Mathematics - Area Under Curves Question with Solution | TestHub

Given curve is $y^2=4\left(x-2\right)$ and line $y=2x-8$

Finding the point of intersection of $y=2x-4$ and $y^2=4x$.

$\Rightarrow x=\frac{y+8}{2}=\frac{{\displaystyle y^2+8}}{4}$

$\Rightarrow 2y+16=y^2+8$

$\Rightarrow y^2-2y-8=0$

$\Rightarrow \left(y-4\right)\left(y+2\right)=0$

$\Rightarrow y=4,-2$

$\Rightarrow x=6,3$

Area of the bounded region is given by, $A={\int }_{-2}^4\left(\frac{y+8}{2}-\left(\frac{y^2}{4}+2\right)\right)dy$

$\Rightarrow A={\left[\frac{y^2}{4}+2y-\frac{y^3}{12}\right]}_{-2}^4$

$\Rightarrow A=4+8-\frac{16}{3}-1+4-\frac{2}{3}$

$\Rightarrow A=9 \mathrm{sq}. \mathrm{units}$.