JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Area Under Curves

The region is bounded below by $y=|x-1|$ and above by the circle $x^2+y^2=5$. The curves intersect when $(x-1{)}^2=5-x^2$, giving $x^2-x-2=0$, so $x=2$ or $x=-1$.

The region consists of triangle ABC (vertices at (1,0), (2,1), (-1,2)) plus the circular segment BCD. The triangle has area $\frac{1}{2}|1(1-2)+2(2-0)+(-1)(0-1)|=\frac{1}{2}|-1+4+1|=2$.

The circular segment from $x=-1$ to $x=2$ on the circle $x^2+y^2=5$ (upper half) has area equal to a quarter circle minus a triangle: $\frac{\pi (\sqrt{5}{)}^2}{4}-\frac{1}{2}(\sqrt{5}{)}^2=\frac{5\pi }{4}-\frac{5}{2}$.

Total area = $2+\frac{5\pi }{4}-\frac{5}{2}=\frac{5\pi }{4}-\frac{1}{2}$.