The area (in sq. units) of the region enclosed between the parabola y 2 = 2 x and the line x + y = 4 is…

$y^2=2x \Rightarrow y^2=2\left(4-y\right)$

$\Rightarrow y^2+2y-8=0 \Rightarrow y=-4,2$

The point of intersection of $y^2=2x$ and $x+y=4$ are $\left(8,-4\right)$ and $\left(2,2\right)$

Area $={\int }_{-4}^2\left(\left(4-y\right)-\frac{y^2}{2}\right)dy ={\left(4y-\frac{y^2}{2}-\frac{y^3}{6}\right)}_{-4}^2$

$=\left(8-2-\frac{4}{3}\right)-\left(-16-8+\frac{32}{3}\right)$

$=\frac{54}{3}=18$ Sq.units.