The area of the region bounded by the curves | vedclass

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(B) The curves $y=3x+1$ and $y=4x+1$ intersect when $3x+1 = 4x+1$,which implies $x=0$.Thus,the region is bounded by $x=0$ and $x=2$.The required area

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$2 	ext{ sq. units}$

Vedclass · Answer

(B) The curves $y=3x+1$ and $y=4x+1$ intersect when $3x+1 = 4x+1$,which implies $x=0$.Thus,the region is bounded by $x=0$ and $x=2$.The required area is given by the integral of the upper curve minus the lower curve from $x=0$ to $x=2$.$	ext{Required area} = \int_0^2 [(4x+1) - (3x+1)] \, dx$$= \int_0^2 x \, dx$$= \left[ \frac{x^2}{2} ight]_0^2$$= \frac{2^2}{2} - \frac{0^2}{2} = \frac{4}{2} = 2 	ext{ sq. units}$.

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

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