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Errors in Left and Right Approximation in Calculus?

  1. Feb 13, 2012 #1
    Moderator's note: This thread is a perfect example of what not to do in the homework help forums. It is unacceptable for the opening poster not to work through the problem and to demand answers. It is inappropriate for the helper to give out those answers, or tell the poster exactly what to do. If it were not already too late, I would have closed/deleted this thread. Since it is too late, I'm leaving it up with this warning that it's an example of what not to do.

    1. The problem statement, all variables and given/known data
    How do you calculate an error using left and right approx? How do you find the exact answer to that integral?
    Here's the homework question.
    How many subdivisions would you need to use so that the error in the left and right approximations of the function f(x)=x^3 from x=0 to 10 was 0. 01?
    So, I know how to calculate the area of right and left but how do I find the real answers first?
    And how do i make the error equal to 0.01?
    Thanks. Calculus is hard so please help.


    2. Relevant equations
    f(x)=x^3 from x=0 to 10

    3. The attempt at a solution
    here is the calculation of my left and right approximation using 10 subdivisions.
    L10= f(0)+f(1)+f(2)+f(3)+f(4)+f(5)+f(6)+f(7)+f(8)+f(9)
    = 0+ .25+ 4+20.25+ 64+156.25+324+600.25+1042+1640.3
    =3851.3
    R10= f(1)+f(2)+f(3)+f(4)+f(5)+f(6)+f(7)+f(8)+f(9)+f(10)
    .25+ 4+20.25+ 64+156.25+324+600.25+1042+1640.3+ 2500
    =6351.3
    if you get the average of the two you have 3400.87
    so, Now what?
     
    Last edited by a moderator: Feb 13, 2012
  2. jcsd
  3. Feb 13, 2012 #2

    lanedance

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    say you subdivide into n equal intervals, the approximations are
    Left
    [tex]L = \sum_{i=0}^{n-1}f(x_i)\Delta x[/tex]

    Right
    [tex]R = \sum_{i=1}^{n}f(x_i)\Delta x [/tex]

    Now in this case as x^3 is concave up you know that
    [tex]L < \int_0^{10}x^3dx < R [/tex]

    So the error in each approximation is less than R-L, if we use that as an approximation of the error or at least as a bound on it, then we get
    [tex]e < R-L = \sum_{i=1}^{n}f(x_i)\Delta x-\sum_{i=0}^{n-1}f(x_i)\Delta x = (f(x_n) -f(x_0))\Delta x [/tex]
     
  4. Feb 13, 2012 #3
    What???I still don't understand!!!we havent learned those equations yet.
     
  5. Feb 13, 2012 #4
    What does "e" mean? just please tell me instructions on how to get the answer. please
     
  6. Feb 13, 2012 #5

    lanedance

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    i pretty much have, but note we don't tell you how to get an answer here, but rather help you find it yourself

    the formula is just a quick way of writing what you are attempting, so consider you L10, this has n=10 subdivsions with [itex] \Delta x = \frac{10}{10}= 1[/itex], whilst the x_i is the value of x at the left side of a subdivision
    [tex]
    L_{10}= f(0)+f(1)+f(2)+f(3)+f(4)+f(5)+f(6)+f(7)+f(8)+f(9)
    [/tex]
    [tex]
    = f(0).1+f(1).1+f(2)+f(3).1+f(4).1+f(5).1+f(6).1+f(7).1+f(8).1+f(9).1
    [/tex]
    [tex]
    = \sum_{i=0}^{9} f(x_i).1
    [/tex]
    [tex]
    = \sum_{i=0}^{10-1} f(x_i) \Delta x
    [/tex]

    whihc is exactly what we had above with n=10

    As for the "e" in the above post this is an upper bound for your error
     
  7. Feb 13, 2012 #6
    I still don't get it. Can you just please tell me how to find the ACTUAL answer for the f(x) x^3 from 0 10? i will just find the error myself.
     
  8. Feb 13, 2012 #7

    lanedance

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    to desribe "e" better, assume we kow the value of the integral
    [tex] I = \int_0^{10}x^3dx[/tex]

    Now for a given n subdivsions as discussed before we know
    [tex] L(n)< I< R(n)[/tex]

    consider the error in the right approximation
    [tex] e_R(n) = R(n) - I >0 [/tex]

    consider the error in the right approximation
    [tex] e_L(n) = I-L(n)>0 [/tex]

    note if we sum the error in each
    [tex] e_L(n) + e_R(n) = R(n) - L(n) >0 [/tex]
     
  9. Feb 13, 2012 #8

    lanedance

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    do you mean adctually calulating the integral? It is as follows
    [tex] I = \int_0^{10} x^3 dx = (\frac{x^4}{4})|_0^{10} = (\frac{10^4}{4})-0 =2500 [/tex]
     
  10. Feb 13, 2012 #9

    lanedance

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    I have read back over this and you are not approximating the integral correctly, for example f(1) should be:
    [tex]
    f(1) = 1^3 = 1
    [/tex]

    where as you appear to be using the form of the integral which is not correct
    [tex]
    F(x) = \frac{x^4}{4}
    [/tex]
    [tex]
    F(1) = \frac{1}{4}=0.25
    [/tex]
     
  11. Feb 13, 2012 #10
    You are killing my brain!
    My question is simple. how do u calculate the actual answer?
    Look at this website
    https://docs.google.com/viewer?a=v&...XNBcXP&sig=AHIEtbSxVTswFleaJW05ZNeAXh5TeRCCGg



    Notice that in the second page, there is an integral equation where it got ln 2=0.693 as the answer. That 0.693 is just what I need. And that is the actual answer that I need to compare my errors with.
     
  12. Feb 13, 2012 #11

    lanedance

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    If you want a quick way to do this problem I suggest you try and make sense of mny first post

    Furthermore it sounds like you're a little confused about what the left and right aproximations are, so I would suggest you try and draw the problem close to scale, for say n=5 subdivisions
     
  13. Feb 13, 2012 #12

    lanedance

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    The integral is as follows
    [tex] I = \int_0^{10} x^3 dx = (\frac{x^4}{4})|_0^{10} = (\frac{10^4}{4})-0 =2500 [/tex]

    Apologies as i think we have been one post out of sync, there is some useful stuff in what i have written so suggest you still take a look
     
  14. Feb 13, 2012 #13
    So 2500 is the Actual answer right?
     
  15. Feb 13, 2012 #14

    lanedance

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    correct
     
  16. Feb 13, 2012 #15
    Thank you.
     
  17. Feb 13, 2012 #16

    lanedance

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    no worries going back to post 2, for n equal subdivisons or [itex] \Delta x = \frac{10}{n}[/itex] you know the sum of the absolute error of the right and left approximations is

    [tex]e_L(n) + e_R(n) = (f(10) -f(0))\Delta x = (10^3-0) \frac{10}{n} = \frac{10^4}{n} = \frac{10000}{n}[/tex]

    and in fact as n gets big you can make the assumption through similar triangles that

    [tex]e_L(n) \approx e_R(n) \approx \frac{5000}{n}[/tex]
     
  18. Feb 13, 2012 #17

    lanedance

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    In fact as n gets big you could say e_L(n) ~ e_R(n) and so each one contributes half the error or e_L(n) ~ e_R(n) ~ 5000/n, so its worth comparing this with your estimates
     
  19. Feb 13, 2012 #18
    Wait. I could just use 1000/n to be able to get the error of 0.01?
     
  20. Feb 13, 2012 #19
    Wait. I could just use 1000/n to be able to get the error of 0.01?
     
  21. Feb 13, 2012 #20

    lanedance

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    No, why would you change the number to 1000?

    You could howerver use the formula I prescribed and solve for n
     
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