Reimann sum solution check

[bcddd214]

lim(n??)??_(i=1)^n?(1+2i/n)^3 (2/n) ?
=2/n ?_(i=1)^n?(1+2i/n)(1+2i/n)(1+2i/n) ?
2/n ?_(i=1)^n?(1+4i/n+(2i^2)/n^2 )(1+2i/n) ?
2/n ?_(i=1)^n?(1+2i/n+4i/n+(2i^2)/n^2 +(8i^2)/n^2 +(4i^3)/n^3 ) ?
2/n ?_(i=1)^n?(1+6i/2n+(10i^2)/(2n^2 )+(4i^3)/n^3 ) ?
2/n (?_(i=1)^n?1+?_(i=1)^n 6i/2n+?_(i=1)^n (10i^2)/(2n^2 )+?_(i=1)^n (4i^3)/n^3 ?)
2/n (?_(i=1)^n?1+3/n ?_(i=1)^n i+5/n^2 ?_(i=1)^n i^2 +4/n^3 ?_(i=1)^n i^3 ?)
2/n (4/n^3 [(n^2 (n+1)^2)/4]+5/n^2 [n(n+1)(2n+1)/6]+3/n [n(n+1)/2]+1n)
8/4 [(n^2 (n+1)^2)/n^4 ]+10/6 [n(n+1)(2n+1)/n^3 ]+6/2 [n(n+1)/n^2 ]+2
24/12+20/12+36/12+24/12=104/12=52/6

All better now?

 

[galactus]

I doubt if anyone wants to blunder through that. Do not copy and paste word documents. They come out looking like Babylonian cuneiform.

 

[bcddd214]

I doubt if anyone wants to blunder through that. Do not copy and paste word documents. They come out looking like Babylonian cuneiform.

There, fixed it.

 

[galactus]

Not really. When math is written like this it is laborious to read through. Try it in Latex. Have you noticed how math can be displayed in a nice format?. Interested?. Click on 'quote' at the upper right corner to see how I done this.
I took the time to convert your post. I had a rough time interpreting the next to last line. Please correct me if it is not correct. While I was doing it, I did notice a fraction addition error. \(\displaystyle \frac{2i}{n}+\frac{4i}{n}\neq \frac{6i}{2n}\).
As you probably know, this is not how you add fraction. It should be \(\displaystyle \frac{2i}{n}+\frac{4i}{n}=\frac{6i}{n}\)


\(\displaystyle \lim_{n\to \infty}\sum_{i=1}^{n}\left(1+\frac{2i}{n}\right)^{3}\cdot \frac{2}{n}\)

\(\displaystyle =\frac{2}{n}\sum_{i=1}^{n}\left(1+\frac{2i}{n}\right)\left(1+\frac{2i}{n}\right)\left(1+\frac{2i}{n}\right)\)

\(\displaystyle \frac{2}{n}\sum_{i=1}^{n}\left(1+\frac{4i}{n}+(\frac{2i^{2}}{n^{2}} )(1+\frac{2i}{n})\right)\)
\(\displaystyle \frac{2}{n}\sum_{i=1}^{n}\left(1+\frac{2i}{n}+\frac{4i}{n}+\frac{2i^{2}}{n^{2}} +\frac{8i^{2}}{n^{2}} +\frac{4i^{3}}{n^{3}} \right)\)

\(\displaystyle \frac{2}{n}\sum_{i=1}^{n}\left(1+\frac{6i}{2n}+\frac{10i^{2}}{2n^{2}}+\frac{4i^{3}}{n^{3}}\right)\)

\(\displaystyle \frac{2}{n}\left(\sum_{i=1}^{n}1+\sum_{i=1}^{n}\frac{61}{2n}+\sum_{i=1}^{n}\frac{10i^{2}}{2n^{2}}+\sum_{i=1}^{n}\frac{4i^{3}}{n^{3}}\right)\)

\(\displaystyle \frac{2}{n}\left(\sum_{i=1}^{n}1+\frac{3}{n}\sum_{i=1}^{n}i+\frac{5}{n^{2}}\sum_{i=1}^{n}i^{2} +\frac{4}{n^{3}}\sum_{i=1}^{n}i^{3}\right)\)

\(\displaystyle \frac{2}{n}\left(\frac{4}{n^{3}}\left[\frac{n^{2}(n+1)^{2}}{4}\right]+\frac{5}{n^{2}}\left[\frac{n(n+1)(2n+1)}{6}\right]+\frac{3}{n}\left[\frac{n(n+1)}{2}\right]+n\right)\)

\(\displaystyle \frac{8}{4}\left[\frac{n^{2}(n+1)^{2}}{n^{4}}\right]+\frac{10}{6}\left[\frac{n(n+1)(2n+1)}{n^{3}}\right]+\frac{6}{2}\left[\frac{n(n+1)}{n^{2}}\right] \left[\frac{n(n+1)}{n^{2}} \right]+2\)

\(\displaystyle \frac{24}{12}+\frac{20}{12}+\frac{36}{12}+\frac{24}{12}=\frac{104}{12}=\frac{52}{6}\)

This Riemann sum represents \(\displaystyle \int_{1}^{3}x^{3}dx\).

Unfortunately, the solution is not \(\displaystyle \frac{52}{6}\). You're aiming for 20.