anonymous
  • anonymous
Hey guys, I am looking for some help with calculus. It is in regard to rieman sums, and I would really appreciate any help understanding. I will post links below and explain further.
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
The graph shown is y=x I just dont see how those steps were derived
anonymous
  • anonymous
And this is part of it also
anonymous
  • anonymous
Is therom 5 something that must be memorized? Or is there any logic to it?

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kc_kennylau
  • kc_kennylau
|dw:1388473083421:dw|
anonymous
  • anonymous
What does k represent?
kc_kennylau
  • kc_kennylau
k is from 1 to n, if you notice the summation sign in your first photo
anonymous
  • anonymous
But why is that over n the height?
anonymous
  • anonymous
And is the width 1/n just because we are taking it as n approaches infinity? so therefor 1/n is approach infinitely small?
anonymous
  • anonymous
kc_kennylau
  • kc_kennylau
yes, and sorry I'm not too familiarized in this topic... furthermore i'm busy :'(
kc_kennylau
  • kc_kennylau
sorry :'(
anonymous
  • anonymous
Thanks for the input anyways kc!
myininaya
  • myininaya
Ok I can try drawing a picture for you and posting it. But first let's see if you can see the pattern first. so the very first number is a_1=(1-1)/n the next number is a_2=(2-1)/n=1/n <---this is how for we are away from 0 or a_1 since a_1 is 0 a_3=(3-1)/n .... now going to the kth interval where we have the x-value being a_k=(k-1)/n ... going all the way to the nth interval we have a_n=(n-1)/n because we are plugging in the left end point of that interval and not the right endpoint because we are doing left-endpoint rule we are taking all the left endpoints of all the n intervals beginning at 0
myininaya
  • myininaya
Now to find the heights of the rectangles we do f(a_k)
anonymous
  • anonymous
By a_1 does that mean a/1?
myininaya
  • myininaya
|dw:1388474479972:dw|
myininaya
  • myininaya
a_1 usually means a subscript 1
myininaya
  • myininaya
the base of each rectangle is 1/n
anonymous
  • anonymous
Alright, okay so f(k_n) would be f(k-1/n)
anonymous
  • anonymous
and thats the height at any given k
myininaya
  • myininaya
you mean f(a_k)?
myininaya
  • myininaya
If so yes
myininaya
  • myininaya
but since this function is f(x)=x then f((k-1)/n)=(k-1)/n
anonymous
  • anonymous
Alright I think it is starting to make sense, but when I try an example I still just dont exactly see it. Ill post if you could take a look,
myininaya
  • myininaya
Ok, so looking at this example, can you tell me what puzzles you and I will see if I can explain it?
anonymous
  • anonymous
Sure just a moment
anonymous
  • anonymous
Well first of all, do we know that it is left hand because of the n-1 on top of the sigma?
anonymous
  • anonymous
If that is so, for left handed questions like this can we always do (1/n)f(k/n)?
anonymous
  • anonymous
So we just set that equal to the function that is given?
myininaya
  • myininaya
yep they started at 0 and when to n-1 so left endpoint
anonymous
  • anonymous
And what does the k=xn really represent?
myininaya
  • myininaya
We are trying to find what f(x) is
myininaya
  • myininaya
\[\int\limits_{a}^{b}f(x) dx=\sum_{i=0}^{n-1} \Delta x f(a+i \cdot \Delta x)\]
myininaya
  • myininaya
where delta x =(b-a)/n
anonymous
  • anonymous
Right, and Xi= a+ i deltax/n
myininaya
  • myininaya
the easiest thing to do is assume a equals 0 so we can just look at i*delta x
anonymous
  • anonymous
Oh i see, and how about the 1?
myininaya
  • myininaya
well they chose the base to be 1/n which is delta x
myininaya
  • myininaya
if we have (b-0)/n=1/n then b has to be?
anonymous
  • anonymous
1
myininaya
  • myininaya
now this is one way to do the problem you don't have to choose it this way and the integral will still have the same value and yes b=1 for this example
myininaya
  • myininaya
we can take this same problem and go a different way though
myininaya
  • myininaya
would you like to?
anonymous
  • anonymous
Yes sure!
myininaya
  • myininaya
I still want to choose a to be 0 because yeah that is just plain easiest!
myininaya
  • myininaya
but with if we choose the intervals to have width 2/n instead of 1/n
myininaya
  • myininaya
then b would have to be what?
anonymous
  • anonymous
then b would be two
anonymous
  • anonymous
So are you also saying that you can choose some of the conventions, as long as they match up?
myininaya
  • myininaya
yes so but now we have \[\frac{2}{n} f(2 \cdot \frac{k}{n}) \text{ instead of } \frac{1}{n} f(\frac{k}{n}) \]
myininaya
  • myininaya
yep
myininaya
  • myininaya
like are answer will look different but it will still hold the same value
myininaya
  • myininaya
our*
myininaya
  • myininaya
\[\text{ so we want } \frac{2}{n} f(\frac{2k}{n})=\frac{1}{k+5n} \arctan(\frac{k+2n}{k+n})\]
myininaya
  • myininaya
Solve for f(2k/n) by multiplying both sides by n/2
anonymous
  • anonymous
son/2 (k +5n)?
myininaya
  • myininaya
\[f(\frac{2k }{n})=\frac{n}{2k+10n} \arctan(\frac{k+2n}{k+n})\]
anonymous
  • anonymous
Okay yea , and then how do you deal with the 2k/n?
myininaya
  • myininaya
Now we want to know what f(x) is right?
anonymous
  • anonymous
right
myininaya
  • myininaya
so what is k if 2k/n=x
myininaya
  • myininaya
basically solve that for k so we can figure out what to replace k with so that we will just have x inside and not that other crap
anonymous
  • anonymous
Okay so we bring in an X ?
anonymous
  • anonymous
xn/2=k
myininaya
  • myininaya
because we are trying to find out what integral notation looks like for this summation notation
myininaya
  • myininaya
looks good so we will replace all the k's we see with that
myininaya
  • myininaya
\[f(x)=\frac{n}{xn+10n}\arctan(\frac{\frac{xn}{2}+2n}{\frac{xn}{2}+n})\] \[f(x)=\frac{n(1)}{n(x+10)}\arctan(\frac{xn+4n}{xn+2n})=\frac{1}{x+10}\arctan(\frac{n(x+4)}{n(x+2)})\] \[f(x)=\frac{1}{x+10}\arctan(\frac{x+4}{x+2})\] so this is what our f(x) looks like from choosing a= 0 and b=2 there isn't a unique answer to your question the answer can totally vary
myininaya
  • myininaya
http://www.wolframalpha.com/input/?i=integrate%281%2F%28x%2B10%29*arctan%28%28x%2B4%29%2F%28x%2B2%29%29%2C+x%3D0..2%29 http://www.wolframalpha.com/input/?i=integrate%281%2F%28x%2B5%29*arctan%28%28x%2B2%29%2F%28x%2B1%29%29%2C+x%3D0..1%29 see the answers look different but hold the same value
myininaya
  • myininaya
basically you get to choose a and b
myininaya
  • myininaya
and nothing else
anonymous
  • anonymous
Wow great. Thanks a lot for your help! Do you know the exact names of these types of problems? Im looking to find some practice problems
myininaya
  • myininaya
This is just some applications to the definition of reimann sums let me see if i can find you some problems one moment
myininaya
  • myininaya
http://freedom.mysdhc.org/teacher/1541derflingerk/documents/Calculus/FOV1-001B4535/Integration%20via%20Sigma(2).pdf these don't look at hard but you can try these 4.3 9-12
myininaya
  • myininaya
as*
myininaya
  • myininaya
they choose a and b for you
anonymous
  • anonymous
Alright, thanks!
myininaya
  • myininaya
there c_i represents a+i*delta x by the way
myininaya
  • myininaya
If you get bored of those attempt this one I made up: Write this as an integral: \[\lim_{n \rightarrow \infty} \sum_{i=0}^{n-1} \frac{i}{i+8n} \cos(\frac{i+3n}{n})\]
anonymous
  • anonymous
Try practicing on http://www.saab.org/calculus.cgi under Select What Kind Of Problems select Calculus I( Integrals(Substitution, FTC)

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