Question

The area bounded between the parabola $y^2=4x$ and the line $y=2x-4$ is equal to

• A. $15sq.units$
• B. $\frac{17}{3}sq.units$
• C. $\frac{19}{3}sq.units$
• D. $9sq.units$

Answer 1

Answer: (D)

The point of intersection of $y^2=4x$ and $y=2x-4$ is

$(2x-4{)}^2=4x$
$\Rightarrow x^2-5x+4=0$
$\Rightarrow (x-1)(x-4)=0$
$\Rightarrow x=1,4$
$\Rightarrow y=-2,4$
$\therefore$ Required area
$={\int }_{-2}^4\left(\frac{y+4}{2}\right)dy-{\int }_{-2}^4\frac{y^2}{4}dy$
$=\frac{1}{2}{\left[\frac{y^2}{2}+4y\right]}_{-2}^4-\frac{1}{4}{\left[\frac{y^3}{3}\right]}_{-2}^4$
$=\frac{1}{2}[8+16-(2-8)]-\frac{1}{12}[64+8]$
$=\frac{1}{2}[30]-\frac{1}{12}(72)$
$=15-6=9$ sq unit