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

The curve $y = 2\sqrt{1 - x^2}$ can be rewritten as:

$y^2 = 4(1 - x^2)$

$x^2 + \frac{y^2}{4} = 1$

This is the upper half of an ellipse. For $x \in [0, 1]$, the region is in the first quadrant.

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The area is given by:

$A = \int_0^1 2\sqrt{1 - x^2} \, dx$

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For integrals containing $\sqrt{1 - x^2}$, use the substitution $x = \sin\theta$.

$x = \sin\theta$

$dx = \cos\theta \, d\theta$

When $x = 0$: $\theta = 0$

When $x = 1$: $\theta = \frac{\pi}{2}$

$\sqrt{1 - x^2} = \sqrt{1 - \sin^2\theta} = \cos\theta$

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$A = \int_0^{\pi/2} 2\cos\theta \cdot \cos\theta \, d\theta$

$A = 2\int_0^{\pi/2} \cos^2\theta \, d\theta$

Using the identity $\cos^2\theta = \frac{1 + \cos(2\theta)}{2}$:

$A = 2\int_0^{\pi/2} \frac{1 + \cos(2\theta)}{2} \, d\theta$

$A = \int_0^{\pi/2} (1 + \cos(2\theta)) \, d\theta$

$A = \left[\theta + \frac{\sin(2\theta)}{2}\right]_0^{\pi/2}$

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At $\theta = \frac{\pi}{2}$:

$\frac{\pi}{2} + \frac{\sin(\pi)}{2} = \frac{\pi}{2} + 0 = \frac{\pi}{2}$

At $\theta = 0$:

$0 + 0 = 0$

Therefore:

$A = \frac{\pi}{2}$