JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Area Under Curves

Sketch the region bounded by the parabola $y^2=4x$ and the line $x+y=1$ in the first quadrant, and integrate along $y$, where the left edge is $x=\frac{y^2}{4}$ and the right edge is $x=1-y$. They meet where $\frac{y^2}{4}=1-y$, i.e. $y^2+4y-4=0$, giving the positive root $y=2\sqrt{2}-2$. The area is ${\int }_0^{2\sqrt{2}-2}\left[(1-y)-\frac{y^2}{4}\right]dy={\left[y-\frac{y^2}{2}-\frac{y^3}{12}\right]}_0^{2\sqrt{2}-2}=\frac{8\sqrt{2}-10}{3}=\frac{8}{3}\sqrt{2}-\frac{10}{3}$. Matching $a\sqrt{2}+b$ gives $a=\frac{8}{3}$ and $b=-\frac{10}{3}$, so $a-b=\frac{8}{3}+\frac{10}{3}=6$.