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jennychan12

  • 2 years ago

Integral Question. See below,

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  1. Dido525
    • 2 years ago
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    O_o .

  2. jennychan12
    • 2 years ago
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    If \[\int\limits_{0}^{6} (x^2-2x+2)dx\] is approximated by 3 inscribed rectangles of equal width, then the approximation is... 1) 24 B) 26 C) 28 D)76 E) 48

  3. jennychan12
    • 2 years ago
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    sorry typed something wrong. and that 1 is supposed to be an A

  4. jennychan12
    • 2 years ago
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    so i got the integrated area to be 48, so answers either a b or c

  5. Dido525
    • 2 years ago
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    Wait.

  6. Dido525
    • 2 years ago
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    Wait, what kind of rectangles? Left, right, or midpoint?

  7. jennychan12
    • 2 years ago
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    the problem doesnt say. i'm assuming left

  8. jennychan12
    • 2 years ago
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    cuz it says inscribed

  9. Dido525
    • 2 years ago
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    Well you are right. It's either A, B or C :D . Now can you do it out? :) .

  10. Dido525
    • 2 years ago
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    \[\Delta(x)=\frac{ 6-0 }{ 3 }\]

  11. jennychan12
    • 2 years ago
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    i'm confused my friend says to use riemann sum. what's the formula for that?

  12. Dido525
    • 2 years ago
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    \[\Delta(x) = 2\]

  13. Dido525
    • 2 years ago
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    In this particular example, a Riemann sum is basically the sum of the individual rectangles.

  14. Dido525
    • 2 years ago
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    When you learn about definite intergals you will have a good idea about what they actually mean.

  15. jennychan12
    • 2 years ago
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    is it \[\frac{ b-a }{ n } \sum_{i = 1}^{n} f(x) \] ??

  16. Dido525
    • 2 years ago
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    No wait, that's right.

  17. Dido525
    • 2 years ago
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    So yeah you get the idea. It's the sum of all the rectangles multiples by the change in x,(i.e the width) .

  18. jennychan12
    • 2 years ago
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    that's what i did but i got 28 and the answer's 26

  19. Dido525
    • 2 years ago
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    Then you have to use diffrent kinds of rectangles because your answer is correct.

  20. jennychan12
    • 2 years ago
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    26 or 28 ?

  21. Dido525
    • 2 years ago
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    It's 26.

  22. Dido525
    • 2 years ago
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    But if you use left rectangles then it's 28.

  23. jennychan12
    • 2 years ago
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    why? sorry can u explain?

  24. jennychan12
    • 2 years ago
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    oh.

  25. jennychan12
    • 2 years ago
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    so right hand it's 26 or..?

  26. Dido525
    • 2 years ago
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    So this is a very stupid question because it dosen't specify the type of rectangles.

  27. Dido525
    • 2 years ago
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    Well right hand is 76 actually.

  28. Dido525
    • 2 years ago
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    But it dosen't specify what type of rectangle. Left, right or midpoint.

  29. jennychan12
    • 2 years ago
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    where'd the 26 come from? midpoint?

  30. Dido525
    • 2 years ago
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    No, midpoint is 46.

  31. Dido525
    • 2 years ago
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    But shall I explain left and right rectangles?

  32. Dido525
    • 2 years ago
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    If you use left, you get 28. If you use right you get 76.

  33. Dido525
    • 2 years ago
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    Shall I explain?

  34. jennychan12
    • 2 years ago
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    wait so where did the 26 come from. please explain. i'm so lost. :(

  35. Dido525
    • 2 years ago
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    26 isn't the correct answer.

  36. jennychan12
    • 2 years ago
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    cuz this other person used 2(f(1)+f(2)+f(4)) and got 26 but i don't understand where the f(1) came from...

  37. jennychan12
    • 2 years ago
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    oh.

  38. Dido525
    • 2 years ago
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    Wait hold on.

  39. Dido525
    • 2 years ago
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    Yeah, 28 should be correct. I disagree with the answer.

  40. Dido525
    • 2 years ago
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    If you use 3 subintervals you CANNOT get 26 in anyway.

  41. Dido525
    • 2 years ago
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    The formula would be (2)(f(0)+f(2)+f(4))

  42. jennychan12
    • 2 years ago
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    yeah i did that... and got 28

  43. Dido525
    • 2 years ago
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    Which is absolutely correct.

  44. Dido525
    • 2 years ago
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    It CANNOT be 26 if you use left rectangles.

  45. Dido525
    • 2 years ago
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    if you use right rectangles it's (2)(f(2)+f(4)+f(6)) which is 76.

  46. jennychan12
    • 2 years ago
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    yeah i got 28 and 76 but i chose 28. so my teacher is wrong. hmph.

  47. Dido525
    • 2 years ago
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    Yep, you teacher is wrong.

  48. jennychan12
    • 2 years ago
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    thanks for clarifying. :)

  49. Dido525
    • 2 years ago
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    In reality the exact value of that integral is 48 but you will learn that later.

  50. jennychan12
    • 2 years ago
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    yeah i know. i calculated it.

  51. Dido525
    • 2 years ago
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    Kk.

  52. yun2thejae
    • 2 years ago
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    Good 'ol Riemann's screwing with everybody's head.

  53. Dido525
    • 2 years ago
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    I just hope I don't have to do it after my first year in uni :/ . AP calc helped so much though XD .

  54. jennychan12
    • 2 years ago
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    ohhhhhhh i think i got it they used average value

  55. yun2thejae
    • 2 years ago
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    LOL

  56. oldrin.bataku
    • 2 years ago
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    $$A\simeq2(2^2-2(2)+2+4^2-2(4)+2+6^2-2(6)+2)\\\ \ \ =2(4-4+2+16-8+2+36-12+2)\\\ \ \ =2(38)=76$$ (using right end-points)$$A\simeq2(0^2-2(0)+2+2^2-2(2)+2+4^2-2(4)+2)=2(2+4-4+2+16-8+2)\\\ \ \ =2(14)=28$$ (using left end-points)$$A\simeq2(1^2-2(1)+2+3^2-2(3)+2+5^2-2(5)+2)\\\ \ \ =2(1-2+2+9-6+2+25-20+2)\\\ \ \ =2(13)=26$$ (using midpoints)

  57. Dido525
    • 2 years ago
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    ^ Reason why they they should specify the question.

  58. jennychan12
    • 2 years ago
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    my friend just assumed that inscribed meant left end points

  59. Dido525
    • 2 years ago
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    @oldrin.bataku : Please don't just give the answer next time. The point of this place is to help other people learn. Not give them the answer.

  60. Dido525
    • 2 years ago
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    @jennychan12 : That's a flase assumption.

  61. Dido525
    • 2 years ago
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    false*

  62. Dido525
    • 2 years ago
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    @oldrin.bataku : If you use midpoints, it's 46 not 26.

  63. yun2thejae
    • 2 years ago
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    Lol Oldrin and i have been spending plenty of time explaining problems to different users. I would give him a break xD. I was bored and decided to use this site for the first time today and, oddly enough, I can't stop attempting to solve problems. Probably because I don't want to forget my math whilst im on winter break.

  64. Dido525
    • 2 years ago
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    I know but it's just that's it's against the TOS so I try not to. I make that mistake too. It's cool ^_^ .

  65. oldrin.bataku
    • 2 years ago
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    @Dido525 you're right, 2(5) = 10 so it's actually 2(23) = 46 for the approximate area.

  66. oldrin.bataku
    • 2 years ago
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    If it requires inscribed rectangles then the approximation must use left end-point... mid- and right- will not be entirely under the parabola.

  67. Dido525
    • 2 years ago
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    That's why we have the trapozoidal rule ;) . Ans simpsons rule :P .

  68. Dido525
    • 2 years ago
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    and*

  69. Dido525
    • 2 years ago
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    But yes @oldrin.bataku is correct. If they MUST be inscribed then you MUSt use left rectangles.

  70. Dido525
    • 2 years ago
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    MUST*

  71. oldrin.bataku
    • 2 years ago
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    @Dido525 if you go into a Scientific Computation course you'll learn about plenty more numerical methods e.g. Runge-Kutta or Adams-Moulton :-p Euler's and Verlet are really common for physics problems.

  72. yun2thejae
    • 2 years ago
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    Conversation is too much for this PoSci major xD.

  73. Dido525
    • 2 years ago
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    I looked at Runge-kutta but it makes no sense XD . I am going much deeper into math next year so hopefully then :) .

  74. Dido525
    • 2 years ago
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    Sorry that we turned this page into a discussion @jennychan12 .

  75. jennychan12
    • 2 years ago
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    it's aiight.

  76. agent0smith
    • 2 years ago
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    Nobody drew a damn diagram :P

  77. Dido525
    • 2 years ago
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    Sigh...

  78. yun2thejae
    • 2 years ago
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    Diagram was drawn in the other thread.

  79. jennychan12
    • 2 years ago
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    ^ haha yes

  80. Dido525
    • 2 years ago
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    |dw:1355899568667:dw|

  81. Dido525
    • 2 years ago
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    Something like that.

  82. jennychan12
    • 2 years ago
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    if u wanna see http://openstudy.com/study#/updates/50d15940e4b069abbb70d3c0

  83. agent0smith
    • 2 years ago
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    that's the same thread as this one...

  84. yun2thejae
    • 2 years ago
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    LOL

  85. Dido525
    • 2 years ago
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    |dw:1355899651047:dw| That would be incorrect since it does not enclose the curve.

  86. jennychan12
    • 2 years ago
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    yeah i got really confused and tried to do another one to resolve my confusion... :D cuz that other person made me confused

  87. oldrin.bataku
    • 2 years ago
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    |dw:1355899646424:dw|

  88. Dido525
    • 2 years ago
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    |dw:1355899717852:dw|

  89. Dido525
    • 2 years ago
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    If you use midpoints it still dosen't enclose the curve fully.

  90. agent0smith
    • 2 years ago
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    Inscribed rectangles *should* mean rectangles within the curve, but it sounds like that didn't work, so the damn question is broken.

  91. Dido525
    • 2 years ago
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    Damn question.

  92. Dido525
    • 2 years ago
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    It's is 28 though without a doubt.

  93. oldrin.bataku
    • 2 years ago
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    More than likely your teacher made an error.

  94. jennychan12
    • 2 years ago
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    yeah

  95. agent0smith
    • 2 years ago
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    inscribed rectangles would give 26, but i feel like that was already covered in this mess somewhere

  96. yun2thejae
    • 2 years ago
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    You haven't done "work" problems in your calc class yet have you?

  97. agent0smith
    • 2 years ago
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    Wait, why isnt the answer 26? 3 inscribed rectangles = area of 26.

  98. jennychan12
    • 2 years ago
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    what do u mean? kinetic energy? i do that in physics....

  99. jennychan12
    • 2 years ago
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    @agent0smith can u explain???

  100. yun2thejae
    • 2 years ago
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    there are "work" problems you use calculus for. Pray that you do not have to do them xD

  101. Dido525
    • 2 years ago
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    If you use 3 inscribed rectangles it's 28 not 26.

  102. yun2thejae
    • 2 years ago
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    The previous thread came up with 26. this one says 28. let's meet half way and say it's 27...oh wait.

  103. jennychan12
    • 2 years ago
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    lol

  104. agent0smith
    • 2 years ago
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    The height of the first is 1, length is 2 height of the second is 2, length is 2 height of the third is 10, length is 2. 2+4+20

  105. agent0smith
    • 2 years ago
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    How'd you get 28, @Dido525

  106. jennychan12
    • 2 years ago
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    left endpoint is 28

  107. Dido525
    • 2 years ago
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    =(2) * ( f(0) + f(2) + f(4) ) = (2) * ( 2 + 2 + 10 ) = 28

  108. agent0smith
    • 2 years ago
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    Inscribed have to be all within the curve, no edges jutting out: it's not f(0), it's f(1), the height of the first rectangle.

  109. agent0smith
    • 2 years ago
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    That is not an inscribed rectangle.

  110. jennychan12
    • 2 years ago
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    that's what that other person did in that other thread 2(f(1)+f(2)+f(4)) = 26 but why would be f(1)??? where does the 1 even come from?

  111. Dido525
    • 2 years ago
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    But f(0) is inscribed.

  112. agent0smith
    • 2 years ago
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    I'll draw it jenny. One sec.

  113. Dido525
    • 2 years ago
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    I want to see this too.

  114. agent0smith
    • 2 years ago
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    dido, look at the graph: http://www.google.com/search?q=x%5E2-2x%2B2&aq=f&oq=x%5E2-2x%2B2&sugexp=chrome,mod=18&sourceid=chrome&ie=UTF-8 and then look where a rectangle at f(0) would fall... it's not inscribed. it goes over the curve.

  115. Dido525
    • 2 years ago
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    Isn't that the same thing as inscribed?

  116. Dido525
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    Inscribed meaning within.

  117. agent0smith
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    No, inscribed means it falls completely within the curve, no jutting out over it.

  118. jennychan12
    • 2 years ago
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    okay. i understand that f(1) = 1 but.... oh wait. i see where you're going with this....

  119. Dido525
    • 2 years ago
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    |dw:1355900439887:dw|

  120. Dido525
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    But rectangles will never cover a curve fully, it's impossible.

  121. Dido525
    • 2 years ago
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    Like, it covers it fully but it's an overestimate.

  122. agent0smith
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  123. Dido525
    • 2 years ago
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    Ohh shoot! O_o .

  124. agent0smith
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    Those are inscribed rectangles, they underestimate the area greatly of course.

  125. Dido525
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    Ahh, NOW this makes sense.

  126. agent0smith
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    hahah :)

  127. Dido525
    • 2 years ago
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    But if it's an underestimate, it's not inscribed is it?

  128. jennychan12
    • 2 years ago
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    ok yeah i see it too.

  129. agent0smith
    • 2 years ago
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    I googled inscribed rectangles just to be sure.

  130. Dido525
    • 2 years ago
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    Wait, why did you do a midpoint for the first rectangle?

  131. Dido525
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    Instead of a left one?

  132. agent0smith
    • 2 years ago
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    Because that's the only way to have it inscribed. http://en.wikipedia.org/wiki/Inscribed_figure

  133. Dido525
    • 2 years ago
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    NOW this all makes sense :D .

  134. agent0smith
    • 2 years ago
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    Inscribed rectangles can be either left or right endpoint

  135. agent0smith
    • 2 years ago
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    or midpoint

  136. agent0smith
    • 2 years ago
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    @jennychan12 you see how it's 26 now? the height of the first is f(1), next is f(2), then f(4)

  137. Dido525
    • 2 years ago
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    I would give you my medal if I could :P .

  138. agent0smith
    • 2 years ago
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    haha :) this whole confusion was all over the meaning of the word inscribed :D

  139. Dido525
    • 2 years ago
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    Agreed! :D . But hey, this might be n my final so it was good for me too :) .

  140. agent0smith
    • 2 years ago
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    haha good :) hopefully this clears things up Jenny, before you went and told your teacher he/she was wrong because a bunch of people on the internet said so! :D

  141. Dido525
    • 2 years ago
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    She is offline :/ .

  142. Dido525
    • 2 years ago
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    @jada1701877 : You should have given that to agent.

  143. jennychan12
    • 2 years ago
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    thanks so much! and yeah, i went off at like 11 cuz i went to sleep. :)

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