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Can you simply replace an integral with a summation?


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#1 SFNQuestions

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Posted 14 February 2017 - 01:03 AM

It doesn't seem right, but I could have sworn I remember seeing instances in statistics where a summation was literally just swapped with an integral sign without anything else changing, except adding a dx. Can you really just swap a sum and an integral? I mean, the sum of a linear variable k is a second degree polynomail, and the integral of a continuous variable k would be a 2nd degree polynomial, so maybe there's something to it.


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#2 Xerxes

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Posted 14 February 2017 - 06:43 PM

The answer is yes, under certain circumstances.

The conventional way to define the Riemann definite intgral of a functionf(x) over a close interval [a,b] is to divide this interval into a number of non-overlapping interval

[x_0,x_1),[x_1,x_2),....,[x_k,x_{k+1}),....,[x_{n-1},x_n] where a \equiv x_0 <x_1 <.....<x_n \equiv b.

You form the so-called Riemann sum \sum\nolimits_{k=0}^{n-1} f(\xi_k)(x_{k+1}-x_k) where \xi_k denotes a point in the interval [x_k,x_{k+1}).

Now you let the number of intervals increase without bound, so that x_{k+1}-x_k \to 0, then provided the limit of the Riemann sum exists, then this goes over to the integral

\int\nolimits_a^b f(x)\,dx
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#3 HallsofIvy

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Posted 17 February 2017 - 12:13 PM

You can ​approximate  ​an integral by a Riemann sum.  Is that what you mean by "replace"?


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