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

Question

Aicharya · Accepted Answer

$\begin{aligned} 	ext { Required area } & =\cdot \int_0^3[4 x+1-(3 x+1)] \mathrm{d} x \ & =\int_0^3 x \mathrm{~d} x \ & =\left[\frac{x^2}{2}ight]_0^3 \ & =\frac{9}{2} 	ext { sq. units }\end{aligned}$

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

Solution