$\int_0^{a} x^\frac{1}{n}dx$ without antiderivative for $n>0$ My exercise is to find $\int_0^{a} x^\frac{1}{n}dx$ without antiderivatives for $n>0$. 
The first thing I did is plot of some of the $x^\frac{1}{n}$ for the first twenty n. This is what I got.

I wanted to make sure that the functions were non-negative and monotonic and I am sure they are now. So I know the question makes sense. I know the integrals of $x^{1}$ and $x^\frac{1}{2}$ and so I tried to find a pattern and prove by induction but didn't get far. Am I on the right track or is there a better way?
 A: As you know, each power function $f_{n}(x) = x^{1/n}$ is monotone on $[0, \infty)$, hence integrable on every interval $[a, b] \subset [0, \infty)$. If you insist on evaluating the integral
$$
\int_{0}^{b} x^{1/n}\, dx
\tag{1}
$$
using limits of Riemann sums, it's probably easiest not to use equal-length partitions of $[0, b]$, but instead to fix $0 < a < b$ and use geometric partitions
$$
x_{i} = a\left(\frac{b}{a}\right)^{\frac{i}{N}},\ i = 0, 1, \dots, N,
$$
of $[a, b]$, for which the left- and right-hand Riemann sums are finite geometric series. The left-hand sum with $N$ subintervals, for example, is

 \begin{align*}\sum_{i=0}^{N-1} f(x_{i}) (x_{i+1} - x_{i}) &= \sum_{i=0}^{N-1} \left[a\left(\frac{b}{a}\right)^{\frac{i}{N}}\right]^{\frac{1}{n}} \left[a\left(\frac{b}{a}\right)^{\frac{i+1}{N}} - a\left(\frac{b}{a}\right)^{\frac{i}{N}}\right] \\ &= \dots \\ &= a^{\frac{n+1}{n}} \left[\left(\frac{b}{a}\right)^{\frac{1}{N}} - 1\right] \sum_{i=0}^{N-1} \left[\left(\frac{b}{a}\right)^{\frac{1}{N}\left(\frac{n+1}{n}\right)}\right]^{i}.\end{align*}

(You'll still have to evaluate an indeterminate limit as $N \to \infty$, perhaps using l'Hôpital's rule. There's no free lunch.)
Once you have the formula for
$$
\int_{a}^{b} x^{1/n}\, dx,
$$
use continuity of the integral to take the limit as $a \to 0^{+}$ to obtain (1).
A: Okay, here's a solution without antiderivatives(*).  Take the substitution $u=x^{(1+n)/n}$, $du=\frac{1+n}{n}x^{1/n}dx$.  Hence $$\int_0^ax^{1/n}dx=\int_0^{a^{(1+n)/n}}\frac{n}{n+1}du=\frac{n}{n+1}\int_0^{a^{(1+n)/n}}1~du$$
Now, integrate that last using Riemann sums (or just take the exact answer, since it's a horizontal line), getting $\frac{n}{n+1}a^{(1+n)/n}$.
(*) Assuming you can prove the substitution rule without using the fundamental theorem of calculus.
