Correct Answer - Option 2 : 8π square units.
Concept:
- The area under the function y = f(x) from x = a to x = b and the x-axis is given by the definite integral \(\rm \left|\int_a^b f(x)\ dx\right|\), for curves which are entirely on the same side of the x-axis in the given range.
- If the curves are on both the sides of the x-axis, then we calculate the areas of both the sides separately and add them.
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Definite integral: If ∫ f(x) dx = g(x) + C, then \(\rm \int_a^b f(x)\ dx = [ g(x)]_a^b\) = g(b) - g(a).
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\(\rm \int {\sqrt{a^2-x^2}}\ dx = \frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C\).
Calculation:
Let's first find the points where the curve meets the x-axis (y = 0).
⇒ \(\rm y=\sqrt{16-x^2}=0\)
⇒ x = ±4
Now, since the curve \(\rm y=\sqrt{16-x^2}\) is entirely on one side of the x-axis in the given range x = -4 to x = 4, we have:
The required area = \(\rm \int_{-4}^{\ \ 4} \sqrt{4^2-x^2}\ dx\).
= \(\rm \left[\frac{x}{2}\sqrt{4^2-x^2}+\frac{4^2}{2}\sin^{-1}\frac{x}{4}\right]_{-4}^{\ \ \ 4}\)
= \(\rm \left[\frac{4}{2}\sqrt{4^2-4^2}+\frac{4^2}{2}\sin^{-1}\frac{4}{4}\right] - \left[\frac{-4}{2}\sqrt{4^2-(-4)^2}+\frac{4^2}{2}\sin^{-1}\frac{-4}{4}\right]\)
= \(\rm 8\frac{\pi}{2}+8\frac{\pi}{2}\)
= 8π square units.
