(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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(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 $x=2$.$\text{Required area} = \int_0^2 [(4x+1) - (3x+1)] \, dx$$= \int_0^2 x \, dx$$= \left[ \frac{x^2}{2} \right]_0^2$$= \frac{2^2}{2} - \frac{0^2}{2} = \frac{4}{2} = 2 \text{ sq. units}$.

Class 12 Mathematics Application of Integration Area bounded by region of single curve

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

MHT CET 2024 All MHT CET

A
$1 \text{ sq. units}$
B
$2 \text{ sq. units}$
C
$3 \text{ sq. units}$
D
$4 \text{ sq. units}$

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Application of Integration

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Area bounded by region of single curve

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