Question

When converting Riemann Sums to Definite Integrals, under what conditions are you allowed to change the limits of integration? I don't mean changing it from 0 to 1 to like 0 to 5, I mean like 0 to 1 to like 1 to 2.

Answer 1

You can always change the limits of integration. The summation index is just a dummy variable.

Answer 2

dummy number*

Answer 3

This assumes you shift all of the terms in the summation by the same amount though.

Answer 4

So, if f(x) is what you're integrating, and f(x) = x^3, then changing the limits wouldn't be the same?

Answer 5

Because I've seen cases where it doesn't equate to the same thing, just trying to find some examples right now.

Answer 6

You can change the limits. Just make sure you shift x^3 byt the same amount.

Answer 7

Here i'll post this.

Answer 8

http://tutorial.math.lamar.edu/Classes/DE/PowerSeries.aspx

Answer 9

Go a bit down to shifting indices.

Answer 10

Hmm, I checked Paul's notes, but not quite sure if Power Series and Riemann sums are quite the same.
But at the example, would this be correct?
\[\lim_{n \rightarrow \infty}\sum_{i = 1}^{n}\frac1n\left(1 + \frac in\right)^{3} = \int_1^2x^3dx\]

Answer 11

No they're not the same thing but the logic is the same.

Answer 12

Or would it be \[\lim_{n \rightarrow \infty}\sum_{i = 1}^{n}\frac1n\left(1 + \frac in\right)^{3} = \int_0^1(x + 1)^3dx\]?

Answer 13

The latter I believe.

Answer 14

Ok, that's what I thought, but there are cases where I've seen the first one used, and I always end up guessing when picking one or the other. How do I know which form to use, specifically the constant?

Answer 15

I'm going to flat out say I can't answer this part. Sorry :/ . No idea.

Answer 16

Alright, no problem :)

Answer 17

Well, unless someone else can give me an answer, I'm going to assume that changing the limits of integration is only when the constant has literally no relationship to the actual variables. Like \(\lim_{n \rightarrow \infty}\sum_{i = 1}^n\frac 1n((\frac in + 1)^3 + 4) = \int_4^5(x + 1)^3dx\)