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August 29th, 2017, 12:35 PM   #1
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Use the definition of the integral as a limit of a Riemann sum

Equation: "integral from 0 to 6 of -(x^2)+36"

I know how to find it the easy way...

Ex.) I know to take the integral -(x^3)/3+36x and evaluate it from 0 to 6 which = (-216/3) + 216 == [144]

So I know the answer, I just don't understand what my teacher wants. I know there's an "n(n+1)(2n+1) that's supposed to be thrown in there.

ANY help would be greatly appreciated!
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August 29th, 2017, 01:04 PM   #2
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What the teacher wants if for you to divide the interval (0,6) into a lot of evenly spaced intervals and set up the Riemann sum over these intervals. Evaluate the sum and then let the number of intervals become infinite and see that the sum converges to the integral.
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August 29th, 2017, 01:13 PM   #3
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Quote:
Originally Posted by mathman View Post
What the teacher wants if for you to divide the interval (0,6) into a lot of evenly spaced intervals and set up the Riemann sum over these intervals. Evaluate the sum and then let the number of intervals become infinite and see that the sum converges to the integral.
Thanks for the information, but I have no idea how to do that. He's never gone over a single problem like that and it's due for homework tomorrow. Do you know any good places to read up on this method of solving?
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August 29th, 2017, 02:53 PM   #4
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Using 6$n$ intervals, each of width 1/$n$, the integral can be approximated as
$\displaystyle \sum_{k=1}^{6n} (-(k/n)^2 + 36)(1/n) = (-(6n)(6n + 1)(12n + 1)/(6n^2) + 36(6n))/n$,
which tends to -6(12) + 216 = 144 as $n \to \infty$.
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September 3rd, 2017, 11:01 AM   #5
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Quote:
Originally Posted by skipjack View Post
Using 6$n$ intervals, each of width 1/$n$, the integral can be approximated as
$\displaystyle \sum_{k=1}^{6n} (-(k/n)^2 + 36)(1/n) = (-(6n)(6n + 1)(12n + 1)/(6n^2) + 36(6n))/n$,
which tends to -6(12) + 216 = 144 as $n \to \infty$.
Thanks for your help. I have no idea what you did, but I guess I'll learn soon!
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