Mathematics - Area Under Curves Question with Solution | TestHub

The given curve is $y=\sqrt{16-x^2}$, which represents the upper half of the circle $x^2+y^2=16$ with center $(0,0)$ and radius $4$.

The region bounded by the curve, $x=0$, $x=4$, and above the X-axis lies entirely in the first quadrant.

The required area is given by the integral:

$A={\int }_0^4\sqrt{16-x^2}dx$

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

$A={\left[\frac{x}{2}\sqrt{16-x^2}+8{\sin }^{-1}\left(\frac{x}{4}\right)\right]}_0^4$

$A=\left(0+8{\sin }^{-1}(1)\right)-\left(0+8{\sin }^{-1}(0)\right)$

$A=8\times \frac{\pi }{2}=4\pi$ sq. units.

Answer: $4\pi$