If the area (in sq. units) of the region x ,   y :   y 2 ≤ 4 x ,   x + y ≤ 1 ,…

The given equations are,

$y^2=4x \& x+y=1$

$\Rightarrow y^2=4\left(1-y\right)$

$\Rightarrow y^2+4y-4=0 \Rightarrow y=2\sqrt{2}-2$

$\therefore$ Required Area $=\underset{0}{\overset{2\sqrt{2}-2}{\int }}\left[\left(1-y\right)-\frac{y^2}{4}\right]dx={\left(y-\frac{y^2}{2}-\frac{y^3}{12}\right)}_0^{2\sqrt{2}-2}$

$=\left[\left(2\sqrt{2}-2\right)-\frac{{\left(2\sqrt{2}-2\right)}^2}{2}-\frac{{\left(2\sqrt{2}-2\right)}^3}{12}-0\right]$

$=\left(2\sqrt{2}-2\right)\left[1-\left(\sqrt{2}-1\right)-\frac{{\left(\sqrt{2}-1\right)}^2}{3}\right]=-\frac{10}{3}+\frac{8\sqrt{2}}{3}$

$\Rightarrow -\frac{10}{3}+\frac{8\sqrt{2}}{3}=a\sqrt{2}+b \Rightarrow a=\frac{8}{3}, b=-\frac{10}{3}$
$\therefore a-b=\frac{8}{3}+\frac{10}{3}=6$