Reimann Sum - Infinity

[Jason76]

Post Edited

Find the area of:

Given \(\displaystyle y = x^{2}\) on the interval \(\displaystyle [0,5]\)

using Right Reimann Sum with infinite triangles.

Setup Info

Interval: \(\displaystyle [a,b]\)

\(\displaystyle \Delta x = \dfrac{b - a}{n}\)

\(\displaystyle \sum_{i = 1}^{n} i^{2} = \dfrac{2n^{3} + 3n^{2} + n}{6}\) - special formula for \(\displaystyle \sum_{i = 1}^{n} f(i^{2})\)

\(\displaystyle \lim_{n \rightarrow \infty} \sum_{i = 1}^{n} \Delta[f(i\Delta x)]\) - Right Reimann Sum Formula

Actual Problem

Interval: \(\displaystyle [0,5]\)

\(\displaystyle i = i\)

\(\displaystyle n = n\)

\(\displaystyle \Delta x = \dfrac{5 - 0}{n} = \dfrac{5}{n}\)

\(\displaystyle \lim_{n \rightarrow \infty} \sum_{i = 1}^{n}\dfrac{5}{n}[f(i\dfrac{5}{n})]\)

\(\displaystyle \lim_{n \rightarrow \infty} \sum_{i = 1}^{n}\dfrac{5}{n}[f(\dfrac{5i}{n})]\)

\(\displaystyle \lim_{n \rightarrow \infty} \sum_{i = 1}^{n}\dfrac{5}{n}(\dfrac{5i}{n})^{2}\)

\(\displaystyle \lim_{n \rightarrow \infty} \sum_{i = 1}^{n}\dfrac{5}{n}(\dfrac{25i^{2}}{n^{2}})\)

\(\displaystyle \lim_{n \rightarrow \infty} \dfrac{5}{n}(\dfrac{25}{n^{2}}) \sum_{i = 1}^{n} i^{2}\)

\(\displaystyle \lim_{n \rightarrow \infty} \dfrac{125}{n^{3}})\sum_{i = 1}^{n} i^{2}\)

\(\displaystyle \lim_{n \rightarrow \infty}\dfrac{125}{n^{3}}( \dfrac{2n^{3} + 3n^{2} + n}{6})\)

\(\displaystyle \lim_{n \rightarrow \infty}\dfrac{250n^{3}}{6n^{3}} + \dfrac{375n^{2}}{6n^{3}} + \dfrac{125n}{6n^{3}} \)

Dividing everything by \(\displaystyle 6n^{3}\) (highest index in the denominator) should yield the answer of \(\displaystyle \dfrac{250}{6} = \dfrac{124}{3}\) square units but it only sort of does.

 

[daon2]

edit: okay you fixed it. simplify to this:

\(\displaystyle \displaystyle A = \lim_{n\to\infty} \sum_{i=1}^{n} f\left(\dfrac{5i}{n}\right) \cdot \dfrac{5}{n} = \lim_{n\to\infty} \dfrac{125}{n^3}\sum_{i=1}^{n} i^2\)

Now make your substitution for \(\displaystyle i^2\) and take limits.

 

[Jason76]

edit: okay you fixed it. simplify to this:

\(\displaystyle \displaystyle A = \lim_{n\to\infty} \sum_{i=1}^{n} f\left(\dfrac{5i}{n}\right) \cdot \dfrac{5}{n} = \lim_{n\to\infty} \dfrac{125}{n^3}\sum_{i=1}^{n} i^2\)

Now make your substitution for \(\displaystyle i^2\) and take limits.

I understand that part. The end part of "dividing by the highest index in the denominator" is the problem.

 

[DrPhil]

I understand that part. The end part of "dividing by the highest index in the denominator" is the problem.

Since there is only one term in the denominator, \(\displaystyle 6n^3\), that is pretty clearly the highest-power term. Ignoring lower powers of n is moot.

 

[Jason76]

Since there is only one term in the denominator, \(\displaystyle 6n^3\), that is pretty clearly the highest-power term. Ignoring lower powers of n is moot.

I know what the highest term is, but when dividing the terms by it, then it doesn't come out right.

 

[stapel]

I know what the highest term is, but when dividing the terms by it, then it doesn't come out right.

Please show your steps which led to your statement. Thank you!

 

[Jason76]

Please show your steps which led to your statement. Thank you!

\(\displaystyle \lim_{n \rightarrow \infty}\dfrac{250n^{3}}{6n^{3}} + \dfrac{375n^{2}}{6n^{3}} + \dfrac{125n}{6n^{3}} \)

\(\displaystyle \lim_{n \rightarrow \infty} \dfrac{6n^{3}/250n^{3}}{1} + \dfrac{6n^{3} /375n^{2}}{1} + \dfrac{6n^{3}/125n}{1}\)

The could have something to do with:

\(\displaystyle \dfrac{b/a}{1} = \dfrac{a}{b}\)

Then you realize that infinity plugged in the bottom produces zero.

 

[DrPhil]

\(\displaystyle \lim_{n \rightarrow \infty}\dfrac{250n^{3}}{6n^{3}} + \dfrac{375n^{2}}{6n^{3}} + \dfrac{125n}{6n^{3}} \)

\(\displaystyle \lim_{n \rightarrow \infty} \dfrac{6n^{3}/250n^{3}}{1} + \dfrac{6n^{3} /375n^{2}}{1} + \dfrac{6n^{3}/125n}{1}\)...NO these are NOT equal - you have inverted each term - not valid, and not useful

The could have something to do with:

\(\displaystyle \dfrac{b/a}{1} = \dfrac{a}{b}\) ...also FALSE

Then you realize that infinity plugged in the bottom produces zero.

\(\displaystyle \displaystyle \lim_{n \to \infty}\dfrac{125}{n^{3}}\left( \dfrac{2n^{3} + 3n^{2} + n}{6}\right)\)

What is the highest power in the numerator? What is the highest (only!) power in the denominator? What is the ratio?

 

[Jason76]

\(\displaystyle \lim_{n \rightarrow \infty} \dfrac{6n^{3}/250n^{3}}{1} + \dfrac{6n^{3} /375n^{2}}{1} + \dfrac{6n^{3}/125n}{1}\)

Should be

\(\displaystyle \lim_{n \rightarrow \infty} \dfrac{250n^{3}/6n^{3}}{1} + \dfrac{375n^{2}/6n^{3}}{1} + \dfrac{125n/6n^{3}}{1}\)

 

[DrPhil]

\(\displaystyle \lim_{n \rightarrow \infty} \dfrac{6n^{3}/250n^{3}}{1} + \dfrac{6n^{3} /375n^{2}}{1} + \dfrac{6n^{3}/125n}{1}\)

Should be

\(\displaystyle \lim_{n \rightarrow \infty} \dfrac{250n^{3}/6n^{3}}{1} + \dfrac{375n^{2}/6n^{3}}{1} + \dfrac{125n/6n^{3}}{1}\)

If you insist on writing the terms separately instead of over the common denominator, what is the power of \(\displaystyle n\) in each term?

\(\displaystyle \displaystyle \dfrac{n^3}{n^3} = \cdot \cdot \cdot\)

\(\displaystyle \displaystyle \dfrac{n^2}{n^3} = \cdot \cdot \cdot\)

\(\displaystyle \displaystyle \dfrac{n^1}{n^3} = \cdot \cdot \cdot\)

 

[Deleted member 4993]

\(\displaystyle \lim_{n \rightarrow \infty}\dfrac{250n^{3}}{6n^{3}} + \dfrac{375n^{2}}{6n^{3}} + \dfrac{125n}{6n^{3}} \)

\(\displaystyle = \displaystyle \lim_{n \rightarrow \infty} \left [ \dfrac{250}{6} \ * \ \dfrac{n^3}{n^3} + \dfrac{375}{6} \ * \ \dfrac{n^2}{n^3} + \dfrac{125}{6} \ * \ \dfrac{n}{n^3} \right ] \)

\(\displaystyle = \displaystyle \lim_{n \rightarrow \infty} \left [ \dfrac{125}{3} * 1 \ \ + \ \dfrac{125}{2} \ * \ \dfrac{1}{n} + \dfrac{125}{6} \ * \ \dfrac{1}{n^2} \right ] \)

\(\displaystyle = \dfrac{125}{3} + \dfrac{125}{2} \ * \ 0 + \dfrac{125}{6} \ * \ 0 \)

\(\displaystyle = \dfrac{125}{3} \)

\(\displaystyle \lim_{n \rightarrow \infty} \dfrac{6n^{3}/250n^{3}}{1} + \dfrac{6n^{3} /375n^{2}}{1} + \dfrac{6n^{3}/125n}{1}\)

The could have something to do with:

\(\displaystyle \dfrac{b/a}{1} = \dfrac{a}{b}\)

Then you realize that infinity plugged in the bottom produces zero.

When do you plan to learn algebra and carry out these steps correctly?!!

 

[Jason76]

\(\displaystyle \lim_{n \rightarrow \infty} \dfrac{250n^{3}/6n^{3}}{1} + \dfrac{375n^{2}/6n^{3}}{1} + \dfrac{125n/6n^{3}}{1}\)

The first term evaluates to \(\displaystyle \dfrac{250}{3}\) over \(\displaystyle 1\). The middle and last terms evaluate to \(\displaystyle \dfrac{375/2}{n}\) over \(\displaystyle 1\), and \(\displaystyle \dfrac{125/6}{n^{2}}\) over \(\displaystyle 1\). These both evaluate to zero when infinity is plugged in. Of course, stuff over \(\displaystyle 1\) is simply itself.

 

[Deleted member 4993]

\(\displaystyle \lim_{n \rightarrow \infty} \dfrac{250n^{3}/6n^{3}}{1} + \dfrac{375n^{2}/6n^{3}}{1} + \dfrac{125n/6n^{3}}{1}\)

The first term evaluates to \(\displaystyle \dfrac{250}{3}\) over \(\displaystyle 1\). The middle and last terms evaluate to \(\displaystyle \dfrac{375/2}{n}\) over \(\displaystyle 1\), and \(\displaystyle \dfrac{125/6}{n^{2}}\) over \(\displaystyle 1\). These both evaluate to zero when infinity is plugged in. Of course, stuff over \(\displaystyle 1\) is simply itself.

No!

The first term evaluates to \(\displaystyle \dfrac{125}{3}\) over \(\displaystyle 1\). The middle and last terms evaluate to \(\displaystyle \dfrac{125/2}{n}\) over \(\displaystyle 1\), and \(\displaystyle \dfrac{125/6}{n^{2}}\) over \(\displaystyle 1\). These both evaluate to zero when infinity is plugged in. Of course, stuff over \(\displaystyle 1\) is simply itself.

 

[Jason76]

No!

The first term evaluates to \(\displaystyle \dfrac{125}{3}\) over \(\displaystyle 1\). The middle and last terms evaluate to\(\displaystyle \dfrac{125/2}{n}\) over \(\displaystyle 1\), and \(\displaystyle \dfrac{125/6}{n^{2}}\) over \(\displaystyle 1\). These both evaluate to zero when infinity is plugged in. Of course, stuff over \(\displaystyle 1\) is simply itself.

My mistake, I didn't reduce the terms.