If I integrate the function $\int_{0}^a\frac {1}{n}\left ( \frac{r}{a}\right)^n dr$ where $a$ is the radius of a sphere and $r \lt a$ is the 'radius where I integrate', $n \ge 0$. Can I view $a$ as a constant because the radius doesn't change? Or would be the integral more complicated?
- I was calculating an electric force of a charged sphere and I made the integral more simple just for my math-question purposes.
Maybe I can take $a$ as a constant because I am concerned about choosing a disk with the radius (distance from an origin) $r$ and then I integrate it from the origin to the radius $a$ and the integral would look like this:
$\int_{0}^a\frac {1}{n} \left( \frac{r}{a}\right)^n dr = \left(\frac {1}{a}\right)^n \int_{0}^a\frac {1}{n} (r)^n dr = \frac {{a}^{n+1}}{a^n{(n+1)n}}$
- but is weird I am plugging $a$ in the end again.