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 August 29th, 2017, 12:35 PM #1 Member   Joined: Feb 2017 From: East U.S. Posts: 40 Thanks: 0 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 ==  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! August 29th, 2017, 01:04 PM #2 Global Moderator   Joined: May 2007 Posts: 6,835 Thanks: 733 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. August 29th, 2017, 01:13 PM   #3
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 Originally Posted by mathman 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? August 29th, 2017, 02:53 PM #4 Global Moderator   Joined: Dec 2006 Posts: 21,028 Thanks: 2259 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 $n \to \infty$. September 3rd, 2017, 11:01 AM   #5
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 Originally Posted by skipjack 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!  Tags definition, integral, limit, riemann, sum Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post hyperbola Calculus 2 September 28th, 2015 01:14 PM azeret Calculus 3 January 19th, 2011 11:43 PM nizzeberra Number Theory 2 November 19th, 2009 05:37 AM a.sundar23 Calculus 1 February 23rd, 2009 07:22 PM cheloniophile Real Analysis 1 November 23rd, 2008 05:30 PM

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