Integrating Using Limit Definition (Riemann Sums)

[Scribendi]

Hello!

A quick thank you in advance for anyone who would be kind enough to give an aspiring English-and-Art major some assistance. (No idea how I ended up in a Calculus class. Lord, help me. Tell me to paint you a mural, I've got your back. "2 + 2?" What? What language are you speaking?) This problem will be rather elementary to you intellects on this board, so please don't laugh at me--I'm already embarrassed.

Apologies if I'm a bit slow.

Anyway! To the problem:

"Evaluate the definite integral by the limit definition."

Yes, I know what you're thinking. "Oh, hardyhar. Silly Scribendi, the answer is obviously 26." Can you please help me use limit definition? I got stuck midway.

Here's what I have so far:

Sorry for the MS-Paintesque images. The blue is just a personal reminder of the summation formulas I'm probably gonna end up using.

Thank you again for any help!

Sincerely,

Scribendi

 

[BigGlenntheHeavy]

Re: Integrating Using Limit Definition

You're evaluting a definite integral, ergo;

\(\displaystyle \int_{1}^{3}3x^{2}dx \ = \ x^{3}]_{1}^{3} \ = \ 27-1 \ = \ 26.\)

 

[Scribendi]

Re: Integrating Using Limit Definition

Yes, I appreciate the feedback and I realize the answer is 26 using basic integration.

However, the directions explicitly state I must use "limit definition", sigma notation and all, which I am pathetically lost at.

Regardless, thank you for your time BigGlenntheHeavy. C: (And I apologize, I should have asked specifically for help on Riemann Sums.)

 

[daon]

\(\displaystyle \sum_i \frac{6n^2 + 24in +24i^2}{n^3} = \frac{6}{n}\sum_i1 + \frac{24}{n^2}\sum_ii + \frac{24}{n^3}\sum_ii^2\)

\(\displaystyle = \frac{6}{n} \cdot n + \frac{24}{n^2}\cdot\frac{n(n-1)}{2} + \frac{24}{n^3}\cdot\frac{n(n+1)(2n+1)}{6}\)

 

[BigGlenntheHeavy]

Ok, Scribendi, I'll use the right endpoint limit for you as this will be a good exercise in LaTex.

\(\displaystyle We \ know \ \int_{1}^{3}3x^{2}dx \ = \ 26. \ Let \ f(x) \ = \ 3x^{2}, \ prove \ it \ by \ limit \ notation.\)

\(\displaystyle Right \ endpoint, \ area \ = \ \lim_{n\to\infty}S(n) \ = \ \lim_{n\to\infty} \sum_ {i=1}^{n}f(M_i)\Delta x\)

\(\displaystyle Now, \ \Delta x \ = \ \frac{b-a}{n} \ = \ \frac{3-1}{n} \ = \ \frac{2}{n} \ and \ M_i \ = \ (1+\frac{2i}{n}).\)

\(\displaystyle Now, \ \lim_{n\to\infty} \sum_ {i=1}^{n}f(M_i)\Delta x \ = \ \lim_{n\to\infty} \sum_{i=1}^{n}f[(1+\frac{2i}{n})]\frac{2}{n}\)

\(\displaystyle = \ \lim_{n\to\infty} \frac{2}{n} \sum_{i=1}^{n}3(1+\frac{2i}{n})^{2} \ = \ \lim_{n\to\infty} \frac{6}{n} \sum_{i=1}^{n}1+\frac{4i}{n}+\frac{4i^{2}}{n^{2}}.\)

\(\displaystyle = \ \lim_{n\to\infty} \frac{6}{n}[\sum_{i=1}^{n}1+\frac{4}{n}\sum_{i=1}^{n}i+\frac{4}{n^{2}}\sum_{i=1}^{n}i^{2}] .\)

\(\displaystyle = \ \lim_{n\to\infty} \frac{6}{n}[n+\frac{4}{n}\frac{[n(n+1)]}{2}+\frac{4}{n^{2}}\frac{[n(n+1)(2n+1)]}{6}]\)

\(\displaystyle = \ \lim_{n\to\infty}\frac{6}{n}[n+2n+2+\frac{2(n+1)(2n+1)}{3n}]\)

\(\displaystyle = \ \lim_{n\to\infty}\frac{6}{n}[3n+2+\frac{4n}{3}+2+\frac{2}{3n}]\)

\(\displaystyle = \ \lim_{n\to\infty}[18+\frac{12}{n}+8+\frac{12}{n}+\frac{12}{3n^{2}}] \ = \ 26.\)

Note: This is the right endpoint, the left endpoint f(m)sub i will also give you the same limit, 26.

Additional note: Always use the right endpoint, unless instructed otherwise, as it is less protracted.

 

[Scribendi]

Ah, I see.

Thank you so very much for your kindness, Daon and BigGlenntheHeavy. Apologies for the trouble especially, BigGlenn; I'm truly grateful that you took the time to demonstrate the problem step-by-step. It certainly cleared up my muddled thought process and you've given me an excellent example to refer to should I have any qualms with Riemann Sums in the future.