The area of the region enclosed by the curves $y=x^2-4 x+4$ and $y^2=16-8 x$ is :

Question

Aicharya · Accepted Answer

$\begin{aligned} 	ext { Area } & =\int_0^2\left(\sqrt{16-8 x}-\left(x^2-4 x+4ight)ight) d x \ & \left.=\frac{-(16-8 x)^{3 / 2}}{12}-\frac{x^3}{3}+2 x^2+4 xight]_0^2 \ & =\frac{8}{3}\end{aligned}$

The area of the region enclosed by the curves $y=x^2-4 x+4$ and $y^2=16-8 x$ is :

Solution