The area of the region bounded by the circle x^2 + y^2 = 1 is

The correct answer is Option (2) → $\pi$ sq units

We have, $x^2 + y^2 = 1^2$

$\Rightarrow y^2 = 1 - x^2 \Rightarrow y = \sqrt{1 - x^2} \quad [∵r = \pm 1]$

In the given circle, area in each quadrant is same.

Therefore area of circle $= 4 \times \text{Area in first quadrant}$

$∴\text{Area enclosed by circle} = 4 \int_{0}^{1} \sqrt{1 - x^2} \, dx = 4 \int_{0}^{1} \sqrt{1^2 - x^2} \, dx$

$= 4 \left[ \frac{x}{2} \sqrt{1^2 - x^2} + \frac{1^2}{2} \sin^{-1} \frac{x}{1} \right]_{0}^{1}$

$= 4 \left[ \frac{1}{2} \cdot 0 + \frac{1}{2} \cdot \frac{\pi}{2} - 0 - \frac{1}{2} \cdot 0 \right]$

$= 4 \cdot \frac{\pi}{4} = \pi \text{ sq. units}$