Question

Give a geometric explanation of why

${\int }_0^{\frac{2\pi }{n}}{r}^{2}d\theta =\pi r^2$

for any positive real number r and any positive integer $n$Would the equation also hold for non-integer values of $n$?

Short answer

The equation${\int }_0^{\frac{2\pi }{n}}{r}^{2}d\theta =\pi r^2$ holds true for any non-integer values of$n$

Step-by-step solution

Given information

Now consider integral${\int }_0^{\frac{2\pi }{n}}{r}^{2}d\theta$

The objective is to find to prove that the integral is equal to πr2 

$\pi r^2$for non-integer values of $n$

Take the integral now.

$\frac{n}{2}{\int }_0^{\frac{2\pi }{n}}{r}^{2}d\theta =\frac{n{r}^2}{2}{\int }_0^{\frac{2\pi }{n}}1d\theta$

Prove that any positive real number r and any positive integer  Would the equation also hold for non-integer values of n?

If $r$is a positive real number and $n$is a non-negative integer, then That is, both $n$and $r$are constants. As a result, they are removed, and the remaining value is integrated within the stated limitations.

$$\begin{gathered} {\int }_0^{\frac{2\pi }{n}}{r}^{2}d\theta =\frac{n{r}^2}{2}{\left[\theta \right]}^2 \\ =\frac{n{r}^2}{2}\left[\frac{2\pi }{n}-0\right] \\ =\pi r^2 \end{gathered}$$

Hence, the equation holds true for any non-integer values of$n$