Can integration be defined using rectangles of equal width?

Question

Typically, the definition of $\int_a^b f(x)\,dx$ is stated along these lines.

Let $a = x_0 < x_1 < \cdots < x_{n-1} < x_n = b$ be a partition of the interval $[a,b]$, for $i=1,\dots,n$ let $\tilde x_i \in [x_{i-1}, x_i]$, and let $\Delta x = \max\limits_{i=1,\dots,n}(x_{i}-x_{i-1})$. Then $$\int_a^b f(x)\,dx := \lim_{\Delta x\to0} \sum_{i=1}^n f(\tilde x_i)\, (x_i - x_{i-1}).$$

The specifics of the definition are not important, however what I notice invariably in all the definitions I've come across is the use of a partition. Is this critical to the definition? Can I instead simplify the definition by considering rectangles of equal width? Surely if $\int_a^b f(x) \, dx$ exists, will be equal to $$\lim_{n\to\infty}\sum_{i=1}^n \frac{b-a}{n}\, f\left(a + \frac{b-a}{n}\,i\right),$$ which does not require the explicit use of partitions.

I am asking this question so that I can present a somewhat more formal definition of an integral to A-level (high school) students so that I can get across the idea that an integral is a limit of a sum (this helps me answer various typical questions that students may have, such as why areas below the $x$-axis contribute negatively to the integral); however I feel that the use of partitions unnecessarily complicates the definition for this purpose.

Answer 1

For continuous functions $f$ the two definitions of the Riemann integral are equivalent. That is true also for many other functions. Other answers here show that if the general Riemann integral exists then your definition converges to the same value.

However, there are cases where your kind of integral converges but the general Riemann definition does not converge. For example, if $f(x)$ is $1$ for rational values of $x$ and $0$ for irrational values, the Riemann integral between $x=0$ and $x=1$ does not converge since the lower sums are all zero and the upper sums are all one. The Lebesgue integral does converge, of course. Your definition converges since all your values of $x_i$ are rational. That limitation does not apply to the general Riemann integral.

This picky detail can probably be skipped at the high school level. Just be careful not to state that your definition always works. Perhaps say that it works for all "reasonable functions" or "real-life functions" or "functions you are likely to encounter."

Answer 2

Yes, you can do that. Note that if you do it, then you are not elliminating partitions from the picture. Instead, you will be dealing only with a specific type of partitions: those in which all intervals have the same length.

A proof of the fact that you can work with those partitions alone only can be found here.

Answer 3

If you call $A$ the are calculated using your second expression, then, the error you make when you calculate $A$ by using $n$ intervals of size $(b-a)/n$ is

$$ {\displaystyle \left\vert \int _{a}^{b}f(x)\,dx-A\right\vert \leq {\frac {M(b-a)^{2}}{2n}}}, $$

where $M$ is the maximum value of $f'(x)$ in $(a,b)$. If this number exists, then the RHS goes to zero when $n\to\infty$. So yes, you can do that