JEE Main 2014 · 6 Apr · Shift 1, Mathematics question 76

The parabola $y^2=1-x$ meets the circle where $x^2=x$, i.e. at $x=0$ and $x=1$.

• For $0\le x\le1$, $1-x\le1-x^2$, so the parabola is the binding curve.
• For $x<0$, the circle is.

$$\text{Area}=\underbrace{\frac\pi2}_{\text{left half-disc}}+\int_0^12\sqrt{1-x}\,dx=\frac\pi2+\frac43.$$