Riemann Sums infinite strips

[grant369]

Use the riemann sums model to estimate the area under the curve f(x) = x^2, between x =2 and x = 10, using an infinite number of strips. Be sure to include appropriate diagriams and full explanation of the method of obtaining all numerical values, full working and justification.

Does anybody have any idea about this one?
thanks guys

 

[hotcommodity]

To use an infinite number of strips to estimate the area, you must first realize that taking the integral of a function is taking the sum of the area of each strip from i=1 to infinity, where i is the strip. Does that make sense so far?

 

[grant369]

i think i follow

 

[Gib Z]

He's basically saying evaluate the integral:

$$\int^{10}_2 x^2 dx = \frac{10^3}{3} - \frac{2^3}{3}$$ In case your wondering.

How ever, the way the question is stated, i think you have to do it this way:

Use a model of riemann sums, say..left hand sums. Make an equation using the sums that calculates the area for a given number of the rectangles. Then take the limit of the number of rectangles to infinity.

 

[HallsofIvy]

Use the riemann sums model to estimate the area under the curve f(x) = x^2, between x =2 and x = 10, using an infinite number of strips. Be sure to include appropriate diagriams and full explanation of the method of obtaining all numerical values, full working and justification.

Does anybody have any idea about this one?
thanks guys

That's very poorly stated. A Riemann sum does NOT have an "infinite number of strips". Any Riemann sum has a finite number of strips- the integral is then the limit as the numbeer of strips goes to infinity. I suspect what is expected is a formula for the Riemann sum for "n" strips, then calculate the limit as n goes to infinity.

Actually, there exist an infinite number of different Riemann sums since you can choose the bases to be any size (as long as they all go to 0 in the limit) and can choose you x^* to be any point within a strip. I would recommend that you divide [2,10] into n equal length strips ( 8/n) and then take x^* within each strip to be the left left edge (2, 2+8/n, 2+ 16/n, ..., up to 10- 8/n). With a little algebra that should reduce to an sum that you know a general formula for. Since x² is increasing within [2, 10], that will give a sum that is always a little less than actual area under the curve. If you choose your x^* to always be the right hand endpoint (2+8/n, 2+ 16/n, ..., up to 10) that would always be larger than the actual area under the curve. You could impress your teacher by doing both and showing that they give the same thing in the limit (which therefore must be the "area under the curve").

 

[grant369]

I understand how you have made equal length strips but i still do not understand how to find the area with infinite strips.
sorry if I am being vague..

 

[HallsofIvy]

In my example, I suggested dividing the interval from 2 to 10 into n intervals each having length (10-2)/n= 8/n and using the left hand endpoint as the x-value, x^*, at which you evaluate the function. That is x^* will take on values of 0, 8/n, 16/n, 24/n, ..., 10- 8/n= (10n-8)/n. f(x*)= x² so each rectangle will have height 0, 64/n², 256/n², ... and thus each rectangle will have area 0, (64/n²)(8/n)= 512/n³, (256/n²)(8/n)= 2048/n³, ... The approximate area under the curve is the sum of those. Since you probably do not want to add up all n areas, you will want to find the formula for the area of the "i^th" rectangle: it has base 8/n and height (8i/n)² with i going from 0 to n-1. You should be able to find a formula for the sum of such a series (do you know the sum of the first n integers, the sum of the first n squares, etc.?) and then take the limit as n goes to infinity.

I'm sure all this is covered in your textbook- review. Also, because this is clearly homework, I am moving this thread to the homework section.

 

[jackson_sun]

This question is similar to a question i am currently undertaken in my course study...could anyone help me with the assumptions that have been assumed in this question (or for any riemann sums with infinite strips for that matter)?
Thankyou
P.S. if anyone has any ideas about the strengths and limitations of this model compared with other models such as the trapezoidal rule, monte carlo, definte integral etc that would also be of great assistance

 

[Gib Z]

The infinite Riemann sums model is what the basic definite integral is defined upon. The trapeziodal rule is just an approximation method, and with the limit to infinity, converge to the same value as a riemann sum.

Search Lebsegue integral for a better one.

 

[Gib Z]

O and assumptions when taking the Riemann Integral..that the integral exists? That the function is continuous over the bounds of the integral.

 

[jackson_sun]

With strengths and limitations I was hoping to discover more about the strengths of using riemann sums over the other methods...and the limitations of the method compared to using the others...
thanks

 

[Gib Z]

Oh ok. Riemann strips are Much easier computations than say, the trapeziodal rule, Even if they arent as accurate. Since, if you take an infinite number of strips, they will add up to the same anyway, its easier to use Riemann strips. Rectangles are easier to work out that rectangles plus triangles :)

 

[HallsofIvy]

I'm not sure what you mean by "strengths and limitations". Are you talking about a numerical integration?I can't imagine anyone using Riemann sums to actually do a numerical integration- use Simpson's rule at least.

But that's not the point of Riemann sums. Riemann sums are used as a way to define the definite integral. That have the nice property that they make it easy to show that the definite integral actually gives the "area under the curve". I also think that knowing the Riemann sum definition of definite integrals helps in setting up the correct integral in applications.

 

[Gib Z]

Numerically Riemann Mid-point sums are quite good. I usually take a weighted average with the trapezoidal rule and Riemann midpoints, as people will know they give the exact areas in some cases and good approximations in others, even with only 3 subintervals.

 

[jackson_sun]

By strengths i mean...what are the advantages of using the riemann sums model over say the trapezoidal rule or monte carlo or definite integral...and limitations the contrary being what are the disadvantages about the model and what is limited in its use

 

[Gib Z]

I have already stated them, the definite integral is, for your knowledge, just Riemann sums with an infinite number of strips! Exactly the same! The definite integral may be easier if you have an elementary derivative, but that is not always the case. Riemann sums with finite strips are easier calculations than the trapezoidal rule, though if used wrong will be less accurate.

 

[jackson_sun]

sorry gib z...i was actually referring to Halls of Ivy...i should have quoted him. Sorry

 

[Gib Z]

Its ok don't be sorry lol.

 

[HallsofIvy]

With strengths and limitations I was hoping to discover more about the strengths of using riemann sums over the other methods...and the limitations of the method compared to using the others...
thanks

It's still not clear to me what your question is! Are you talking about strengths and weaknesses of methods of deciding how to set up an integral or for doing a numerical integral?

If you are talking about doing a numerical calculation of an integral, while a "Riemann sum" (I wouldn't call it that) using rectangles is simpler for each individual piece, Simpson's rule is so much more accurate for the same amount of calculation that I wouldn't consider using anything else.

If you are talking about deciding how to set up an integral for a specific application, simplicity is everything and "accuracy" (since you will be doing the same integral in the end) I would only consider Riemann sums.

The problem I have is that I don't consider "Riemann sums" as anything like the "trapezoid rule" or "Simpson's rule"- they are intended for completely different things.

 

[jackson_sun]

what about compared to the monte carlo method of integration