Question

The area of the region included between the parabola $y=\frac{3x^2}{4}$ and the line 3x – 2y + 12 = 0 is

• A. $\frac{4}{3}$ sq.unit
• B. $\frac{2}{3}$ sq.unit
• C. $\frac{21}{2}$ sq.unit
• D. 27 sq. unit

Answer 1

The correct option is D 27 sq. unit



$y=\frac{3x^2}{4},3x–2y+12=0\Rightarrow 3x–\frac{3x^2}{4}+12=0$
$\Rightarrow 6x–3x^2+24=0\Rightarrow x^2–2x–8=0$
$\Rightarrow (x–4)(x+2)=0\Rightarrow x=–2,4$
Area = ${\int }_{-2}^4(\frac{3x+12}{2}-\frac{3x^2}{4})dx={[\frac{3x^2}{4}+6x-\frac{x^3}{4}]}_{-2}^4$
$=12+24–16–3+12–2=27$