Can you simply replace an integral with a summation?

[SFNQuestions]

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.

[Xerxes]

The answer is yes, under certain circumstances.

 

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

 

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

 

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

 

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

 

[math]\int\nolimits_a^b f(x)\,dx[/math]

[Country Boy]

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

[SFNQuestions]

The answer is yes, under certain circumstances.

 

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

 

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

 

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

 

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

 

[math]\int\nolimits_a^b f(x)\,dx[/math]

That's just the details of the formal definition of an integral, that doesn't really answer anything.

 

 

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

No I mean flat out swap, as in the only thing that changes is the summation sign turning into an integral sign and vice versa. The only time I've seen it is when you have the integral of a sum, and then you can switch them, but that's not what I am referring to.

Edited February 25, 2017 by SFNQuestions

[zztop]

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.

Yes, you can, this is used routinely for calculating series limits.