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Partitions for Riemann sum

  1. Dec 24, 2013 #1
    So my textbook asks to show [itex]\int^{3}_{1} x^{2}dx = \frac{26}{3}[/itex].
    They let the partition P = {[itex]x_{0},.....,x_{n}[/itex]}, and define the upper Riemann sum as U(P) = [itex]\sum^{i=1}_{n} x_{i}Δx_{i}[/itex] and lower sum as
    L(P) = [itex]\sum^{i=1}_{n} x_{i-1}Δx_{i}[/itex]

    I understand this part, but the next part is where I'm confused.

    For each index i, 1[itex]\leq[/itex]i[itex]\leq[/itex]n,
    [itex]3x^{2}_{i-1}\leq x^{2}_{i-1} + x_{i-1}x_{i}+x^{2}_{i}\leq3x^{2}_{i}[/itex]

    It's probably something I'm overlooking by where does the middle term come from and the 3 ??
     
  2. jcsd
  3. Dec 25, 2013 #2

    Office_Shredder

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    The inequality comes from the fact that [itex] x_{i-1} \leq x_i [/itex] for all i, and therefore
    [tex] x_{i-1}^2 \leq x_{i-1}^2 [/tex]
    and
    [tex] x_{i-1}^2 \leq x_{i-1}x_i [/tex]
    and
    [tex] x_{i-1}^2 \leq x_{i}^2 [/tex]
    so adding these all together give
    [tex] 3 x_{i-1}^2 \leq x_{i-1}^2 + x_{i-1} x_i + x_{i}^2 [/tex]
     
  4. Dec 25, 2013 #3
    I think they want to express this, because we have
    $$3x_{i-1}^{2}\leq x_{i-1}^{2} + x_{i-1} x_{i}^{2} +x_{i}^{2}\leq 3 x_{i}^{2}, $$
    so, $$ 3x_{i-1}^{2}(x_{i}-x_{i-1})\leq (x_{i-1}^{2} + x_{i-1} x_{i} +x_{i}^{2}) (x_{i}-x_{i-1}) \leq 3 x_{i}^{2} (x_{i}-x_{i-1}) ,$$
    namely,$$3 x_{i-1}^{2}\Delta{x_{i}}\leq ( x_{i}^{3} - x_{i-1}^{3} )\leq 3x_{i}^{2}\Delta{x_{i}}.$$
    Then, we can get
    $$\sum_{i=1}^{n} x_{i-1}^{2} \Delta x_{i} \leq \frac{1}{3}( x_{n}^{3} - x_{0}^{3} ) \leq \sum_{i=1}^{n} x_{i}^{2} \Delta x_{i} ,$$
    that is to say,
    $$L(P)\leq \frac{26}{3} \leq U(P).$$
    Because ##f(x)=x^{2}## is Riemann-integrable on ##[1,3]##, let ##I=\int _{1}^{3}f(x)dx\, , \lambda = \max \limits_{1\leq i \leq n}(\Delta{x_{i}})\rightarrow 0 ##, so
    $$\lim_{\lambda\rightarrow 0}{U(P)}=L=I=l= \lim_{\lambda \rightarrow 0}{L(P)}.$$
    According to the former reasoning, both of ##L## and ##l## equal ##\frac{26}{3}##, so ##I=\frac{26}{3}##.
     
    Last edited: Dec 25, 2013
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