## RobertSn 2 years ago 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.

1. RobertSn

The graph shown is y=x I just dont see how those steps were derived

2. RobertSn

And this is part of it also

3. RobertSn

Is therom 5 something that must be memorized? Or is there any logic to it?

4. kc_kennylau

|dw:1388473083421:dw|

5. RobertSn

What does k represent?

6. kc_kennylau

7. RobertSn

But why is that over n the height?

8. RobertSn

And is the width 1/n just because we are taking it as n approaches infinity? so therefor 1/n is approach infinitely small?

9. RobertSn

@kc_kennylau ?

10. kc_kennylau

yes, and sorry I'm not too familiarized in this topic... furthermore i'm busy :'(

11. kc_kennylau

sorry :'(

12. RobertSn

Thanks for the input anyways kc!

13. 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

14. myininaya

Now to find the heights of the rectangles we do f(a_k)

15. RobertSn

By a_1 does that mean a/1?

16. myininaya

|dw:1388474479972:dw|

17. myininaya

a_1 usually means a subscript 1

18. myininaya

the base of each rectangle is 1/n

19. RobertSn

Alright, okay so f(k_n) would be f(k-1/n)

20. RobertSn

and thats the height at any given k

21. myininaya

you mean f(a_k)?

22. myininaya

If so yes

23. myininaya

but since this function is f(x)=x then f((k-1)/n)=(k-1)/n

24. RobertSn

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,

25. myininaya

Ok, so looking at this example, can you tell me what puzzles you and I will see if I can explain it?

26. RobertSn

Sure just a moment

27. RobertSn

Well first of all, do we know that it is left hand because of the n-1 on top of the sigma?

28. RobertSn

If that is so, for left handed questions like this can we always do (1/n)f(k/n)?

29. RobertSn

So we just set that equal to the function that is given?

30. myininaya

yep they started at 0 and when to n-1 so left endpoint

31. RobertSn

And what does the k=xn really represent?

32. myininaya

We are trying to find what f(x) is

33. myininaya

$\int\limits_{a}^{b}f(x) dx=\sum_{i=0}^{n-1} \Delta x f(a+i \cdot \Delta x)$

34. myininaya

where delta x =(b-a)/n

35. RobertSn

Right, and Xi= a+ i deltax/n

36. myininaya

the easiest thing to do is assume a equals 0 so we can just look at i*delta x

37. RobertSn

Oh i see, and how about the 1?

38. myininaya

well they chose the base to be 1/n which is delta x

39. myininaya

if we have (b-0)/n=1/n then b has to be?

40. RobertSn

1

41. 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

42. myininaya

we can take this same problem and go a different way though

43. myininaya

would you like to?

44. RobertSn

Yes sure!

45. myininaya

I still want to choose a to be 0 because yeah that is just plain easiest!

46. myininaya

but with if we choose the intervals to have width 2/n instead of 1/n

47. myininaya

then b would have to be what?

48. RobertSn

then b would be two

49. RobertSn

So are you also saying that you can choose some of the conventions, as long as they match up?

50. myininaya

yes so but now we have $\frac{2}{n} f(2 \cdot \frac{k}{n}) \text{ instead of } \frac{1}{n} f(\frac{k}{n})$

51. myininaya

yep

52. myininaya

like are answer will look different but it will still hold the same value

53. myininaya

our*

54. myininaya

$\text{ so we want } \frac{2}{n} f(\frac{2k}{n})=\frac{1}{k+5n} \arctan(\frac{k+2n}{k+n})$

55. myininaya

Solve for f(2k/n) by multiplying both sides by n/2

56. RobertSn

son/2 (k +5n)?

57. myininaya

$f(\frac{2k }{n})=\frac{n}{2k+10n} \arctan(\frac{k+2n}{k+n})$

58. RobertSn

Okay yea , and then how do you deal with the 2k/n?

59. myininaya

Now we want to know what f(x) is right?

60. RobertSn

right

61. myininaya

so what is k if 2k/n=x

62. 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

63. RobertSn

Okay so we bring in an X ?

64. RobertSn

xn/2=k

65. myininaya

because we are trying to find out what integral notation looks like for this summation notation

66. myininaya

looks good so we will replace all the k's we see with that

67. 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

68. myininaya
69. myininaya

basically you get to choose a and b

70. myininaya

and nothing else

71. RobertSn

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

72. myininaya

This is just some applications to the definition of reimann sums let me see if i can find you some problems one moment

73. 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

74. myininaya

as*

75. myininaya

they choose a and b for you

76. RobertSn

Alright, thanks!

77. myininaya

there c_i represents a+i*delta x by the way

78. myininaya

If you get bored of those attempt this one I made up: Write this as an integral: $\lim_{n \rightarrow \infty} \sum_{i=0}^{n-1} \frac{i}{i+8n} \cos(\frac{i+3n}{n})$

79. eliassaab

Try practicing on http://www.saab.org/calculus.cgi under Select What Kind Of Problems select Calculus I( Integrals(Substitution, FTC)