Infinite riemann sums discrepancy

In summary: You should post the method you used to compute the sum, or perhaps some steps from your simplification, and then we can probably spot the error for you.\lim_{n \to \infty }\sum_{1}^{n}\frac{6}{n}(1+\frac{12i}{n} + \frac{36i^2}{n^2} - 4 - \frac{24i}{n}+2)\lim_{n \to \infty }\sum_{1}^{n}[\frac{6}{n} + \frac{72i}{n^2} + \frac{216i^2}{n^3} - \
  • #1
computerex
68
0
Hello. I have to solve some integrals using both the standard theorem of calculus and infinite Riemann sums.

[itex] \int_{1}^{7} (x^2-4x+2) dx = \lim_{n \to \infty } \sum f(x_i)\Delta x_i = \lim_{n \to \infty } \sum (x_i^2 - 4x_i + 2)6/n [/itex]

Evaluating the definite integral results in an answer of 30 units.. But it seems I have done something wrong in trying to do it with infinite Riemann sums:

step1.jpg

step2.jpg


Taking the limit results in 42, not 30. Note I may have prematurely omitted some terms that will evaluate to 0 after taking the limit to save space.. Can someone tell me what I am doing wrong?
 
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  • #2
Why exactly did you sub in [itex]1+\frac{6i}{n}[/itex] for [itex]x^2[/itex], but not for [itex]x[/itex]?
 
  • #3
I caught that minutes after posting.. But I still am not getting the right answer. Can someone please calculate the integral so that I can compare it with my work to see where I went wrong?
 
  • #5
Screwdriver said:
The value of the integral is 30.

http://www.wolframalpha.com/input/?i=integrate+x^2+-+4x+%2B+2+from+1+to+7

You must have gone wrong somewhere in the computation of the sum:

[tex]\sum_{i=0}^{n} \frac{6}{n}[(1 + \frac{6i}{n})^2 - 4 (1 + \frac{6i}{n}) + 2][/tex]

Thank you for the reply.. But
computerex said:
Evaluating the definite integral results in an answer of 30 units.. But it seems I have done something wrong in trying to do it with infinite Riemann sums

I already know how to evaluate definite integrals, I need to know where my mistake is in computing the sum, which is tedious and error prone.. I already know how to set everything, it is an error in simplification. I would highly appreciate it if someone could post the solution.
 
  • #6
computerex said:
I would highly appreciate it if someone could post the solution.

I'd love to, but that's not the way it works around here. The best I (or anyone else) is allowed to do is guide you towards the solution as per the forum rules. You should post the method you used to compute the sum, or perhaps some steps from your simplification, and then we can probably spot the error for you.
 
  • #7
[tex]\lim_{n \to \infty }\sum_{1}^{n}\frac{6}{n}(1+\frac{12i}{n} + \frac{36i^2}{n^2} - 4 - \frac{24i}{n}+2)[/tex]

[tex]\lim_{n \to \infty }\sum_{1}^{n}[\frac{6}{n} + \frac{72i}{n^2} + \frac{216i^2}{n^3} - \frac{24}{n} - \frac{144i}{n^2} + \frac{12}{n}] = 0 + 72+432-0-144+0=360[/tex]
 
  • #8
Do you see the error? Since we are taking limit to infinity I don't bother evaluating everything, I just get the coefficients..
 
  • #9
Okay, wait a minute. You're expanding this:

[tex]\sum_{i=0}^{n} \frac{6}{n}[(1 + \frac{6i}{n})^2 - 4 (1 + \frac{6i}{n}) + 2][/tex]

It should give you this:

[tex]\sum_{i=0}^{n} [\frac{216 i^2}{n^3} - \frac{72 i}{n^2} - \frac{6}{n}][/tex]

You have to find this sum first before taking the limit. To do this, split it up and factor out constants to get this:

[tex]\frac{216 }{n^3} \sum_{i=0}^{n} [i^2] - \frac{72 }{n^2} \sum_{i=0}^{n} - \frac{6}{n} \sum_{i=0}^{n} [1] [/tex]

Do you know how to find these sums?
 
  • #10
Screwdriver said:
Do you know how to find these sums?

Sorry for the delayed response, but does this seem right?
[tex]
\frac{216 }{n^3} \sum_{i=0}^{n} [i^2] - \frac{72 }{n^2} \sum_{i=0}^{n} - \frac{6}{n} \sum_{i=0}^{n} [1] = \frac{216(n)(n+1)(2n+1)}{6n^3} - \frac{72(n)(n+1)}{2n^2} - 6[/tex]
 
  • #11
computerex said:
Sorry for the delayed response, but does this seem right?
[tex]
\frac{216 }{n^3} \sum_{i=0}^{n} [i^2] - \frac{72 }{n^2} \sum_{i=0}^{n} - \frac{6}{n} \sum_{i=0}^{n} [1] = \frac{216(n)(n+1)(2n+1)}{6n^3} - \frac{72(n)(n+1)}{2n^2} - 6[/tex]


Yes, now take the limit n->infinity.
 
  • #12
Dick said:
Yes, now take the limit n->infinity.

Yep. Through out the entire day I have been subtracting from 432 as opposed to 432/6... Thank you all for helping me :)

And this is a testament to why I hate these stupid problems. Conceptual understanding is nice.. But my calc teacher who has been teaching the subject since before I was born (1970) also made numerous mistakes in doing examples of these problems in front of the class.. No reason as to why we should resort to first principles just to torture the students.

I think being able to set up the problem is good enough.
 

What is infinite Riemann sum discrepancy?

Infinite Riemann sum discrepancy is a mathematical concept that measures the difference between the exact value and the calculated value of an infinite Riemann sum. It is a measure of how accurately the Riemann sum approximates the area under a curve.

How is infinite Riemann sum discrepancy calculated?

Infinite Riemann sum discrepancy is calculated by taking the difference between the exact value and the calculated value of an infinite Riemann sum. This can be done using various mathematical methods, such as taking the limit of the Riemann sum as the number of partitions approaches infinity.

What factors affect the value of infinite Riemann sum discrepancy?

The value of infinite Riemann sum discrepancy is affected by the function being integrated, the interval of integration, and the number of partitions used to calculate the Riemann sum. It can also be affected by the method used to calculate the Riemann sum.

Why is infinite Riemann sum discrepancy important?

Infinite Riemann sum discrepancy is important because it is a measure of the accuracy of a Riemann sum approximation. It can help determine the reliability of using a Riemann sum to estimate the area under a curve and can also provide insight into the behavior of a function.

How can infinite Riemann sum discrepancy be minimized?

Infinite Riemann sum discrepancy can be minimized by using more partitions in the Riemann sum calculation, using more accurate methods for calculating the Riemann sum, and choosing appropriate functions and intervals for integration. In some cases, it may also be helpful to use a different method of integration altogether.

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