A = \frac{16}{2} - 0 = 8 	ext{ sq. units}$.Alternatively,using the formula for the area of a triangle with base on the line $x=4$:Base length $= 13 - 9 = 4$.Height $= 4 - 0 = 4$.Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 4 	imes 4 = 8 	ext{ sq. units}$.

The area of the region bounded by the lines $y=2x+1$,$y=3x+1$ and $x=4$ is

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