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Homework Help: Prove that the integr of x² = (1/3)(b³-a³)

  1. Jul 18, 2008 #1
    1. The problem statement, all variables and given/known data

    I should have to prove that [tex]\int[/tex][tex]^{b}_____________{a}[/tex] x² dx = (1/3)(b³-a³)

    3. The attempt at a solution

    I know that f is continuous on [a,b] and that there exists a unique number I so that
    L[tex]_{f}[/tex](P)[tex]\leq[/tex] I [tex]\leq[/tex] U[tex]_{f}[/tex](P) .

    So the first thing to do is to find the lower and upper sum which are:
    L[tex]_{f}[/tex](P) = f(x0)[tex]\Delta[/tex]x1[/tex]+f(x1)[tex]\Delta[/tex]x2
    + ... + f(xn-1)[tex]\Delta[/tex]xn

    U[tex]_{f}[/tex](P) = f(x1)[tex]\Delta[/tex]x1+f(x2)[tex]\Delta[/tex]x2
    + ... + f(xn)[tex]\Delta[/tex]xn

    Now that we 've found the Upper and Lower sums, we must try to find the number I.

    My textbook says the following:
    " For each index i, 1<=i<=n we have:
    3xi-1² [tex]\leq[/tex]xi-1²+xi-1xi+xi²[tex]\leq[/tex]3xi² "

    Now where does the author get the middle term from? Why can't one just the take the mean like this (, or does the middle term means something else?) : f((b+a)/2) ?

    But apparently this doesn't lead to the end result of the middle term , i.e. (1/3)(b³-a³).

    How does one reach this end result from the Upper and Lower sums?

    Thank for your help!
    Last edited: Jul 18, 2008
  2. jcsd
  3. Jul 18, 2008 #2
    Hmm should 've used subscripts instead of superscripts... Apologies!
  4. Jul 18, 2008 #3


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    Homework Helper

    That middle term looks very much like one of the factors in a difference of two cubes:

    [tex]a^3 - b^3 = (a - b) \cdot (a^2 + ab + b^2) [/tex]
  5. Jul 18, 2008 #4
    Thank you for the quick reply but the thing that bothers me is how one reaches
    inequality 3xi-1<=xi²+xi-1xi<=3xi

    And why did the author multiply the first and the third factor by 3? I cannot give any reason to do so unless perhaps you work from the result (=(1/3)(b³-a³)) backwards? So, how would the author know how to proceed further from: Uf(P) and Lf(P) ?

    Thank you.
  6. Jul 18, 2008 #5


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    Homework Helper

    I suspect a certain measure of hokum in this proof, in the sense that the party arguing knew what they wanted to prove before they proved it.

    I haven't sorted all of this out yet, but the inequality

    [tex]3x_{i-1}^2 \leq x_{i-1}^2 + x_{i-1}x_{i} +x_{i}^2 \leq 3x_{i}^2 [/tex]

    comes from the idea that there are three terms in the middle, two of which are plainly larger than [tex]x_{i-1}^2[/tex] and two of which are smaller than [tex]x_{i}^2[/tex]. If you now divide through by 3 and multiply all terms by [tex](x_{i} - x_{i-i})[/tex], which is the [tex]\Delta x[/tex], you have the factors for the terms of the Riemann sum for the lower and upper sum on each end and the factors for [tex]\frac{1}{3}(x_{i}^3 - x_{i-1}^3)[/tex] in the middle. So you have the term you're after bracketed in the inequality, thus the sum over i on this inequality will bracket [tex]\frac{1}{3}(b^3 - a^3)[/tex], since all the middle terms in that sum will cancel out.

    Not having your book in front of me, I don't know the full context of this. It's a clever argument, but also one where it helps to know what you're intending to come up with. (You see that at times in proofs in textbooks.)
    Last edited: Jul 18, 2008
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