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anonymous
 5 years ago
Use the midpoint rule: n=3, integral from 5 to 0, (2x6x^2)dx
anonymous
 5 years ago
Use the midpoint rule: n=3, integral from 5 to 0, (2x6x^2)dx

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0*grunt* having to do summation finally, you do it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.05 to 0 is 5 so with n=3 your dx will be 5/3. This means that your first midpoint will be at (5/3)/2 or 5/6 and all subsequent ones will be dx multiples away. So using this you can then calculate the rieman sum 5/3[{2(5/6)6(5/6)^2}+{2(15/6)6(15/6)^2}+{2(25/6)6(25/6)^2}]. You can do the math on that one i dont have a calculator near me. explanation to follow.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0THANK YOU SOOOOOOO MUCH!!!!!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0All i did was apply the rule for the mid point rule which is Q[F(a+Q)+F(a+2Q)+F(a+3Q)...all the way up to F(a+NQ) where N is the amount of divisions they want, Q is the range of the integral divided by N (in your case 5/3) and a is the first midpoint that you use (in your case it was 5/6 because it was half the distance of the base but this is not for all cases because you could have the same base with a range of 10 to 5 but the first point would be 105/6) http://en.wikipedia.org/wiki/Riemann_sum#Middle_sum
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