When would we want to use uneven subintervals in a Riemann integral? The formal definition of a Riemann Integral is written such that you can have uneven subintervals and it still works.  Why do we need to generalize to the case of uneven subintervals?  Why not insist our subintervals are always of equal length, and $\Delta x$ is the same for all of them?
Here's the full definition from Wiki:

The Riemann Integral of $f$ equals $s$ if: For all $\epsilon > 0$, there exists a $\delta > 0$ such that for any tagged partition $x_0 \dots x_n$ and $t_0 \dots t_{n-1}$ whose mesh is less than $\delta$ we have
$$\left|  \left(\sum_{i=0}^{n} f(t_i) (x_{i+1}-x_{i})\right) - s \right| < \epsilon$$

Notice that they have to reference the "mesh" in this definition, i.e. the length of the longest subinterval.  Wouldn't we get a simpler definition by just requiring equal subintervals? E.g [my version]

The Riemann Integral of $f$ equals $s$ if: For all $\epsilon > 0$, there exists a $\delta > 0$ such that for any tagged partition $x_0 \dots x_n$ and $t_0 \dots t_{n-1}$ [with equal subintervals] less than $\delta$ we have
$$\left|  \left(\sum_{i=0}^{n} f(x_i)\Delta x_i \right) - s \right| < \epsilon$$

Or is there a use for having different subinterval lengths?
 A: First example
Suppose we wish to calculate $\int_0^1 \sqrt{x} dx$. This is trivial if we already have the fundamental theorem of calculus in our disposal, but let's see what happens if we try to calculate this integral directly from the definition.
If we take a partition with equal intervals, that is, for every $n\in\mathbb{N}$ we take $x_k = \frac{k}{n}$ for $k\in \{0,1,\ldots,n\}$, we get the Riemann sum
$$\sum_{k=1}^n \Delta x_k f(x_k) = \sum_{k=1}^n \frac{1}{n}\sqrt{\frac{k}{n}}=\frac{1}{n\sqrt{n}}\sum_{k=1}^n \sqrt{k}$$
and since there's no formula for the sum of square roots it appears we are stuck.
Since we're bothered by the the square roots, it would be a nice idea to try a partition with unequal intervals given by $x_k = \frac{k^2}{n^2}$, because now the Riemann sum becomes
$$\sum_{k=1}^n \Delta x_k f(x_k) = \sum_{k=1}^n (x_k - {x_{k-1}})\sqrt{\frac{k^2}{n^2}}=\sum_{k=1}^n \left(\frac{k^2}{n^2}-\frac{(k-1)^2}{n^2}\right)\frac{k}{n}=\frac{1}{n^3}\sum_{k=1}^n 2k^2 - k$$
so we got rid of the square roots. By the well known formulas for $\sum k$ and $\sum k^2$ (see Faulhaber's formula) we get that the Riemann sum equals
$$\frac{1}{n^3}\left(2\cdot\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}\right)$$
so that taking the limit as $n\to\infty$ we get $\int_0^1 \sqrt{x} dx=\frac{2}{3}$.

Second example
Suppose we wish to find
$$\lim_{n\to\infty} \frac{1}{n^2}\sum_{k=1}^n (2k-1) \cos\left(\frac{\pi k^2}{2n^2}\right)$$
Noticing that $2k-1 = k^2 - (k-1)^2$, the limit equals
$$\lim_{n\to\infty} \sum_{k=1}^n \frac{k^2 - (k-1)^2}{n^2} \cos\left(\frac{\pi k^2}{2n^2}\right) = \frac{2}{\pi} \lim_{n\to\infty} \sum_{k=1}^n \left(\frac{\pi k^2}{2n^2} - \frac{\pi (k-1)^2}{2n^2}\right)\cos\left(\frac{\pi k^2}{2n^2}\right)$$
Denoting $x_k = \frac{\pi k^2}{2n^2}$ we see that this is
$\frac{2}{\pi} \lim_{n\to\infty} \sum_{k=1}^n \left(x_k - x_{k-1}\right)\cos\left(x_k \right)$. Aside for $\frac{2}{\pi}$ this is the limit of a Riemann sum for $f(x)=\cos x$ on the interval $[0,\frac{\pi}{2}]$, so it equals $\int_0^{\frac{\pi}{2}} \cos x dx$ since we know $f(x)=\cos x$ is integrable. By the fundamental theorem of calculus $\int_0^{\frac{\pi}{2}} \cos(x)dx = 1$, so that the limit of the whole expression equals $\frac{2}{\pi}$.
Without knowing that Riemann sums of integrable functions converge to the same value (the integral) regardless of how we partition the domain of integration, this limit would probably be very hard to calculate.
