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Math Help - Integration help. Riemann sums

  1. #1
    Junior Member Khaali91's Avatar
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    Integration help. Riemann sums

    Use X*(subscript k) as the left endpoint of each sub interval to find the area under the curve y=F(x) over [a,b]

    f(x)= x^3 a=2 b=6 __________________
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    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by Khaali91 View Post
    Use X*(subscript k) as the left endpoint of each sub interval to find the area under the curve y=F(x) over [a,b]

    f(x)= x^3 a=2 b=6 __________________

    Click the link below for a good explanation of using estimating rectangles. It would be very difficult for someone to teach you how to do this on a forum.

    Pauls Online Notes : Calculus I - Area Problem
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  3. #3
    o_O
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    Recall: \int_a^b f(x) dx  =  \lim_{n \to \infty} \sum_{i=1}^n f(x_i^*) \Delta x

    where \Delta x = \frac{b-a}{n} = \frac{6-2}{n} = \frac{4}{n}

    Edit: Sorry I used right endpoints even though I said I was using left endpoints! I've corrected it:
    Since we're using left end points, we consider: x_i^* = x_{i-1} \ = \ a + (i-1)\Delta x \ = \ 2 + \frac{4(i-1)}{n}

    So, using the formula:
    \begin{aligned} \int_2^6 x^3 dx & =  \lim_{n \to \infty} \sum_{i=1}^n \left(2 + \frac{4(i-1)}{n}\right)^3 \frac{4}{n} \\ & = \lim_{n \to \infty} \frac{4}{n} \sum_{i=1}^n \left(4 + \frac{48(i-1)}{n} + \frac{96(i-1)^2}{n^2} + \frac{64(i-1)^3}{n^3} \right) \\ & = \lim_{n \to \infty} \frac{4}{n} \left( \sum_{i = 1}^n 4 + \frac{48}{n}\sum_{i = 1}^n (i-1) + \frac{96}{n^2}\sum_{i = 1}^n (i-1)^2 + \frac{64}{n^3}\sum_{i = 1}^n(i-1)^3\right)  \end{aligned}

    A bit messy but entirely doable. As you can see, right endpoints are usually nicer =|
    Last edited by o_O; February 10th 2009 at 06:22 PM.
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  4. #4
    Junior Member Khaali91's Avatar
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    Quote Originally Posted by o_O View Post
    Recall: \int_a^b f(x) dx  =  \lim_{n \to \infty} \sum_{i=1}^n f(x_i^*) \Delta x

    where \Delta x = \frac{b-a}{n} = \frac{6-2}{n} = \frac{4}{n}

    Edit: Sorry I used right endpoints even though I said I was using left endpoints! I've corrected it:
    Since we're using left end points, we consider: x_i^* = x_{i-1} \ = \ a + (i-1)\Delta x \ = \ 2 + \frac{4(i-1)}{n}

    So, using the formula:
    \begin{aligned} \int_2^6 x^3 dx & =  \lim_{n \to \infty} \sum_{i=1}^n \left(2 + \frac{4(i-1)}{n}\right)^3 \frac{4}{n} \\ & = \lim_{n \to \infty} \frac{4}{n} \sum_{i=1}^n \left(4 + \frac{48(i-1)}{n} + \frac{96(i-1)^2}{n^2} + \frac{64(i-1)^3}{n^3} \right) \\ & = \lim_{n \to \infty} \frac{4}{n} \left( \sum_{i = 1}^n 4 + \frac{48}{n}\sum_{i = 1}^n (i-1) + \frac{96}{n^2}\sum_{i = 1}^n (i-1)^2 + \frac{64}{n^3}\sum_{i = 1}^n(i-1)^3\right)  \end{aligned}

    A bit messy but entirely doable. As you can see, right endpoints are usually nicer =|
    Thank You very much!
    But is there a way to eliminate the Limit and the Sigma notation? My teacher briefly swooped over this and moved on so i never fully understood the concept.
    From what i remember there are theorems such as [n(n+1)]/2
    Is that right?
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  5. #5
    o_O
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    Yes. First we have to take care of the sums. Then we have to take the limit and then we finally get a numerical answer.

    So, things to note about summations:
    • \sum_{i=1}^n 1 = \overbrace{1 + 1 + 1 + \cdots + 1}^{n \text{ times}} = n
    • \sum_{i=1}^n c a_i = c \sum_{i=1}^n a_i (Basically, you can pull out the constant)
    • \sum_{i=1}^n i \ = \ 1 + 2 + \cdots + n \ = \ \tfrac{1}{2} n(n+1)
    • \sum_{i=1}^n i^2 \ = \ 1^2 + 2^2 + \cdots + n^2 \ = \ \tfrac{1}{6} n(n+1)(2n+1)
    • \sum_{i=1}^n i^3 \ = \ 1^3 + 2^3 + \cdots + n^3 \ = \ \tfrac{1}{4}n^2(n+1)^2


    Now, looking at that last expression I gave you: \lim_{n \to \infty} \frac{4}{n} \left( 4\sum_{i = 1}^n 1 + \frac{48}{n}\sum_{i = 1}^n (i-1) + \frac{96}{n^2}\sum_{i = 1}^n (i-1)^2 + \frac{64}{n^3}\sum_{i = 1}^n(i-1)^3\right)

    Notice how the summations don't exactly look like the bulleted ones. Let's consider \sum_{i=1}^n (i-1)^2.

    Expand it: \sum_{i=1}^n (i-1)^2 \ = \ 0^2 + 1^2 + 2^2 + \cdots + n^2 \ = \ 1^2 + 2^2 + \cdots + n^2 \ = \ \sum_{i=1}^n i^2

    which matches our bulleted formula. Basically, we opened up the shorthand notation and rewritten it.

    See if you can carry on from here.
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