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 $2\left(\frac{3 x^{2}}{4}\right)=3x+12$
Thus, the required area $=\displaystyle \int _{0}^{4} \left(\frac{3 x + 12}{2} - \frac{3 x^{2}}{4}\right) d x$
$=\left[\frac{3}{4} x^{2} + 6 x - \frac{x^{3}}{4}\right]_{0}^{4}$
$=\frac{3}{4}\left(16\right)+24-16=12+8=20$ sq. units