The area (in sq. units) of the region in the first quadrant bounded by $y = 3\sqrt{1-x^2},x ∈ [0,1]$ and the x-axis is equal to

$\frac{3π}{4}$

The correct answer is Option (2) → $\frac{3π}{4}$

Given: The curve is $y = 3\sqrt{1 - x^2}$, with $x \in [0, 1]$.

Required: Area bounded by the curve and x-axis in the first quadrant.

Formula: Area = $\int_0^1 3\sqrt{1 - x^2} \, dx$

This is a standard integral: $\int \sqrt{1 - x^2} \, dx = \frac{1}{2} \left( x \sqrt{1 - x^2} + \sin^{-1}x \right)$

So, $\int_0^1 3\sqrt{1 - x^2} \, dx = 3 \int_0^1 \sqrt{1 - x^2} \, dx = 3 \cdot \frac{\pi}{4} = \frac{3\pi}{4}$