Question

Prove that:
${\int }_0^4\sqrt{16-x^2}dx$

Answer 1

${\int }_0^4\sqrt{16-x^2}dx={\int }_0^4\sqrt{4^2-x^2}dx=[\frac{x\sqrt{16-x^2}}{2}+\frac{4^2}{2}{\text{sin}}^{-1}(\frac{x}{4}){]}_0^4=[0+8{\text{sin}}^{-1}(1)-0-8{\text{sin}}^{-1}(0\left)\right]=8\times \frac{\pi }{2}=4\pi$