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Using the Riemann Sum

  1. Jul 2, 2006 #1
    Use the Riemann sum and a limit to evaluate the exact area under the graph of [itex] y = 2x^2 + 4 [/itex] on [0, 1]

    I know how to do this normally but now they ask to do it w/ a limit and I'm not sure how.

    (LaTex corrected by HallsofIvy.)
     
    Last edited by a moderator: Jul 2, 2006
  2. jcsd
  3. Jul 2, 2006 #2
    Ok, well i guess i can't do it at all.

    I have the answer, but no idea how to get there.

    I think the formula is [tex] \frac {b-a}{n} \sum^n_{i=1}F(Xi) [/tex]

    So when I plug it in I get [tex] \frac {1-0}{n} \sum^n_{i=1}[2(x_{i})^2 +4] [/tex]

    Is this right so far, and if so, what's next?
     
    Last edited: Jul 2, 2006
  4. Jul 2, 2006 #3

    HallsofIvy

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    How could you do it without a limit? The "Riemann sum" itself can give an approximation to the area under the curve but the actual area, i.e. integral, is, by definition, the limit of the Riemann sums as the number of divisions of the x-axis goes to infinity.

    If the interval [0,1] is broken into n equal length divisions, then each will have length 1/n. Taking the right endpoint of each division as x, x= i/n, i= 1 to n so the Riemann sum becomes
    [tex]\Sigma_{i=1}^n (\frac{2i^2}{n^2}+ 4)\frac{1}{n}[/tex]
    [tex]= 2n\Sigma_{i=1}^n i^2+ \frac{1}{n}\Sigma_{i=1}^n 4[/tex]
    Of course
    [tex]\Sigma_{i=1}^n 4 = 4n[/tex]
    Do you know a formula for
    [tex]\Sigma_{i=1}^n i^2[/tex]?
     
  5. Jul 2, 2006 #4
    yeah, i think it's [tex] \frac{n (n+1)(2n+1)}{6}[/tex]

    I'm not really sure I understand what you wrote above that though.
     
  6. Jul 2, 2006 #5
    bump for a test tomorrow
     
  7. Jul 2, 2006 #6

    StatusX

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    What you are doing is summing over the areas of a series of n rectangles of width 1/n, spaced so as to partition the interval [0,1] and with height equal to the value of the function at the corresponding point on the x-axis (say, the point at the left end of the rectangle, although exactly how you do this becomes unimportant when you take the limit). Then if you let n become very large, these rectangles capture more detail of the variation of the function and the sum of their areas becomes a better and better approximation to the area underneath the function on this range, ie, to the definite integral. Do you understand this part?

    Taking the limit just means determining what happens as n gets very large. For example, the expression (n+1)/n becomes very close to 1 as n gets very large, and in a mathematically precise way we can say that the limit of this expression as n goes to infinity is 1. Similarly, once you find an expression for the area of the n rectangles described above, you need to find out what its limit is as n goes to infinity, and this should be equal to the value you get using familiar integration methods. Is your problem understanding one of these concepts or computing the explicit formula for the area of the rectangles?
     
    Last edited: Jul 2, 2006
  8. Jul 3, 2006 #7

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    By the way, all you need for these types of problems is:

    [tex] \sum_{k=1}^n k^p = \frac{1}{p+1} n^{p+1} + ...[/tex]

    Where we have ignored all smaller powers of n, since they will drop out once you take the limit. This can be easily proved by induction.
     
    Last edited: Jul 3, 2006
  9. Jul 3, 2006 #8

    HallsofIvy

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    What I said before was that dividing the interval from 0 to 1 into n intervals, each interval has length 1/n. The x-coordinate of the point at the right end of each interval is i/n (i= 1 to n) and the value of the function there is [itex]2\frac{i^2}{n}^2+ 4[/itex] so the area of the rectangle having that base and height is [itex]\left(\frac{1}{n}\right)\left(2\frac{i^2}{n^2}+ 4\right)[/itex]. The sum of all of the rectangles, the approximate "area under the curve" then is
    [tex]\left(\Sigma_{i=1}^n (\frac{2i^2}{n^2}+ 4\right)\frac{1}{n}[/tex]
    [tex]= \frac{2}{n}\Sigma_{i=1}^n i^2+ \frac{1}{n}\Sigma_{i=1}^n 4[/tex]
    Since [itex]\Sigma_{i=1}^n 4= 4n[/itex] and, as you say,
    [tex]\Sigma_{i=1}^n i^2= \frac{n (n+1)(2n+1)}{6}[/tex]
    that sum is
    [tex]\frac{2}{n^3}\frac{n(n+1)(2n+1)}{6}+ \frac{4n}{n}[/tex]
    You can see clearly that the last term is just 4, no matter what n is.
    You can also see, perhaps less clearly, that the first term is
    [tex]\frac{4n^3+ \cdot\cdot\cdot}{6n^3}[/itex]
    Where the dots indicate powers of n less than 3. What is the limit of that as n goes to infinity?
     
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