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Using Upper, Lower Sum, Prove the Following

  1. Mar 13, 2013 #1
    1. The problem statement, all variables and given/known data
    [tex]f(x)=x^{2}[/tex] from [a,b]

    Prove that [tex]F(x)=\frac{b^{3}}{2}-\frac{a^{3}}{2}[/tex]


    2. Relevant equations



    3. The attempt at a solution
    Using the definition of an integral, we get:
    [tex]U(f,P)=\sum^{n}_{i=1}M_{i}(t_{i}-t_{i-1})[/tex]
    [tex]L(f,P)\sum^{n}_{i=1}m_{i}(t_{i}-t_{i-1})[/tex]

    for the function x2, how do we find M?

    I'm somewhat confused, a nudge in the right directions is welcomed.
     
  2. jcsd
  3. Mar 13, 2013 #2

    CompuChip

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    Divide the interval [a,b] into n pieces and let t be the boundaries of these pieces; so t0 = a, t1 = a + (b - a) / n, t2 = a + 2 (b - a) / n, etc.
    Mi is the maximum of f on the piece [ti-1, ti] and mi is the minimum.

    It may become clearer if you draw a picture in which you draw the graph, the division of [a, b] and the rectangular areas that make up U(f, P) or L(f, P)
     
  4. Mar 13, 2013 #3
    Ah, that's right, the partition is a=t0<t1<....<tn=b

    So we get this:
    [tex]\sum^{n}_{i=1}M_{i}(a+\frac{b-a}{n})[/tex]

    But I still am not understanding where we get M and m from. We need to find the maximum minimum points on the interval? Or is there an easier way?
     
  5. Mar 13, 2013 #4

    HallsofIvy

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    As long as a and b are positive, [itex]x^2[/itex] is an increasing function so the maximum value on each interval, [itex]M_i[/itex], is at the right end and the minimum value on each interval, [itex]m_i[/itex], is at the left end. Further, if you divide [a, b] into n intervals, each interval has length [itex]t_i- t_{i-1}= (b- a)/n[/itex], the left endpoint is a+ (b- a)i/n for i= 0 to n-1, and the right endpoint is a+ (b- a)i/n for i= 1 to n. That is, [itex]M_i= (a+ (b-a)i/n)^2[/itex], for i= 1 to n and [itex]m_i= (a+ (b-a)i/n)^2[/itex] for i= 0 to n-1.
    [itex]M_i(t_i- t_{i-1})= ((b-a)/n)(a+ (b-a)i/n)^2[/itex] for i= 1 to n and [itex]m_i(t_i- t_{i-1})= ((b- a)/n)(a+ (b- a)i/n)^2[/itex]
    Multiply those out and sum. It will help if you know that [itex]\sum_{i=1}^n c= nc[/itex], [itex]\sum_{i=1}^n i= n(n+ 1)/2[/itex] and [itex]\sum_{i=1}^n i^2= [n(n+1)(2n+1)]/6[/itex]. Those should be in your text.
     
    Last edited by a moderator: Mar 15, 2013
  6. Mar 13, 2013 #5
    You had some formatting problems XD

    I think I get what you're saying though, it makes sense. So, in the interval [a,b], the highest value of f(x) will be at b. If I understand what you're saying, M will be equal to ti2, and m will be t2i-1?

    There seems to be some weird latex problem.
     
    Last edited by a moderator: Mar 15, 2013
  7. Mar 13, 2013 #6
    I think I got it.

    [tex]U(f,P)=\sum^{n}_{i=1}M_{i}(t_{i}-t_{i-1})[/tex]
    [tex]L(f,P)=\sum^{n}_{i=1}m{i-1}(t_{i}-t_{i-1})[/tex]

    [itex]m_{i}=t^{2}_{i-1}; M_{i}=t^{2}_{i}[/itex]

    [tex]=\sum^{n}_{i=1}t^{2}_{i}(\frac{b-a}{n})[/tex]
    [tex]=\sum^{n}_{i=1}t^{2}_{i-1}(\frac{b-a}{n})[/tex]

    [tex]t^{2}_{i}=i^{2}\frac{b^{2}-a^{2}}{n^{2}}[/tex]
    [tex]t^{2}_{i-1}=(i-1)^{2}\frac{b^{2}-a^{2}}{n^{2}}[/tex]

    [tex]=\sum^{n}_{i=1}i^{2}\frac{b^{2}-a^{2}}{n^{2}}(\frac{b-a}{n})[/tex]
    [tex]=\sum^{n}_{i=1}(i-1)^{2}\frac{b^{2}-a^{2}}{n^{2}}(\frac{b-a}{n})[/tex]

    Now we just need to simplify, right?
     
    Last edited: Mar 13, 2013
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