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Math Help - approximate the area using Riemann sums

  1. #1
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    approximate the area using Riemann sums

    I need to approximate int((1/3*x^2+2) dx) on the interval [1, 9] using a left Riemann sum, and 4 rectangles of equal base.

    I'm having trouble graphing it, I've used my calculator to see what it looks like but if I try to graph it by hand it won't be as perfect and the rectangles wouldn't fit under the curve. The graphing part is really stressing me out.

    Would the 1st endpoint be at 2/3? by plugging in 1 into (1/3*x^2) and the 2nd be 27 by plugging in 9?

    Each rectangle's... width would be 2...
    something like that?
    Last edited by omgeezy; January 11th 2009 at 01:22 PM.
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  2. #2
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    Quote Originally Posted by omgeezy View Post
    I need to approximate int((1/3*x^2+2) dx) on the interval [1, 9] using a left Riemann sum, and 4 rectangles of equal base.

    I'm having trouble graphing it, I've used my calculator to see what it looks like but if I try to graph it by hand it won't be as perfect and the rectangles wouldn't fit under the curve. The graphing part is really stressing me out.
    The inteval has length 8 so the width of each rectangle \Delta x is

    \Delta x = 8/4 = 2

    When we split up the interval [1,9] into 4 and use the left endpoint, we obtain

    x_1 = 1,\;\;x_2 = 3,\;\;x_3=5,\;\;x_4 = 9

    If we let f(x) = \frac{1}{3}\,x^2+2

    then the area is approximately

    \sum_{i=1}^4 f(x_i) \Delta x = \left( f(x_1) + f(x_2) + f(x_3) + f(x_4) \right) \Delta x
     = \left( f(1) + f(3) + f(5) + f(7) \right) 2
     = \left( \frac{1}{3}\,1^2+2+ \frac{1}{3}\,3^2+2 + \frac{1}{3}\,5^2+2 + \frac{1}{3}\,7^2+2 \right) 2

    Then simplify.
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  3. #3
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    If it isn't too much trouble, what would the graph look like?

    OR is there no need for a graph? maybe that's why

    thanks very much
    Last edited by omgeezy; January 11th 2009 at 01:40 PM.
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  4. #4
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    in the end I have

    2(\frac{2}{3}+5+\frac{31}{3}+\frac{55}{3})=\frac{2  06}{3}

    \frac{206}{3}=68.6667

    right or wrong?
    just checking for any mathematical errors
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  5. #5
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    left sum = 72
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  6. #6
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    Quote Originally Posted by skeeter View Post
    left sum = 72
    What did I miss?
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  7. #7
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    you didn't calculate it correctly. try again.

    2\left(\frac{1^2}{3} + 2 + \frac{3^2}{3} + 2 + \frac{5^2}{3} + 2 + \frac{7^2}{3} + 2 \right) = 72
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  8. #8
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    I see, I missed the ( \frac{1}{3}\,1^2+2) what a stupid mistake
    now that makes sense
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