1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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]
    [tex] x_{i-1}^2 \leq x_{i-1}x_i [/tex]
    [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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted