The area (in sq. units) of the region bounded by \(y = 2\sqrt{1 - x^2}, x \in [0, 1]\) and x-axis is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{\pi}{2}\)
On [0, 1] the curve lies on or above the x-axis, so the area is the definite integral of y:
\(\displaystyle \text{Area} = \int_{0}^{1} 2\sqrt{1 - x^2}\,dx\)
By the standard integral \(\int \sqrt{a^2 - x^2}\,dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C\) with a = 1:
\(\displaystyle = 2\left[\frac{x}{2}\sqrt{1 - x^2} + \frac{1}{2}\sin^{-1}x\right]_{0}^{1}\)
\(\displaystyle = 2\left(0 + \frac{1}{2} \cdot \frac{\pi}{2}\right) - 0 = \frac{\pi}{2}\)
Shortcut: \(y = 2\sqrt{1 - x^2}\) is the upper half of the ellipse \(x^2 + \frac{y^2}{4} = 1\) (a = 1, b = 2). For \(0 \le x \le 1\) the region is a quarter of it, so the area is \(\frac{\pi ab}{4} = \frac{\pi \cdot 1 \cdot 2}{4} = \frac{\pi}{2}\).