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June 12th, 2015, 03:24 PM   #1
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Fundamental Theorem of Calculus Question

I am preparing for an exam and was looking at a past exam problem. I was really stuck on this question. Can anyone elaborate where the integrand x^2 came from? As well as the limits of integration of 0 and 1, or what this whole 1/N Riemann's sum means in relation to the right-hand rule in general. I have no idea where to start.



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Tzad

Last edited by skipjack; June 12th, 2015 at 04:52 PM.
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June 12th, 2015, 04:20 PM   #2
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In the Riemann sum definition of the integral we start by dividing the interval of integration (here from 0 to 1) into N subintervals. One can show that the subintervals don't have to be all the same length, but that is can be done and is simplest. Taking the subintervals to be the same length, each will have length 1/N. We then choose a single point, $\displaystyle x_i$, in each interval and find the area of the rectangle with base 1/N and height $\displaystyle f(x_i)$. If we choose that point to always be the right endpoint of the interval, it will be $\displaystyle x_i= \frac{i}{N}$ so the height of that rectangle will be $\displaystyle \left(\frac{I}{N}\right)$ which, here, is $\displaystyle \frac{i^2}{N^2}$. The area of a rectangle with base $\displaystyle \frac{1}{N}$ and height $\displaystyle \frac{i^2}{N^2}$ is $\displaystyle \frac{1}{N}\frac{i^2}{N^2}$. The total area of all those rectangles is the sum $\displaystyle \frac{1}{N}\sum_{i=1}^N \frac{i^2}{N^2}$.

That sum could be written as $\displaystyle \frac{1}{N^3}\sum_{i= 0}^N i^2$ but I assume they did not do that in order to make it clear that the "$\displaystyle \frac{i^2}{N^2}$" term is the "$\displaystyle x^2$" term.

By the way, You could also do that problem "the other way around", using the fact that $\displaystyle \sum_{i=1}^n i^2= n(2n-1)(n- 2)$. Then the sum $\displaystyle \frac{1}{N^3}\sum_{i=1}^N i^2= \frac{1}{N^3}N(2N- 1)(n- N)= \frac{2N^3+ 6N^2+ N}{6N^3}= \frac{1}{3}+ \frac{1}{n^2}+ \frac{6}{N^2}$ which goes to $\displaystyle \frac{1}{3}$ as N goes to infinity.
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Last edited by Country Boy; June 12th, 2015 at 04:26 PM.
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June 12th, 2015, 04:29 PM   #3
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Do you know the definition of a Reimann Sum?

Let $P = \{x_0,\,x_1,\,x_2,\,...,\,x_n\}$ (where $x_0 < x_1<x_2<...<x_n$) be a partition of $f(x)$ into $n$ subintervals. For each subinterval $[x_{i - 1},\,x_i]$ choose a number $c_i$ in this interval. The Reimann sum is defined as
$$R = \sum_{i = 1}^{n}(x_i - x_{i - 1})f(c_i)$$
Also, if the largest subinterval $[x_{i - 1},\,x_i]$ approaches zero length (it does in your case) then
$$\lim_{n\to\infty}\,R = \int_{x_0}^{x_n}f(x)\,dx$$

In your specific case, you have $f(x) = x^2$, $c_i = x_i = \dfrac{i}{N}$ and $x_{i - 1} - x_i = \dfrac{1}{N}$. Thus your Reimann sum is
$$R = \sum_{i = 1}^N\dfrac{1}{N}\left(\dfrac{i}{N}\right)^2
\\
= \dfrac{1}{N}\sum_{i = 1}^N\dfrac{i^2}{N^2}$$

Now, it is easy to see that $x_N = \dfrac{N}{N} = 1$ and $x_0 = x_1 - \dfrac{1}{N} = \dfrac{1}{N} - \dfrac{1}{N} = 0$.

Thus,
$$\lim_{n\to\infty}\,R = \int_0^1x^2\,dx$$


Sorry if the explanation is too rigorous.
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June 12th, 2015, 05:07 PM   #4
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Quote:
Originally Posted by Country Boy View Post
In the Riemann sum definition of the integral we start by dividing the interval of integration (here from 0 to 1) into N subintervals. One can show that the subintervals don't have to be all the same length, but that is can be done and is simplest. Taking the subintervals to be the same length, each will have length 1/N. We then choose a single point, $\displaystyle x_i$, in each interval and find the area of the rectangle with base 1/N and height $\displaystyle f(x_i)$. If we choose that point to always be the right endpoint of the interval, it will be $\displaystyle x_i= \frac{i}{N}$ so the height of that rectangle will be $\displaystyle \left(\frac{I}{N}\right)$ which, here, is $\displaystyle \frac{i^2}{N^2}$. The area of a rectangle with base $\displaystyle \frac{1}{N}$ and height $\displaystyle \frac{i^2}{N^2}$ is $\displaystyle \frac{1}{N}\frac{i^2}{N^2}$. The total area of all those rectangles is the sum $\displaystyle \frac{1}{N}\sum_{i=1}^N \frac{i^2}{N^2}$.

That sum could be written as $\displaystyle \frac{1}{N^3}\sum_{i= 0}^N i^2$ but I assume they did not do that in order to make it clear that the "$\displaystyle \frac{i^2}{N^2}$" term is the "$\displaystyle x^2$" term.

By the way, You could also do that problem "the other way around", using the fact that $\displaystyle \sum_{i=1}^n i^2= n(2n-1)(n- 2)$. Then the sum $\displaystyle \frac{1}{N^3}\sum_{i=1}^N i^2= \frac{1}{N^3}N(2N- 1)(n- N)= \frac{2N^3+ 6N^2+ N}{6N^3}= \frac{1}{3}+ \frac{1}{n^2}+ \frac{6}{N^2}$ which goes to $\displaystyle \frac{1}{3}$ as N goes to infinity.
Hey,
Thanks for the response, I have a couple of questions.

1. I understand that I'm breaking up the function so that each base is 1/N, I also understand that the height is f(xi). When you say the right point will be xi=i/N, are you saying that's its area? Also, I kind of lost you after you said the height of the rectangle is I/N. I don't see how you got that.

2. I also know in an ordinary Reimman's sum, the 1/N (Or Delta X) is on the right side of the f(xi). Can we rearrange this equation so that it is like that? Or is that wrong?

3. I have no idea what you mean about looking at the problem from "the other way around". It seems really cool and I'd like to know more if you can elobarate on how ∑i^2=n(2n−1)(n−2)

Thanks,
Tzad
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June 12th, 2015, 05:11 PM   #5
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Quote:
Originally Posted by Azzajazz View Post
Do you know the definition of a Reimann Sum?

Let $P = \{x_0,\,x_1,\,x_2,\,...,\,x_n\}$ (where $x_0 < x_1<x_2<...<x_n$) be a partition of $f(x)$ into $n$ subintervals. For each subinterval $[x_{i - 1},\,x_i]$ choose a number $c_i$ in this interval. The Reimann sum is defined as
$$R = \sum_{i = 1}^{n}(x_i - x_{i - 1})f(c_i)$$
Also, if the largest subinterval $[x_{i - 1},\,x_i]$ approaches zero length (it does in your case) then
$$\lim_{n\to\infty}\,R = \int_{x_0}^{x_n}f(x)\,dx$$

In your specific case, you have $f(x) = x^2$, $c_i = x_i = \dfrac{i}{N}$ and $x_{i - 1} - x_i = \dfrac{1}{N}$. Thus your Reimann sum is
$$R = \sum_{i = 1}^N\dfrac{1}{N}\left(\dfrac{i}{N}\right)^2
\\
= \dfrac{1}{N}\sum_{i = 1}^N\dfrac{i^2}{N^2}$$

Now, it is easy to see that $x_N = \dfrac{N}{N} = 1$ and $x_0 = x_1 - \dfrac{1}{N} = \dfrac{1}{N} - \dfrac{1}{N} = 0$.

Thus,
$$\lim_{n\to\infty}\,R = \int_0^1x^2\,dx$$


Sorry if the explanation is too rigorous.
Great explanation, the only problem I'm having is understanding why, $x_0 = x_1 - \dfrac{1}{N} = \dfrac{1}{N} - \dfrac{1}{N} = 0$. And how that translates to $$\lim_{n\to\infty}\,R = \int_0^1x^2\,dx$$

Thanks,
Tzad

Last edited by Tzad; June 12th, 2015 at 05:18 PM.
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June 12th, 2015, 06:14 PM   #6
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Quote:
Originally Posted by Tzad View Post
... the only problem I'm having is understanding why, $x_0 = x_1 - \dfrac{1}{N} = \dfrac{1}{N} - \dfrac{1}{N} = 0$
I got this from $x_i - x_{i - 1} = \dfrac{1}{N}$ and $x_i = \dfrac{i}{N}$(which is wrong in my original post... ooops) so $x_1 - x_0 = \dfrac{1}{N}$, or $x_0 = x_1 - \dfrac{1}{N} = \dfrac{1}{N} - \dfrac{1}{N} = 0$.

Quote:
Originally Posted by Tzad View Post
And how that translates to $$\lim_{n\to\infty}\,R = \int_0^1x^2\,dx$$
If you look at my first post, you'll see the line that says
$$\lim_{n\to\infty}\,R = \int_{x_0}^{x_n}f(x)\,dx$$
In your case, $x_0 = 0$, $x_n = 1$ and $f(x) = x^2$.
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