Question

The area bounded by $y=\frac{3x^2}{4}$ and the line $3x-2y+12=0$ is:

• A. $9$
• B. $18$
• C. $27$
• D. None of these

Answer 1

The correct option is C $27$

Solving both the equation of the curves gives us

$3x-\frac{3x^2}{2}+12=0$
$6x-3x^2+24=0$
$2x-x^2+8=0$
$x^2-2x-8=0$
$(x-4)(x+2)=0$

$x=4$ and $x=-2$

Now the area occupied will be equal to

${\int }_{-2}^4{y}_2-y_1.dx$

$={\int }_{-2}^4\frac{3x+12}{2}-\frac{3x^2}{4}.dx$

$=[\frac{3x^2}{4}+6x-\frac{x^3}{4}{]}_{-2}^4$

$=\left[\right(12+24-16)-(3-12+2\left)\right]$

$=20+7$
$=27$