## anonymous one year ago Riemann Sum...

1. anonymous

2. anonymous

I got as far as figuring out delta x to be 2/n

3. anonymous

And what I preceived the second one to be $(\frac{ 12i ^{2} }{ n ^{2} } + \frac{ 40i }{ n }) (\frac{ 2 }{ n })$

4. anonymous

ah its been so long since ive done this. can you jog my memory when it says the general formulae of $x _{i}$

5. anonymous

You mean the part where $x _{i}=0+i \Delta x$?

6. anonymous

yeah is that what the formulae is for that bit? ah i should know this cause its so fundemental to calc but i only vaguely remember haha

7. anonymous

perhaps @IrishBoy123 can help he's a bit of a god

8. anonymous

Aha, I think so. There was a previous question regarding Riemann Sum but tackling it from the left and the question is a bit different. I could potentially upload that and tell you what I got for if you'd prefer it.

9. anonymous

perhaps that could help but since we have n divisions, then our partitions are 0, $\Delta x, 2\Delta x, ... , (n-2) \Delta x, (n-1) \Delta x,$

10. anonymous

2

11. anonymous

hmm that won't help

12. anonymous

for right hand side, $\Delta x \left[ f( a + \Delta x ) + f(a + 2 \Delta x)+\cdots+f(b) \right].$

13. anonymous

where a=0 b=2

14. anonymous

that is your approx function if you do the right end method

15. anonymous

it looks like they are leading you the way step by step, i'm just unsure what it means by the general formulae of x(i)

16. anonymous

is this like a web this where you enter an answer and you can see if its right or wrong?

17. anonymous

It isn't a web thing, more just like an extended question that the professor gave at the end of lecture. I've been trying to follow a link that my friend shared with me on Facebook. It seems pretty legit despite being a yahoo answer source, but even then as I say that it was hard for me to get my head around the answer: https://answers.yahoo.com/question/index?qid=20110910015823AARrDF4

18. anonymous

sweet, i have something to work with now

19. anonymous

Hope that it'd be more useful to you than it was to me! :P

20. IrishBoy123

https://gyazo.com/ce50525553b52e9dbd904fb7970fc13a i'm with you on this

21. IrishBoy123

if that helps :p

22. anonymous

ok so x(i) is defined as $x _{i}=i \Delta x$

23. IrishBoy123

yes

24. IrishBoy123

$x_i = i \frac{2}{n}$

25. anonymous

you know $\Delta x= \frac{ 2 }{ n }$

26. anonymous

yep!^^

27. anonymous

28. IrishBoy123

i'm pretty rusty on this too so careful with the symbology but i worked it through and got 48 both ways for the integral at least that means we are on track

29. anonymous

haha

30. anonymous

I'm understanding what you two are saying so far so that's good I suppose. :p

31. anonymous

now since you know {x _{i}\]

32. anonymous

$x _{i}$

33. anonymous

sub that into the x variables in the function 3x^2+20x

34. IrishBoy123

@chris00 yes!

35. anonymous

so you get $3x _{i}^2+20x _{i}$

36. anonymous

can you do this long hand @Clarence

37. anonymous

its bringing back memories now ahaha

38. anonymous

you should get $\frac{ 12i^2 }{ n^2 }+\frac{ 60i }{ n }$

39. anonymous

I think I got 40 rather than 60 for that bit...

40. IrishBoy123

$f(x_i) = 3 (\frac{2i}{n})^2 + 20 (\frac{2i}{n})$

41. anonymous

hahah opps

42. IrishBoy123

let's agree it before we move on :-)

43. anonymous

i can math

44. anonymous

:P

45. anonymous

Aha, we all make these mistakes :)

46. IrishBoy123

so we are here, right?? $\frac{ 12i^2 }{ n^2 }+\frac{ 40i }{ n }$

47. anonymous

yeah so $f(x _{i})=\frac{ 12i^2 }{ n^2 }+\frac{ 40i }{ n }$

48. IrishBoy123

yes, i forgot the $$f(x_i)$$ cool next bit

49. anonymous

$f(x _{i}) \Delta x=(\frac{ 12i^2 }{ n^2 }+\frac{ 40i }{ n }) \Delta x$

50. anonymous

and you know what delta x is right clarence?

51. IrishBoy123

exellent!

52. anonymous

Yup, so just input that into the equation to get 2/n at the right end

53. IrishBoy123

54. anonymous

simply sub 2/n into delta x to get: $f(x _{i}) \Delta x=(\frac{ 12i^2 }{ n^2 }+\frac{ 40i }{ n })(\frac{ 2 }{ n })$

55. anonymous

which is what you got!

56. anonymous

Before I got stuck, yes ;)

57. anonymous

now we approximate the area

58. IrishBoy123

my go!! $f(x _{i}) \Delta x=\frac{ 24i^2 }{ n^3 }+\frac{ 80i }{ n^2 }$

59. IrishBoy123

do we agree that?!

60. anonymous

yep!

61. IrishBoy123

cool, now for the real fun bit!

62. anonymous

$A _{app}\approx \sum_{i=0}^{n}(f(x _{i}) \Delta x$

63. anonymous

i think thats by definition...

64. anonymous

|dw:1441541710476:dw| somthing like that. this is not your graph but it just shows that the area approximation of one rectangle is Delta x multiplied by the function governing the rectangle

65. IrishBoy123

yes, we approximate the area by adding up all the little elements $\Sigma_{i = 1}^{\infty} f(x _{i}) \Delta x=\Sigma_{i = 1}^{\infty} \left[ \frac{ 24i^2 }{ n^3 }+ \frac{ 80i }{ n^2 } \right]$ $=\Sigma_{i = 1}^{\infty} \frac{ 24i^2 }{ n^3 }+\Sigma_{i = 1}^{\infty} \frac{ 80i }{ n^2 }$

66. IrishBoy123

and now we use these https://gyazo.com/776b4188ce5e7548a6cad4c61eda1c3a

67. IrishBoy123

$=\frac{24}{n^3}\Sigma_{i = 1}^{\infty} i^2+\frac{ 80 }{ n^2 } \ \Sigma_{i = 1}^{\infty} i$

68. anonymous

beat me too it haha

69. anonymous

should be hunky dory from there @Clarence

70. IrishBoy123

i am latexing like a crazy man here :-))

71. IrishBoy123

@Clarence take your time we can stop here for a while if that suits!

72. anonymous

Never thought that maths could look so pretty... But now I don't even know what to do with the information that I have!

73. anonymous

you are given critical information in the latter part of the question

74. anonymous

$\sum_{i}^{n}i=\frac{ n(n+1) }{ 2 }$

75. anonymous

and $\sum_{i}^{n}i^2=\frac{ n(n+1)(2n+1) }{ 6 }$

76. anonymous

can we not just sub these corresponding summations into our question?

77. IrishBoy123

yes we can!

78. anonymous

where @IrishBoy123 mentioned that $A=\frac{24}{n^3}\Sigma_{i = 1}^{\infty} i^2+\frac{ 80 }{ n^2 } \ \Sigma_{i = 1}^{\infty} i$

79. anonymous

or not to be confused with the infinity clarence, $A=\frac{24}{n^3}\Sigma_{i = 1}^{n} i^2+\frac{ 80 }{ n^2 } \ \Sigma_{i = 1}^{n} i$

80. anonymous

see the relation?

81. anonymous

Aha, yes, I can see that now! Sorry, totally forgot about the original question, I'm not the brightest freshman about if you couldn't already tell :p

82. IrishBoy123

@Clarence if you're doing this stuff, you *are* very bright this is amazing maths as we shall see at the end!!

83. anonymous

so fundamental to calculus.

84. anonymous

can you solve the area now clarence?

85. anonymous

Maybe... :p

86. anonymous

its just simple substitution

87. IrishBoy123

this is a pig latex-wise shall i just do it? it gets easier when we lose all the terms

88. anonymous

go for it!

89. IrishBoy123

ok give me a few minutes to copy and paste it all together

90. IrishBoy123

$A=\frac{24}{n^3} \frac{ n(n+1)(2n+1) }{ 6 }+\frac{ 80 }{ n^2 } \ \frac{ n(n+1) }{ 2 }$ this good?

91. anonymous

That's what I got

92. IrishBoy123

excellent so we now have a formula that approximates the area under the curve. we choose n as the number of elements and we plug it in and find the area let's simplify it remembering we want to set $$n = \infty$$ to get the perfect answer

93. IrishBoy123

$A=\frac{24}{n^3} \frac{ 2n^3 + 3n^2 +n}{ 6 }+\frac{ 80 }{ n^2 } \ \frac{ n^2 +n }{ 2 }$

94. IrishBoy123

everyone agree?

95. anonymous

yep

96. anonymous

Yeah, I got up to that, so far :p

97. IrishBoy123

$A=4 (2 + \frac{3}{n} +\frac{1}{n^2})+40 (1 +\frac{1}{n} )$

98. anonymous

good so far

99. anonymous

@Clarence what happens now if n approaches infinity?

100. IrishBoy123

some more simplifying $A= 48 + \frac{52}{n} +\frac{4}{n^2}$

101. anonymous

haha more simplifying, love it

102. anonymous

103. IrishBoy123

yep! totally unnecesary of course as we're setting n to infinity :-((

104. anonymous

Aha, maths sure loves their infinities :p

105. IrishBoy123

$\int_{0}^{2} \ 3x^2 + 20x \ dx$

106. IrishBoy123

if that's not 48, we are all doomed is it?

107. IrishBoy123

108. anonymous

I got 48 for that..

109. anonymous

sweet as

110. anonymous

what a journey

111. anonymous

:P

112. IrishBoy123

beautiful beautiful stuff thank you both for involving me, really enjoyed it @Clarence are you good to go with this now??

113. anonymous

such a pleasure

114. anonymous

Aha, I think so :)

115. IrishBoy123

if not keep the thread open @chris00 and i will get a notification if there is more to do

116. anonymous

Awesome, thanks! :)

117. amistre64

it appears to me that you ran thru the LEFT side rundown of this

118. amistre64

on an interval from a to b, the left side setup is: $x_i=a+i\frac{b-a}{n}~:~0\le i<n$ this hits all the high points starting at the left end and moving in 'i'ncrements of the partition widths. in this case a=0 your inputs worked out. the right side setup is either an adjustment of the interval of i: 0 < i <= n this way we simply hit the right side of the partitions.

119. amistre64

another alternative to the right side setup is simply to replace the a+ with b-

120. amistre64

hmmm, ok, it was a hard read thru for these tired old eyes, but it does look like you implemented i going from 1 to n, so my observation is moot :) good job yall

121. IrishBoy123

its a really long thread but we used $$x_i = i \frac{2}{n}$$ (and $$\Delta x= \frac{ 2 }{ n }$$) and ran it from i=1. so for n = i = 1 we are up at the right hand end.

122. IrishBoy123

and we were spoon-fed because the question was so well laid out....:-))

123. amistre64

if a had been say 3 (not 0),and b was 5 (not 2), our interval width is still 5-3=2 and we divide that into n partitions.$\Delta x=\frac2n$ then the value of x for the ith iterations would have been: $x_i=3+i\frac2n$ and $f(x_i)=f(3+i\frac2n)$ just wanting to make sure that we remember that xi is not in general the same as: i /\x

124. IrishBoy123

gotcha. that particular *snag* was missing from our problem here. we had $f(x_i)=f(0+i\frac2n) = f(i\frac2n)$ thank you @amistre.