The area (in sq. units) of the region in the first quadrant bounded by \(y = 3\sqrt{1 - x^2}, x \in [0, 1]\) and the x-axis is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{3\pi}{4}\)
For \(0 \le x \le 1\) the curve lies on or above the x-axis, so the area is the definite integral:
\(\displaystyle \text{Area} = \int_{0}^{1} 3\sqrt{1 - x^2}\,dx\)
Using the standard integral \(\int \sqrt{1 - x^2}\,dx = \frac{x}{2}\sqrt{1 - x^2} + \frac{1}{2}\sin^{-1}x + C\):
\(\displaystyle = 3\left[\frac{x}{2}\sqrt{1 - x^2} + \frac{1}{2}\sin^{-1}x\right]_{0}^{1}\)
\(\displaystyle = 3\left(0 + \frac{1}{2}\cdot\frac{\pi}{2}\right) = \frac{3\pi}{4}\)
Shortcut: Squaring gives \(x^2 + \frac{y^2}{9} = 1\), an ellipse with semi-axes 1 and 3. The region is a quarter of this ellipse, so its area is \(\frac{1}{4}\pi(1)(3) = \frac{3\pi}{4}\).