The area (in sq. units) of the region, given by the set x , y ∈ R × R ∣ x ≥ 0 , 2 x 2…

Given curves are $y=2x^2 \& y=4-2x$

So, $2x^2=4-2x$

$\Rightarrow 2x^2+2x-4=0\Rightarrow \left(x-1\right)\left(x+2\right)=0$

So, the $x$-coordinate of the intersection point is $1$ as $x=-2<0$

Required area ${\int }_0^1\left(4-2x-2x^2\right)dx={\overline{)\left[4x-x^2-\frac{2x^3}{3}\right]}}_0^1$

$=4-1-\frac{2}{3}=\frac{7}{3}$