Integrating Powers without much Calculus

You can derive all of the integrals $\int_0^1 x^p dx$ by chopping them in half, and rescaling each half to fit in [0,1] again.
The proof is by induction on p, and by "recovering the integral back". To give you an idea of how this will go, I will just derive $\int_0^1 x^2 dx$ assuming we already have the results for p=0,1 (which are apparent geometrically anyway).

$$\int_0^1 x^2 \textrm{ d}x = \int_0^\frac{1}{2} x^2 \textrm{ d}x + \int_\frac{1}{2}^1 x^2 \textrm{ d}x$$

$$= \frac{1}{2}\int_0^1 (\frac{x}{2})^2 \textrm{ d}x + \frac{1}{2}\int_0^1 (\frac{x+1}{2})^2 \textrm{ d}x$$

$$= \frac{1}{8}\int_0^1 x^2 \textrm{ d}x + \frac{1}{8}\int_0^1 x^2 +2x+1 \textrm{ d}x$$

so $$6 \int_0^1 x^2 \textrm{ d}x = \int_0^1 2x+1 \textrm{ d}x$$

$$6 \int_0^1 x^2 \textrm{ d}x = 2$$

$$\int_0^1 x^2 \textrm{ d}x = \frac{1}{3}$$

A similar argument applies to other values of $p$, by induction. It is a little messy though.

Note that all I am using is linearity of the integral, and that scaling a shape by a factor in one direction should scale the area by that same factor.

I did these computations while reading Tom Leinster's note here https://www.maths.ed.ac.uk/~tl/glasgowpssl/banach.pdf to see if I could actually compute any integrals using his characterization.

This may not be in the spirit of what you want, but... by scaling arguments it suffices to establish that $\int_0^1 x^p dx = \frac{1}{p + 1}$. Consider the following probabilistic argument (not entirely rigorous but very suggestive): the integral describes the probability that if you choose $p + 1$ points uniformly at random in the interval $[0, 1]$, then the first point you chose is the rightmost. (If the first point you chose is $x$, then the probability that each of the remaining $p$ points is to the left of $x$ is $x^p$.)

On the other hand, you can choose the points simultaneously and then decide which one was the first point you chose. You'll end up choosing the rightmost point with probability $\frac{1}{p + 1}$.

A rigorous version of this argument proceeds by partitioning $[0, 1]^{p+1}$ into $p + 1$ parts of the same measure depending on which of the coordinates is the largest and then observing that the measure of one of these parts can be expressed using the above integral. I admit I cannot readily visualize these parts even for $p = 2$...

Here is a very simple proof for nonnegative integer $p$.

By elementary combinatorial reasoning, we have $$ \sum_{j=0}^{n-1} \binom{j}{p} = \binom{n}{p+1}, $$ which is the same as $$ \sum_{j=0}^{n-1} j(j-1)\cdots(j-p+1) = \frac{n(n-1)\cdots(n-p)}{p+1}.$$ After scaling that becomes a lower bound for $\int_0^1 x^p dx$.

Similarly, $$ \sum_{j=0}^{n-1} j(j+1)\cdots(j+p-1) = \frac{(n-1)(n)\cdots(n+p-1)}{p+1},$$ which scales to an upper bound.

Now let $n\to\infty$.