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Homework Help: Trapazoidal Approximation help

  1. May 2, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex] \int^{5}_{1} \frac{x}{x+1} dx [/tex]

    using dx = .5

    2. Relevant equations

    [tex] \sum^{a}_{b} \frac {f(x)+f(x+dx)}{2} dx = [\frac{1}{2}f(x_{0})+f(x_{1})+\cdots+f(x_{n-1})+\frac{1}{2}f(x_{n})]dx [/tex]

    3. The attempt at a solution
    TAI = Trapazoidal Approximation Input value

    [tex] x_{0}=1 \rightarrow f(x_{0})=\frac {1}{2} \rightarrow TAI = \frac{1}{2}\frac{1}{2} =\frac{1}{4} [/tex]

    [tex] x_{1} = 1.5 \rightarrow f(x_{1}) = \frac{1.5}{2.5} \rightarrow TAI = \frac{1.5}{2.5} [/tex]

    [tex] x_{2} = 2 \rightarrow f(x_{2}) = \frac{2}{3} \rightarrow TAI = \frac{2}{3}[/tex]

    [tex] x_{3} = 2.5 \rightarrow f(x_{3})= \frac{2.5}{3.5} \rightarrow TAI = \frac {2.5}{3.5} [/tex]

    [tex] x_{4} = 3 \rightarrow f(x_{4}) = \frac{4}{5} \rightarrow TAI = \frac {4}{5} [/tex]

    [tex] x_{5} = 3.5 \rightarrow f(x_{5})= \frac{3.5}{4.5} \rightarrow TAI = \frac {3.5}{4.5} [/tex]

    [tex] x_{6} = 4 \rightarrow f(x_{6}) = \frac {4}{5} \rightarrow TAI = \frac {4}{5} [/tex]

    [tex] x_{7} = 4.5 \rightarrow f(x_{7}) = \frac{4.5}{5.5} \rightarrow TAI = \frac {4.5}{5.5} [/tex]

    [tex] x_{8} = 5 \rightarrow f(x_{8}) = \frac {5}{6} \rightarrow TAI = \frac{1}{2}\frac{5}{6} = \frac {5}{12} [/tex]

    so now, following the formula I add up all the TAI values, and the 1/2's have already been applied, so:

    [tex] \frac{1}{4} + \frac {1.5}{2.5} + \frac{2}{3} + \frac{2.5}{3.5} + \frac {3}{4}+ \frac{3.5}{4.5} + \frac {4}{5} + \frac {4.5}{5.5} + \frac {5}{12} = 5.843578644 = \frac{20248}{3465} [/tex]

    and then I multiply the sum by dx which is .5:

    [tex] \frac {20248}{3465} * \frac {1}{2} = 2.921789 [/tex]

    the book states that the answer they are looking for is 2.8968

    and for E(dx) (error) I take the second derivative:

    [tex] y = \frac {x}{x+1} [/tex]

    [tex] y' = \frac {(x+1)(1)-(x)(1)}{(x+1)^{2}} [/tex]

    [tex] y' = \frac {x+1-x}{(x+1)^{2}} [/tex]

    [tex] y' = \frac {1}{(x+1)^{2}} [/tex]

    which the same as:

    [tex] y' = (x+1)^{-2} [/tex]

    so:

    [tex] y''= -2(x+1)^{-3} [/tex]

    and the formula for error is:

    [tex] |f''(x)| \leq M [/tex] for [tex] a \leq x \leq b [/tex]

    and:

    [tex] E(dx)= \frac {b-a}{12} M (dx)^{2} [/tex]

    so now:

    [tex] |f''(1)| = .25 [/tex]

    [tex] |f''(5)| = .00926 [/tex]

    so f'' has M at 1 so to find error we do:

    [tex] \frac {5-1}{12}(.25)(.25)^{2} = \frac {1}{192} [/tex]

    the book states that the correct E(dx) is [tex] \frac {1}{48} [/tex]

    I have done this many times over and can not find where I went wrong, especially with the actually summing in the first part.
     
  2. jcsd
  3. May 2, 2010 #2
    For x4 you put down 4/5 instead of 3/4. You wrote the right thing in the horizontal addition, but the sum you come up with is the sum including 4/5 instead of 3/4.

    In the error part, your dx is 0.5, not 0.25.
     
  4. May 3, 2010 #3
    hmm i just mixed up on the post with the 4/5 instead of 3/4, thats what I have written on the paper but i just did the problem again and it all works out, i wonder why i couldnt get it to come out before.... and the error, yeah i got mixed up, thanks for the help!
     
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