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Trapping the exact value of the integral

  1. Feb 9, 2005 #1
    If [tex]f[/tex] is a positive function and [tex]f^{\prime \prime}(x) < 0[/tex] for [tex]a\leq x\leq b[/tex], show that

    [tex]T_n < \int _a ^b f(x)\: dx < M_n[/tex]


    [tex]T_n[/tex] is the Trapezoidal Rule.
    [tex]M_n[/tex] is the Midpoint Rule.

    In my textbook (Calculus: concepts and contexts / James Stewart. --- 2nd ed. --- page 419), there is a figure with the following caption:

    "The area of a typical rectangle in the Midpoint Rule is the same as the trapezoid [tex]ABCD[/tex] whose upper side is tangent to the graph at [tex]P[/tex]. The area of this trapezoid is closer to the area under the graph than is the area of the trapezoid [tex]AQRD[/tex] used in the Trapezoidal Rule."

    In other words, provided that the concavity remains the same, we can trap the exact value of the integral between the trapezoidal and midpoint sums for any number of subdivisions.

    Maybe, it can be used as the answer. However, I think I'd be better to express that in mathematical terms. So, here's what I've got:

    [tex]\frac{h}{2}\left\{ f(a) + 2f(a+h) + 2f(a+2h) + \dots + 2f\left[ a + (n-2)h \right] + 2f\left[ a + (n-1)h \right] + f\left( a + nh \right) \right\}[/tex]​


    [tex]\int _a ^b f(x)\: dx[/tex]​


    [tex]h \left\{ f\left( a + \frac{h}{2} \right) + f\left( a + \frac{3h}{2} \right) + f\left( a + \frac{5h}{2} \right) + \dots + f\left[ a + \frac{(2n-3)h}{2} \right] + f\left[ a + \frac{(2n-2)h}{2} \right] + f\left[ a + \frac{(2n-1)h}{2} \right] \right\}[/tex]​

    which can be written as

    [tex]\frac{1}{2} f(a) + f(a+h) + f(a+2h) + \dots + f\left[ a + (n-2)h \right] + f\left[ a + (n-1)h \right] + \frac{1}{2} f\left( a + nh \right) [/tex]​


    [tex]\int _a ^b f(x)\: dx[/tex]​


    [tex]f\left( a + \frac{h}{2} \right) + f\left( a + \frac{3h}{2} \right) + f\left( a + \frac{5h}{2} \right) + \dots + f\left[ a + \frac{(2n-3)h}{2} \right] + f\left[ a + \frac{(2n-2)h}{2} \right] + f\left[ a + \frac{(2n-1)h}{2} \right][/tex]​

    Am I on the right track? What should I do next?

    Any help is highly appreciated.
    Last edited: Feb 9, 2005
  2. jcsd
  3. Feb 10, 2005 #2
    Hello, I cannot provide much help in terms of getting to the desired result using symbols but I have got the solution for that question. I can post it here or PM it to you if you want. In any case the solution does not necessarily have to be presented symbolically as you have done. The solution my solution manual has is relatively concise. As a hint I think you should draw a diagram, consider a general curve(that is, sketch it on the diagram) and use the fact that f''(x) < 0. This means that the curve is concave down. The diagram on page 419 should help you get started.
  4. Feb 16, 2005 #3
    Hey Benny, I'm sorry I wasn't able to get back to you sooner. My instant notification by email probably didn't go through. Anyway, it seems that the graph in page 419 along with its caption (found in my 1st post) already give the answer. If you have anything else that I miss, please let me know.

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