Find the area enclosed between the circle \(x^2 + y^2 = 1\) and the line \(x + y = 1\) lying in the first quadrant.

Answer: \(\dfrac{\pi - 2}{4}\) sq. units \(\left(= \dfrac{\pi}{4} - \dfrac{1}{2}\right)\)

[Figure: unit circle centred at the origin and the line \(x + y = 1\) through A(1, 0) and B(0, 1); the required region in the first quadrant lies between the chord AB and the arc AB.]

The line meets the circle at \((1, 0)\) and \((0, 1)\). For \(0 \le x \le 1\), the circle gives \(y = \sqrt{1 - x^2}\) and the line gives \(y = 1 - x\), with the arc above the line.

Required area \(= \displaystyle\int_0^1\sqrt{1 - x^2}\,dx - \int_0^1(1 - x)\,dx\)

\(\displaystyle\int_0^1\sqrt{1 - x^2}\,dx = \left[\frac{x}{2}\sqrt{1 - x^2} + \frac{1}{2}\sin^{-1}x\right]_0^1 = \frac{1}{2}\cdot\frac{\pi}{2} = \frac{\pi}{4}\)

\(\displaystyle\int_0^1(1 - x)\,dx = \left[x - \frac{x^2}{2}\right]_0^1 = \frac{1}{2}\)

\(\therefore\) Area \(= \dfrac{\pi}{4} - \dfrac{1}{2} = \dfrac{\pi - 2}{4}\) sq. units.