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Upper and lower bound Riemann sums

  1. Nov 27, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the upper, lower and midpoint sums for $$\displaystyle\int_{-3}^{3} (12-x^{2})dx$$
    $$\rho = \Big\{-3,-1,3\Big\}$$

    3. The attempt at a solution

    For the upper:
    (12-(-1)^2)(-1-(-3)) + (12-(-1))(3-(-1))
    =74

    For the lower:
    (12-(-3)^2)(-1-(-3))+(12-3)(3-(-1))
    =42

    For midpoint sum:
    I used the expression $$\frac{b+a}{n}$$

    (12-(-2)^2)(2) + (12-(2)^2)(4)
    =48

    Are these values correct? Does the process seem good?

    Please explain
    Thank you
     
  2. jcsd
  3. Nov 28, 2015 #2

    Svein

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    I think you have misunderstood slightly. "Upper", "lower" and "midpoint" do not refer to the argument ("x") values, but to the function values.

    In your case you need to calculate f(-3), f(-1) and f(3) and use those values to determine "upper" and "lower".
     
  4. Nov 28, 2015 #3
    @Svein I thought finding upper, lower and midpoints was essentially taking the partition {-3,-1,3} then breaking that into sub-intervals (-3,-1),(-1,3) then referencing the graph to see which is biggest or smallest in that interval. And in (-3,-1) if we are looking for upper that would be -1 then we would evaluate -1 by plugging it back into the original function and multiplying that by the distance between those two points and so on.
     
  5. Nov 28, 2015 #4

    Svein

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  6. Nov 28, 2015 #5

    SammyS

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    You made errors in evaluating your function for the Upper and Lower sums.

    How are upper and lower sums defined in your textbook/course?

    The midpoint sum looks to be correct.
     
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