Question

The area bounded by the parabola $4y=3x^2,$ the line $2y=3x+12$ and the $y$ -axis is

• A. $10$ sq. units
• B. $20$ sq. units
• C. $30$ sq. units
• D. $36$ sq. units

Answer 1

Answer: (B)

The intersection point is $A\left(4,12\right)$ by solving $2\left(\frac{3x^2}{4}\right)=3x+12$
Thus, the required area $={\int }_0^4\left(\frac{3x+12}{2}-\frac{3x^2}{4}\right)dx$
$={\left[\frac{3}{4}{x}^2+6x-\frac{x^3}{4}\right]}_0^4$
$=\frac{3}{4}\left(16\right)+24-16=12+8=20$ sq. units