Question

Find the area cut off from the parabola $4y={3x}^2$ by the straight line $2y=3x+12$.

• A. $25sq.units$
• B. $27sq.units$
• C. $36sq.units$
• D. $16sq.units$

Answer 1

The correct option is B $27sq.units$

Are enclosed by $4y=3x^2$ & $2y=3x+12$

$\to$ first find Interssection point.

$4y=3x^2$

$y=\frac{3x^2}{4}$

Put value of y in equation

$2\times \frac{3x^2}{4}=3x+12$

$=\frac{3x^2}{4}=3x+12$

$=3x^2=6x+24$

$=x^2=2x+8$

$=x^2-2x-8=0$

$=(x-4)(x+2)=0$

$x=4$ or $x=-2$

$x=4\to y=12$

$x=-2\to y=3$

$\to$ Draw enclosed by the curve

$A={\int }_{-2}^4\frac{(3x+12)}{2}-\left(\frac{3x^2}{4}\right)\cdot dx$

$={\int }_{-2}^4(\frac{3x}{2}+6-\frac{3x^2}{4})dx$

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

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

$=[12+24-16]-[3-12+2]$

$=20(-7)$

$=20+7$

$A=27c{m}^2$