The area (in sq. units) of the region bounded by y = 2 √ 1 − x ² , x ∈ [0,1] and x-axis is equal to

The correct option is:
$\frac{\pi }{2}$

Step-by-Step Explanation:

To find the area of the region bounded by the curve $y=2\sqrt{1-x^2}$, the line $x\in [0,1]$, and the $x$-axis, we set up a definite integral representing this area.

The area $A$ under the curve $y\left(x\right)$ from $x=0$ to $x=1$ is given by:
$$A={\int }_0^{1}ydx={\int }_0^{1}2\sqrt{1-x^2}dx$$

We can factor out the constant $2$ from the integral:
$$A=2{\int }_0^1\sqrt{1-x^2}dx$$

Next, we use the standard integral formula for $\int \sqrt{a^2-x^2}dx$ where $a=1$:
$$\int \sqrt{1-x^2}dx=\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}{\sin }^{-1}\left(x\right)$$

Now, applying the limits of integration from $0$ to $1$:
$$A=2{[\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}{\sin }^{-1}(x\left)\right]}_0^1$$

Evaluating this expression at the upper limit $x=1$:
$$\text{Value at 1}=\frac{1}{2}\sqrt{1-1^2}+\frac{1}{2}{\sin }^{-1}\left(1\right)=0+\frac{1}{2}\left(\frac{\pi }{2}\right)=\frac{\pi }{4}$$

Evaluating this expression at the lower limit $x=0$:
$$\text{Value at 0}=\frac{0}{2}\sqrt{1-0^2}+\frac{1}{2}{\sin }^{-1}\left(0\right)=0+0=0$$

Subtracting the lower limit evaluation from the upper limit evaluation, we get:
$$A=2\left(\frac{\pi }{4}-0\right)=2\cdot \frac{\pi }{4}=\frac{\pi }{2}$$

Thus, the area of the bounded region is indeed $\frac{\pi }{2}$ square units.