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Proving x^2 + x is integrable.

  1. Dec 10, 2012 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    I'm curious about how to do something like this. My professor gave us a very simple example of how to prove f(x) = k is integrable on [1,3] for k in ℝ.

    I'm trying to apply the same logic to any function so I can see how it works. So I guess I could state the problem as :

    Suppose [itex]f(x) = x^2 + x[/itex], prove f is integrable on [2,5] by showing I = sup{sp} = inf{Sp} = J.

    2. Relevant equations

    sp is my underestimate and Sp is my overestimate.

    3. The attempt at a solution

    So first off we note :

    mi = inf{ f(x) | x in [2,5] } = 6
    Mi = sup{ f(x) | x in [2,5] } = 30

    Forming our upper and lower sums and then calculating them :

    [itex]s_p = \sum_{i=1}^{n} m_i Δx_i = 6(3) = 18[/itex]

    [itex]S_p = \sum_{i=1}^{n} M_i Δx_i = 30(3) = 90[/itex]

    This is definitely not what's supposed to happen I think as we all intuitively know that f(x) is integrable in this case. Showing it is a bit difficult for me though. Could someone point out what I may have done wrong?

    Thanks in advance.
     
  2. jcsd
  3. Dec 10, 2012 #2

    Ray Vickson

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    These are NOT what are meant by S_p and s_p. You need to use
    [tex] m_i = \inf_{x_i \leq x \leq x_i + \Delta x_i} f(x) ,\;\;\;\;\;
    M_i = \sup_{x_i \leq x \leq x_i + \Delta x_i} f(x), \; i = 1,2 \ldots, n . [/tex]
     
  4. Dec 10, 2012 #3

    pasmith

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    This is your error.

    Let [itex]P[/itex] be a partition of [2,5] with points of division [itex]2 = x_0 < x_1 < \dots < x_{n-1} < x_n = 5[/itex], and let [itex]\delta_i = x_i - x_{i-1}[/itex] for [itex]1 \leq i \leq n[/itex]. Then you should have
    [tex]
    m_i = \inf \{x^2 + x : x \in [x_{i-1},x_i] \} = x_{i-1}^2 + x_{i-1} \\
    M_i = \sup \{x^2 + x : x \in [x_{i-1},x_i] \} = x_i^2 + x_i
    [/tex]
    since [itex]x^2 + x[/itex] is strictly increasing on [2,5].

    Now
    [tex]x_{i-1}(x_i-x_{i-1}) < \frac12(x_i+x_{i-1})(x_i-x_{i-1})
    = \frac12 (x_i^2 - x_{i-1}^2) \\
    x_{i-1}^2(x_i- x_{i-1}) < \frac13(x_i^2 + x_i x_{i-1} + x_{i-1}^2)(x_i - x_{i-1})
    = \frac13 (x_i^3 - x_{i-1}^3)
    [/tex]
    and in the other direction
    [tex]x_{i}(x_i-x_{i-1}) > \frac12(x_i+x_{i-1})(x_i-x_{i-1})
    = \frac12 (x_i^2 - x_{i-1}^2) \\
    x_{i}^2(x_i- x_{i-1}) > \frac13(x_i^2 + x_i x_{i-1} + x_{i-1}^2)(x_i - x_{i-1})
    = \frac13 (x_i^3 - x_{i-1}^3)
    [/tex]
    so
    [tex]
    s(P) = \sum m_i \delta_i =
    \sum x_{i-1}^2(x_i- x_{i-1}) + \sum x_{i-1}(x_i-x_{i-1}) < \sum \frac13 (x_i^3 - x_{i-1}^3) + \sum \frac12 (x_i^2 - x_{i-1}^2)\\ = \frac13(5^3 - 2^3) + \frac12 (5^2 - 2^2)
    [/tex]
    by telescoping and similarly [itex]S(P) = \sum M_i \delta_i > \frac13(5^3 - 2^3) + \frac12 (5^2 - 2^2)[/itex].

    Since these limits are independent of [itex]\max \delta_i[/itex], we must have
    [tex]
    \sup s(P) \leq \frac13(5^3 - 2^3) + \frac12 (5^2 - 2^2) \leq \inf S(P)
    [/tex]
    Now one just has to prove equality.

    It's slightly easier to prove for the Riemann integral: [itex]x^2 + x[/itex] is continuous, so by the intermediate value theorem there exists [itex]z_i \in [x_i,x_{i-1}][/itex] such that
    [tex]z_i^2 + z_i = \frac13(x_i^2 + x_ix_{i-1} + x_{i-1}^2) + \frac12(x_i + x_{i-1})[/tex].
    Then
    [tex]\sum_{i} (z_i^2 + z_i)(x_i - x_{r-1}) = \sum_{i} \frac13(x_i^3-x_{i-1}^3)
    + \frac12 (x_i^2 - x_{i-1}^2) = \frac13(5^3 - 2^3) + \frac12(5^2 - 2^2)[/tex]
    and since the result is independent of [itex]\max \delta_i[/itex] it must be the limit as [itex]\max \delta_i \to 0[/itex].

    Of course I'd never have come up with those inequalities if I didn't already know the answer.
     
  5. Dec 10, 2012 #4

    Zondrina

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    Ahhh that was the trick. I completely forgot to partition my interval into n equal sub intervals ( Derp moment ).

    Then [itex]x_i = 5i/n[/itex] and [itex]x_{i-1} = 5(i-1)/n[/itex] so that [itex]m_i = x_{i-1} = 5(i-1)/n[/itex] and [itex]M_i = x_i = 5i/n[/itex]

    Thus :

    [itex]S_p = \sum_{i=1}^{n} M_i Δx_i = 25/n^2 \sum_{i=1}^{n} i[/itex]

    ( We do a similar case for the lower sum ).

    After evaluating and simplifying that, we get J = inf{ Sp } ≤ Sp and since inf{ Sp } does not rely on n, we can observe what happens as n→∞ which will yield our result for J. We can do the same thing for I and the lower sum.

    Since I = J, we deduce that f is integrable on [2,5] and its common value is 99/2.

    EDIT : Hmmm after writing this out I get I = 25/2 = J which is sorta what I wanted to happen, except I want 99/2, not 25/2.

    Did i partition incorrectly?
     
    Last edited: Dec 10, 2012
  6. Dec 10, 2012 #5

    pasmith

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    I'll just do this for [itex]f(x) = x[/itex] on [itex][a,b][/itex] with [itex]a \geq 0[/itex]; the same idea should work for [itex]f(x) = x^2[/itex].

    For the upper sum, I want to show that for all [itex]\epsilon > 0[/itex] there exists a paritition [itex]P[/itex] such that
    [tex]\frac 12(b^2 - a^2) < S(P) < \frac 12(b^2 - a^2) + \epsilon[/tex]
    So I have
    [tex]S(P) = \sum_i x_i (x_i - x_{i-1})
    = \sum_i \frac{x_i + x_{i-1} + x_i - x_{i-1}}{2}(x_i - x_{i-1})
    = \sum_i \frac12 (x_i^2 - x_{i-1}^2) + \sum_i \frac12 (x_i - x_{i-1})^2 \\
    = \frac12 (b^2 - a^2) + \frac12 \sum_i \delta_i^2
    \leq \frac12 (b^2 - a^2) + n (\max \delta_i)^2
    [/tex]
    So I need [itex]\max \delta_i < \sqrt{\epsilon/n}[/itex]. If I take an equally spaced partition, then [itex]\max\delta_i = (b-a)/n[/itex] and so I need [itex]n > (b-a)^2/\epsilon[/itex], which I can certainly do. This proves that [itex]\inf S(P) = \frac12 (b^2 - a^2)[/itex].
     
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