The area of the region (x, y): x^2 + y^2 ≤ 1 ≤ x + y is

The correct answer is Option (3) → \(\displaystyle\frac{1}{2}\left(\frac{\pi}{2} - 1\right)\) sq. unit

The region is the part of the disc \(x^2 + y^2 \le 1\) that lies on or above the line \(x + y = 1\).

The line meets the circle where

    \(x^2 + (1 - x)^2 = 1\)
    \(2x^2 - 2x = 0\)
    \(x = 0\) or \(x = 1\)

that is, at (0, 1) and (1, 0). For \(0 \le x \le 1\), the region lies between the line \(y = 1 - x\) below and the arc \(y = \sqrt{1 - x^2}\) above:

    \(\displaystyle \text{Area} = \int_{0}^{1} \sqrt{1 - x^2}\,dx - \int_{0}^{1} (1 - x)\,dx\)

Using the standard integral \(\int \sqrt{a^2 - x^2}\,dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C\):

    \(\displaystyle \int_{0}^{1} \sqrt{1 - x^2}\,dx = \left[\frac{x}{2}\sqrt{1 - x^2} + \frac{1}{2}\sin^{-1}x\right]_{0}^{1}\)

    \(\displaystyle = \frac{1}{2}\cdot\frac{\pi}{2} = \frac{\pi}{4}\)

    \(\displaystyle \int_{0}^{1} (1 - x)\,dx = \left[x - \frac{x^2}{2}\right]_{0}^{1} = \frac{1}{2}\)

    \(\displaystyle \text{Area} = \frac{\pi}{4} - \frac{1}{2} = \frac{1}{2}\left(\frac{\pi}{2} - 1\right)\) sq. unit

Shortcut: The region is the quarter disc in the first quadrant minus the triangle with vertices (0, 0), (1, 0) and (0, 1): \(\frac{\pi}{4} - \frac{1}{2}\).