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Midpoint integral approx.

  1. Feb 10, 2005 #1
    (a) Use the Midpoint Rule and the given data to estimate the value of the integral [tex]\int _0 ^{3.5} f(x)\: dx[/tex]

    [tex]x=0.0\quad 0.4\quad 0.8\quad 1.2\quad 1.6\quad 2.0\quad 2.4\quad 2.8\quad 3.2[/tex]

    [tex]f(x)=6.8\quad 6.5\quad 6.3\quad 6.4\quad 6.9\quad 7.6\quad 8.4\quad 8.8\quad 9.0[/tex]

    (b) If it is known that [tex]-4 \leq f^{\prime \prime} (x) \leq 1[/tex] for all [tex]x[/tex], estimate the error involved in the approximation in part (a).

    The answers given in my textbook are:

    (a) [tex]23.44[/tex]
    (b) [tex]0.341\overline{3}[/tex]

    Anyhow, this is what I have:

    [tex]\bar{x}_i=0.2\quad 0.6\quad 1.0\quad 1.4\quad 1.8\quad 2.2\quad 2.6\quad 3.0[/tex]

    Then, I simply figured out the arithmetic mean of subsequent values of [tex]f(x)[/tex] (taken two at a time)

    [tex]f(\bar{x}_i)=6.65\quad 6.4\quad 6.35\quad 6.65\quad 7.25\quad 8.0\quad 8.6\quad 8.9[/tex]

    which makes it possible to obtain

    [tex]M_8 = \Delta x \sum _{i=1} ^{8} f(\bar{x}_i) = \frac{3.2-0}{8}(6.65+ 6.4+ 6.35+ 6.65+ 7.25+ 8.0+ 8.6+ 8.9) = 23.52[/tex]

    For part (b), I just figured out the error bound

    [tex]\left| E_M \right| \leq \frac{K(b-a)^3}{24n^2}[/tex]

    We are given [tex]-4 \leq f^{\prime \prime} (x) \leq 1[/tex], then [tex]\left| f^{\prime \prime} (x) \right| \leq 4 = K[/tex]. So, it follows

    [tex]\left| E_M \right| \leq \frac{4(3.2-0)^3}{24(8)^2} = 0.085\overline{3}[/tex]

    As you can see, my results are a bit different. In part (a), the author of my textbook could have applied some weighted average (I really don't know!!!). In part (b), there is nothing different I could do (so I wonder what is wrong with my calculations).

    Any help is highly appreciated. :smile:
    Last edited: Feb 10, 2005
  2. jcsd
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