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Let $f(x)$ be a continuous function on the interval $[a,b]$, $(a<b)$ and let: $$a=x_0,x_1,x_2,...,x_{i-1},x_i,...,x_{n-1},x_n=b$$ be an arbitrary (randomly selected) partition of the interval $ [a, b] $ , which divides the interval into $ n $ subintervals (subdivisions). Let

$ c_{1}, c_{2}, c_{3}, $ ... $ , c_{n-2}, c_{n-1}, c_{n} $

be the sampling numbers (or sampling points) selected from the subintervals. That is,

$ c_{1} $ is in $ [x_{0}, x_{1}], c_{2} $ is in $ [x_{1}, x_{2}], c_{3} $ is in $ [x_{2}, x_{3}],..., c_{n-2} $ is in $ [x_{n-3}, x_{n-2}], c_{n-1} $ is in $ [x_{n-2}, x_{n-1}] $ and $ c_{n} $ is in $ [x_{n-1}, x_{n}] $ .

Define the mesh of the partition to be the length of the largest subinterval. That is, let

$ \Delta x_{i} = x_{i} - x_{i-1} \ \ $

for $ i = 1, 2, 3, ..., n $ and define

$ mesh = \displaystyle{ \max_{1 \le i \le n} \{ x_{i} - x_{i-1} \}} $ .

The definite integral of $ f $ on the interval $ [a, b] $ is most generally defined to be

$$ \displaystyle{ \int^{b}_{a} f(x) \, dx} \overset{def}{=} \displaystyle{ \lim_{mesh \to 0} \sum_{i=1}^{n} f(c_{i}) \Delta x_{i} } $$

A special case of the above definition uses $ n $ subintervals of equal length and sampling points chosen to be the right-hand endpoints of the subintervals. Thus, each subinterval has length

$ \ \ \ \ \ \ \ \ \Delta x_{i} = \displaystyle{ b-a \over n } $

for $ i = 1, 2, 3, ..., n $ and the right-hand endpoint formula is

$ \ \ \ \ \ \ \ \ c_{i} = \displaystyle{ a + \Big( { b-a \over n } \Big) i } $

for $ i = 1, 2, 3, ..., n $ . The definite integral of $ f $ on the interval $ [a, b] $ can now be alternatively defined by

$$ \displaystyle{ \int^{b}_{a} f(x) \, dx} \overset{def}{=} \displaystyle{ \lim_{n \to \infty} \sum_{i=1}^{n} f(c_{i}) \Delta x_{i} } $$


The part I don't understand is why the second definition is a special case for $n$ subintervals of equal length. Why does their length matter if $n$ approaches infinity?

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    $\begingroup$ In the first definition, we have the limit of a Riemann sum where the intermediate points are arbitrary within each subinterval and the partition points are not necessarily evenly spaced. This is guaranteed to converge for a Riemann integrable function -- which you have since $f$ is continuous and bounded. This implies the limit in the second definition converges to the same value. $\endgroup$
    – RRL
    Mar 9, 2016 at 18:02
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    $\begingroup$ It can also be shown for that for a continuous function that convergence of the second definition limit implies convergence of the first, but this takes some effort. Look up Cauchy integrable versus Riemann-Darboux integrable. $\endgroup$
    – RRL
    Mar 9, 2016 at 18:04

1 Answer 1

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It's relevant because if the intervals are all of equal length, we then know the length of any given interval. We can then use that to actually calculate $c_i$; with that, we can take the limit of the Riemann sum to calculate the definite integral without having to use the fundamental theorem of calculus.

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  • $\begingroup$ Why doesn't the length of subintervals matter in the first case, when we define the definite integral as a limit when $mesh \to 0$? $\endgroup$
    – A6SE
    Mar 9, 2016 at 17:56
  • $\begingroup$ Because the first case is just concerned with rigorously defining the definite integral as a limit of Riemann sums, it's not trying to give a way to actually calculate the integral. $\endgroup$
    – DylanSp
    Mar 9, 2016 at 18:07

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