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

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

Solution


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

Asked in: MHT CET 2024 (15 May Shift 1)

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