In my calculus class we have begun talking about integrals. In particular we have begun talking about Reimann sums and how through the limit of a Reimann sum we can integral. But so far all our examples have been using the right endpoint which has simple formulas like ∆x=$\frac{a-b}{n}$ and xi=a+∆x*i. For the right endpoint we get the equation $\int_{a}^{b} f(x)dx = \lim_{\max \Delta x_i \to 0} \sum_{i=1}^{n} f(x_i^*)\Delta x_i$. This is for the right endpoint. For the left and midpoints I've seen

$\int_{a}^{b} f(x)dx = \lim_{\max \Delta x_i \to 0} \sum_{i=1}^{n} f(x_{i-1})\Delta x_i$

$\int_{a}^{b} f(x)dx = \lim_{\max \Delta x_i \to 0} \sum_{i=1}^{n} f(\frac{x_{i-1} + x_i}{2})\Delta x_i$

Where for left endpoints we have $x_i = a + \frac{(b-a)(i-1)}{n}$ and for midpoints we have $x_i = \frac{1}{2}(x_{i-1} + x_i)$

For the right endpoint things seem quite simple at our level. Just break up the sum into multiple, factor things out, and then we usually just up with some variation of i^n where n is usually 1,2,3, or rarely 4. But when you're introducing fractions or i-1 things seem conceptually trickier. How could we work out a definite integral at those points.

As an in class example we simply had to find the definite integral of $\int_{1}^{2} x^2dx$. Through math we worked our way to $\lim_{n \to \infty} \left( \sum_{i=1}^n \left( 1 + \frac{i}{n} \right)^2 \cdot \frac{1}{n} \right)$. This is through the right hand rule. If we had been using the left hand rule would we say

$\lim_{n \to \infty} \left( \sum_{i=1}^n \left( 1 + \frac{i-1}{n} \right)^2 \cdot \frac{1}{n} \right)$? Or something like that. And the midpoint seems even more complex. How would we go about solving that?


1 Answer 1


Since this type of integral (Riemann integral) is defined as a Riemann sum over some partition of the region of interest, it does not matter how we choose the partitions themselves so long as the number of partitions approaches infinity (which is the case by definition). The right-hand, left-hand, and midpoint rules will yield the same value.

As much as this definition is important to understand, you won't be using it extensively in the future for the evaluation of integrals. This is comparable to how we don't usually use the limit definition of a derivative to evaluate derivatives of functions.

That being said, these sorts of integral evaluations make use of many summation rules, so be sure to know them! Here's one way to evaluate the integral you mention in your question using a left-handed sum: $$ \begin{align} \int^2_1x^2\mathrm{d}x=&\lim_{n \to \infty} \left( \sum_{i=1}^n \left( 1 + \frac{i-1}{n} \right)^2 \cdot \frac{1}{n} \right) \\ =&\lim_{n\to\infty}\left(\frac{1}{n}\sum_{i=1}^n\left(1+\frac{2(i-1)}{n}+\frac{(i-1)^2}{n^2}\right)\right) \\ =&\lim_{n\to\infty}\left(\frac{1}{n}\left(\sum_{i=1}^n1+\sum_{i=1}^n\frac{2(i-1)}{n}+\sum_{i=1}^n\frac{(i-1)^2}{n^2}\right)\right) \\ =&\lim_{n\to\infty}\left(\frac{1}{n}\left(n+\frac{1}{n}\sum_{i=1}^n2(i-1)+\frac{1}{n^2}\sum_{i=1}^n(i-1)^2\right)\right) \\ =&\lim_{n\to\infty}\left(\frac{1}{n}\left(n+\frac{n^2-n}{n}+\frac{(n-1)(2n-1)}{6n}\right)\right) \\ =&\lim_{n\to\infty}\left(1+\frac{n^2-n}{n^2}+\frac{(n-1)(2n-1)}{6n^2}\right) \\ =& 1+1+\frac{1}{3} \\ \int^2_1x^2\mathrm{d}x=& \frac{7}{3}. \end{align} $$

To clarify, the closed forms of the sums on line 5 were found using the following well-known formulas: $$ \sum_{i=1}^nn=\frac{n(n+1)}{2} $$ and $$ \sum_{i=1}^nn^2=\frac{n(n+1)(2n+1)}{6}. $$

Does this clarify things? Let me know if it doesn't or if anything seems incorrect!

  • $\begingroup$ "so long as the number of partitions approaches infinity" technically the requirement is that the width of the largest partition has to approach zero, but having the number of them grow infinitely is a result of that. $\endgroup$
    – ConMan
    Commented Mar 16, 2023 at 3:40
  • $\begingroup$ @Scene I do believe that does clarify things quite a bit. You said that "The right-hand, left-hand, and midpoint rules will yield the same value." As I can see from this where you got the same 7/3 that my prof got using the right hand rule. I assume that midpoint would yield the same value. So is that why all the examples I've seen so far of this process use right hand rule? Because it's a bit easier and yields the same result anyways? $\endgroup$
    – ATR2400
    Commented Mar 16, 2023 at 3:56
  • $\begingroup$ @ATR2400 that's right! Sometimes it may be better to choose one partition over the other so that the algebra is simpler. $\endgroup$
    – Scene
    Commented Mar 16, 2023 at 4:24

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