Question

${\int }_0^4\sqrt{16-x^2}{d}_X=$

• A. $\frac{\pi }{4}$
• B. $\pi$
• C. $16\pi$
• D. $4\pi$

Answer 1

Answer: (D)

$4\pi$

${\int }_0^4\sqrt{16-x^2}dx$
$\int \sqrt{16-x^2}dx=\int \sqrt{4^2-x^2}dx$
$=\frac{x}{2}\sqrt{16-x^2}+\frac{16}{2}si{n}^{-1}\left(\frac{x}{14}\right)+c$
$={\int }_0^4\sqrt{16-x^2}dx[\frac{x}{2}\sqrt{16-x^2}+\frac{16}{2}si{n}^{-1}\left(\frac{x}{4}\right)+c]{\int }_0^4$
$=(\frac{4}{2}\sqrt{16-16}+8si{n}^{-1}\left(\frac{4}{4}\right)+c)-(0+0+c)$
$=8\left(\frac{\pi }{2}\right)=4\pi$
${\int }_0^4\sqrt{16-x^2}dx=4\pi$