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August 29th, 2017, 01:35 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 == [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! 
August 29th, 2017, 02:04 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,661 Thanks: 648 
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, 02:13 PM  #3  
Member Joined: Feb 2017 From: East U.S. Posts: 40 Thanks: 0  Quote:
 
August 29th, 2017, 03:53 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,100 Thanks: 1905 
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, 12:01 PM  #5 
Member Joined: Feb 2017 From: East U.S. Posts: 40 Thanks: 0  Thanks for your help. I have no idea what you did, but I guess I'll learn soon! 

Tags 
definition, integral, limit, riemann, sum 
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