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Riemann Sum-Area

  1. Apr 29, 2006 #1
    I was hoping someone could check my answer?

    Use the limit of a Riemann sum to find the area of the region bounded by the graphs of y=2x^3+1, y=0, x=0, x=2.

    Area=2
     
  2. jcsd
  3. Apr 29, 2006 #2
    The limit of the Riemann Sum is the integral. So evaluate [tex]\int_{0}^{2}2x^3+1dx[/tex]. I don't get 2. How did you do your problem?
     
  4. Apr 30, 2006 #3

    benorin

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    Homework Helper

    A useful formaula is

    [tex]\int_{a}^{b}f(x) \, dx = \lim_{n\rightarrow\infty} \frac{b-a}{n}\sum_{k=1}^{n}f\left( a+\frac{b-a}{n}k\right) [/tex]​

    hence

    [tex]\int_{0}^{2}(2x^3+1) \, dx = \lim_{n\rightarrow\infty} \frac{2}{n}\sum_{k=1}^{n}\left[ 2\left( \frac{2k}{n}\right) ^3 +1\right] [/tex]​
     
  5. May 1, 2006 #4
    Now I am getting two answers-10 for the def integral and 8 for the sum.
     
  6. May 1, 2006 #5

    HallsofIvy

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    10 is obviously the correct answer. How are you taking the limit, as n goes to infinity on
    [tex]\lim_{n\rightarrow\infty} \frac{2}{n}\sum_{k=1}^{n}\left[ 2\left( \frac{2k}{n}\right) ^3 +1\right] [/tex]
    ?

    Do you know a formula for the sum of k3, k2, and k that you are using?
     
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