Limits of Riemann Sums

• Nov 13th 2009, 01:09 PM
ipreferhistory
Limits of Riemann Sums
I have to answer three questions about limits of Riemann sums, and I can't remember how to figure it out. I know it involves sequences with sigma but I don't know where to start now.
the first one I have is written with the integral from 2 to 5 (x^2 - 1)dx
can someone please get me started or show the steps so I can do the rest of my homework?
• Nov 13th 2009, 01:52 PM
tonio
Quote:

Originally Posted by ipreferhistory
I have to answer three questions about limits of Riemann sums, and I can't remember how to figure it out. I know it involves sequences with sigma but I don't know where to start now.
the first one I have is written with the integral from 2 to 5 (x^2 - 1)dx
can someone please get me started or show the steps so I can do the rest of my homework?

As the function is continuous everywhere it is so in the given interval and thus it is Riemann integrable there. Now subdivide $\displaystyle [2,5]\,\, in \,\,n$ subintervals of equal length $\displaystyle \frac{5-2}{n}=\frac{3}{n}$, and choose to evaluate the function $\displaystyle f(x)=x^2-1$ at the left endpoints of each subinterval (we can do that since we know the functions is Riemann int. and thus we can freely choose the partition ,as long as it length parameter tends to zero when $\displaystyle n\rightarrow\infty\$, and the points at each subinterval at which the function's evaluated:

$\displaystyle \int\limits_2^5(x^2-1)dx=\lim_{n\to\infty}\frac{3}{n}\sum\limits_{k=0} ^{n-1}\left[\left(2+\frac{3k}{n}\right)^2-1\right]=\lim_{n\to\infty}\frac{3}{n}\sum\limits_{k=0}^{n-1}\left[4+\frac{12}{n}k+\frac{9}{n^2}k^2-1\right]$ $\displaystyle =\lim_{n\to\infty}\frac{3}{n}\left[4n+\frac{12}{n}\frac{(n-1)n}{2}+\frac{9}{n^2}\frac{n(n+1)(2n+1)}{6}-n\right]=$

$\displaystyle \lim_{n\to\infty}3\left[3+\frac{6(n-1)n}{n^2}+\frac{3(2n+1)(n+1)n}{2n^3}\right]=3(3+6+3)=36$

Tonio
• Nov 13th 2009, 01:55 PM
Quote:

Originally Posted by ipreferhistory
I have to answer three questions about limits of Riemann sums, and I can't remember how to figure it out. I know it involves sequences with sigma but I don't know where to start now.
the first one I have is written with the integral from 2 to 5 (x^2 - 1)dx
can someone please get me started or show the steps so I can do the rest of my homework?

If you need to right the integral as a sum, then this is what you're looking for.

$\displaystyle \int_a^bf(x)dx=\lim_{n->\infty}\Sigma_{i=1}^nf(x_i)\Delta x$

$\displaystyle x_i=a+i\Delta x$

$\displaystyle \Delta x =\frac{b-a}{n}$

Do you understand how to apply this? It's pretty straight forward.
• Nov 22nd 2009, 06:27 AM
Blakjax
I have a similar question

Find a formula for right hand endpoint sum by dividing the interval into n equal subintervals. Then take the limit of these sums as n approaches infinity to calculate area under the curve [a,b]. f(x)=5x^2 over [1,2]
• Nov 23rd 2009, 02:38 AM
mr fantastic
Quote:

Originally Posted by Blakjax

Find a formula for right hand endpoint sum by dividing the interval into n equal subintervals. Then take the limit of these sums as n approaches infinity to calculate area under the curve [a,b]. f(x)=5x^2 over [1,2]

You have been given a fair bit of help. Where are you stuck?
• Nov 23rd 2009, 03:07 AM
Blakjax
unsure
This is what I got, but I'm not sure if it's right.

R=∑(5k^2/n^2 + 5)1/n

5/n^3(n(n+1)(2n+1)/6) + 5

lim ((5n(n+1)(2n+1))/6n^3) + 5

= 20/3....my problem is I don't think it's the right answer
• Nov 23rd 2009, 04:07 PM
mr fantastic
Quote:

Originally Posted by Blakjax
This is what I got, but I'm not sure if it's right.

R=∑(5k^2/n^2 + 5)1/n

5/n^3(n(n+1)(2n+1)/6) + 5

lim ((5n(n+1)(2n+1))/6n^3) + 5

= 20/3....my problem is I don't think it's the right answer

Find $\displaystyle \lim_{n \to +\infty} \frac{5}{n} \sum_{i=1}^{n} \left( 1 + \frac{i}{n}\right)^2$.

Note that by direct integration you expect the above limit to equal 35/3 (so you should be very sure about whether or not you have the correct answer).