## anonymous 5 years ago Find the value of the following definite integral using the limit of a Riemann sum, and express as a common fraction. You should use the formulas for ∑i, ∑i^2, and ∑i^3 given in class. S(Integral from -2 to 1) (x^3-3x)dx

1. anonymous

The actual question is attached

2. anonymous

i attached the actual question

3. amistre64

i see it... so what are the 3 summation formulas?

4. anonymous

let me get them from my notes

5. amistre64

if I recall correctly, the Reiman limits are just taking the area of slices as the thickness of dx goes to zero right?

6. anonymous

yes and i she didnt give us the formulas even though she said she did, let me check the book.

7. amistre64

(f(x+dx) - f(x)) dx; we could also take the "average height" between f(x+dx) and f(x)...

8. amistre64

dx just stands for delta x, any arbitrary number... where dx = (b-a)/n... any of this make sense?

9. anonymous

here is the formulas

10. amistre64

dx = (1--2)/n = (1+2)/n = 3/n

11. anonymous

k i got that far

12. anonymous

a=-2 b=1

13. amistre64

yep, them pictures is what I said lol

14. amistre64

the limit as n-> inf......

15. anonymous

so then I got lim as n goes to infinity of the summation [(-2+3i/n)^3-3(-2+3i/n)](3/n)

16. amistre64

need to work out something like this right?

17. anonymous

yea but that one is less complicated haha

18. amistre64

lol .... use the Esummation for the cubic function

19. anonymous

the answer we are going for is 3/4 for reference

20. anonymous

is my first step correct?

21. amistre64

we can split the addition up into to parts just like regualr integration, because regular integration is the result of this reimann sum stuff, so do x^3 and x sepertaely

22. anonymous

ok

23. amistre64

[E] x^3 dx - [E] 3x dx the 3 can move out [E] x^3 (3/n) - 3 [E] x (3/n)

24. anonymous

$\lim_{n \rightarrow \infty}\sum_{i=1}^{n}[(-2+3i/n)^3-3(-2+3i/n)](3/n)$

25. amistre64

im lost on what that (-2+3i/n) stuff means, help me out :)

26. anonymous

its part of the reimann sum

27. anonymous

the formulas we have to use, that is the n^3

28. anonymous

i^3 my mistake

29. amistre64

$\sum_{} x^{3}(3/n) -3 \sum_{}x(3/n)$ is whats in my head

30. anonymous

well the integral is $\int\limits_{-2}^{1}(x^3-3x)dx$

31. amistre64

yep, and the reiman is proof that we can split that into 2 parts right? $\int\limits_{} x^3 dx -3\int\limits_{}x dx$

32. anonymous

yes that works

33. amistre64

∑ x^3 (3/n)−3 ∑ x (3/n) is its equivalent, tell me if im wrong... but whats our next step after this?

34. amistre64

lets work each one on its own...ok

35. anonymous

plug in x^3 and x

36. anonymous

and thats what i showed above

37. anonymous

as one complete summation multiplied by 3/n

38. amistre64

you say "plug in" but i have no idea how that happens, im pretty much an idiot when it comes to the limit stuff, show me step by step please how we "plug in" to get: lim n→∞ ∑ i=1 n [(−2+3i/n)^3 −3(−2+3i/n)](3/n)

39. anonymous

k let me show you

40. anonymous

hopefully this helps

41. anonymous

i need to break it even further for you

42. amistre64

lol.... well, I got that part :) How do we move from that to the "formula for" [E] i^3 and [E] i I am not familiar with those formulas, yet :)

43. anonymous

its just pulling things out and taking individual summations, should i show you where i am so far?

44. amistre64

yes please ∑ i^3 = (1/4)n4 + (1/2)n3 + (1/4)n2 is what I found online

45. anonymous

$\lim_{n \rightarrow \infty}\sum_{i=1}^{n}3/n[-8+12i/n+12i/n-18i^2/n^2+12i/n-18i^2/n^2-18i^2/n^2-27i^3/n^3-3(-2+3i/n)$

46. anonymous

wow i guess my thing was too long, thats only half of it

47. anonymous

im so confused lol

48. amistre64

:) heres what I found that helps me out

49. anonymous

ok that works, lets start over then from 3/n being delta x

50. amistre64

since 3/n = delta x, we simply multiply that to x^3? to get: 3(x^3)/n ?? is that right?

51. anonymous

hmm sure, keep going because i wont know where you are going till you go further

52. amistre64

$3\sum_{n=1}^{\infty} (x^3)/n$ Does this look right?

53. anonymous

no but keep going

54. amistre64

lol ....if it dont look right, then why should I keep going :)

55. anonymous

im trying to make sense of where you are going

56. amistre64

....... you gotta drive some of the way, im lost here for the moment

57. anonymous

oh k , im trying to think...problem is i cant find the equation that would help you

58. anonymous

let me see...

59. anonymous

where are you lost?

60. anonymous

how i formulated the entire sum?

61. anonymous

A=lim_{N→∞}Δx∑_{j=1}^{N}f(a+jΔx)

62. anonymous

does this help

63. anonymous

i=j

64. amistre64

yup.... im still at the split and trying to apply the formula for [E] x^3 to the left part. the top part of the [E] is gonna be (b-a) right? which we determined to be 3.... but its not sinking in....

65. anonymous

66. anonymous

heres a better look

67. amistre64

ok....so delta x is a constant that gets pulled out to the left ... am I seeing that right?

68. anonymous

yes

69. anonymous

but you see how the i is appearing now. In the equation, it names it j, but the common known term is i

70. amistre64

and the "i" is simply each iteration of the sums....right?

71. anonymous

yes

72. anonymous

the integral should =3/4 after the long process, im trying to get through all the steps

73. anonymous

thats the problem

74. amistre64

yeah....when i read over that reaimann stuff I gloosed at it and figured, its true, so why do it the long way :)

75. anonymous

lol cause my professor doesnt want to make life easy

76. amistre64

f(a + i /x\) the i is each iterations from 1 to infinity. is the a the same as in our b-a interval?

77. anonymous

yes

78. anonymous

-2

79. anonymous

limn→∞∑i=1n[(−2+3i/n)3−3(−2+3i/n)](3/n)

80. amistre64

f(-2 + 3i/n) is a standard interation then, is that correct?

81. anonymous

perfect

82. anonymous

dont forget the cubed

83. amistre64

is the "n" from 3/n is the one we are limiting to infinity right?

84. anonymous

yes perfect

85. amistre64

f^3(x)...yeah yeah

86. anonymous

hey if you dont wanna type the dumb sigma and lim n thing, just say lim and sigma, ill know what you are saying or use the equation feature on here

87. anonymous

dumb summation sign i mean*

88. amistre64

n[(−2+3i/n)3−3(−2+3i/n)](3/n) this is you combining the terms together right??

89. anonymous

the n in front shouldnt be there that was part of the format for the summation

90. amistre64

ok...... (sigma) (-2 +3i/n) (3/n) is the right term then correct?

91. anonymous

summation but yes that is correct

92. anonymous

isnt it -3(-2 +3i/n) (3/n)

93. anonymous

you forgot the -3x

94. amistre64

$\sum_{} (-2+3i/n) (3/n)$ would be the $\sum_{} x$ formula part right? the -3 part of the -3x doesnts need to be included in this process if i recall correctly

95. anonymous

im pretty sure you do because its part of the integral function

96. anonymous

but if you are confident, you understand this better then me so ill take your word for it

97. amistre64

$\int\limits_{?}^{?} -3x = -3\int\limits_{?}^{?}x \rightarrow-3 x^2/2$

98. anonymous

well you are going to have a -8 from the cube equation so its better if you keep everything together when you distribute it right?

99. anonymous

i guess it doesnt matter go ahead, you sound excited because i think you understand it now

100. anonymous

do your thing, ill just learn from you

101. amistre64

keep this split up, dont try to solve it all as one summation, work that parts.... use the formulas for [E]x and [E]x^3 seperately

102. anonymous

it still will work out taking as pieces?

103. amistre64

$\sum{}(-2 +3/n)(3/n)$ fits into n(n+1)/2 somehow, and yes, it will work easier and just as good if we split it up

104. amistre64

think of it as the granny way of doing integrals; its actually proof that we can split integerals up and work them

105. amistre64

forgot to pu an i in there :)

106. anonymous

ok let me write step by step following you on actual paper

107. amistre64

we can split derivatives up and work them, integrals are just eh opposite of derivatives.....

108. amistre64

this computer is having a hard time keeping up with formating the "equation"editor stuff, can we start a new question post?

109. anonymous

yea

110. anonymous

let me make a new one

111. anonymous

k go to the top of the question list