Infinite riemann sums discrepancy

[computerex]

Hello. I have to solve some integrals using both the standard theorem of calculus and infinite Riemann sums.

\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

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:

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?

 

[Screwdriver]

Why exactly did you sub in 1+\frac{6i}{n} for x^2, but not for x?

 

[computerex]

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?

 

[Screwdriver]

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:

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

 

[computerex]

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:

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

Thank you for the reply.. But

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.

 

[Screwdriver]

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.

 

[computerex]

\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} - \frac{24}{n} - \frac{144i}{n^2} + \frac{12}{n}] = 0 + 72+432-0-144+0=360

 

[computerex]

Do you see the error? Since we are taking limit to infinity I don't bother evaluating everything, I just get the coefficients..

 

[Screwdriver]

Okay, wait a minute. You're expanding this:

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

It should give you this:

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

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

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

Do you know how to find these sums?

 

[computerex]

Do you know how to find these sums?

Sorry for the delayed response, but does this seem right?
<br /> \frac{216 }{n^3} \sum_{i=0}^{n} [i^2] - \frac{72 }{n^2} \sum_{i=0}^{n} <i> - \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</i>

 

[Dick]

Sorry for the delayed response, but does this seem right?
<br /> \frac{216 }{n^3} \sum_{i=0}^{n} [i^2] - \frac{72 }{n^2} \sum_{i=0}^{n} <i> - \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</i>

Yes, now take the limit n->infinity.

 

[computerex]

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.