Question

Integrate ${\int }_0^r\sqrt{r^2-x^2}dx$

Answer 1

${\int }_0^r\sqrt{r^2-x^2}dx=?$

As we know that,

$\int \sqrt{a^2-x^2}dx=\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}{\sin }^{-1}\left(\frac{x}{a}\right)+C$

Therefpre,

${\int }_0^r\sqrt{r^2-x^2}dx$

$={[\frac{x\sqrt{r^2-x^2}}{2}+\frac{r^2}{2}{\sin }^{-1}\left(\frac{x}{r}\right)+C]}_0^r$

$=[(\frac{r\sqrt{r^2-r^2}}{2}+\frac{r^2}{2}{\sin }^{-1}\left(\frac{r}{r}\right)+C)-(\frac{0\sqrt{r^2-0^2}}{2}+\frac{r^2}{2}{\sin }^{-1}\left(\frac{0}{r}\right)+C)]$

$=\frac{r^2}{2}\left(\frac{\pi }{2}\right)-0$

$=\frac{\pi r^2}{4}$

Hence the required answer is $\frac{\pi r^2}{4}$.