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anonymous
 4 years ago
Im doing a problem at the moment and im stuck its a right riemann sum problem, equation and how far i got posted below
anonymous
 4 years ago
Im doing a problem at the moment and im stuck its a right riemann sum problem, equation and how far i got posted below

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{3}^{0} (2x^2 + x +3) dx\] \[\Delta x = 3/n\] \[x_{k} = 3 + 3k/n\] \[f(x_k) = (3+ 3k/n)^2  (3 + 3k/n) +1\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0im not sure where to go from there

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think you'd be better off using: \[\huge \int\limits_{a}^{b} f(x) dx = \int\limits_{b}^{a} f(x) dx\] at the start

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then \[f(c _{k}) = 2*k^2 *\Delta x ^3 + k*\Delta x^2 + 3 \Delta x\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and \[S _{n} = 2 \Delta x ^3 \sum_{}^{} k^2 + \Delta x^2 \sum_{}^{} k +3 \Delta x\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so if be better of making the original equation negative then using that formula to find the equation

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0a bit easier, yes... that's probably why they gave it to you in the form that they did... so you could make the easy simplification

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you got it from here?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yea i think so i think what confuses me the most is 3+(3k/n)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0do what I did before you sub.s in 'delta x = b/n'

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[2 * (3+ \frac{ 3k }{ n }^2) * \frac{ 3 }{ n }^3 + (3+ \frac{ 3k }{ n }) * \frac{ 3 }{ n } + 3\frac{ 3 }{ n }\] is this what im looking for or do does k = 1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0start at x=0... move delta x to the right... what's the height of the rectangle you'd draw there?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1354238714258:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0height is f(delta x) width is delta x area is f(delta x) * delta x which is (2*(delta x)^2 + delta x +3))*delta x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1354238854668:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0next step: go another 'delta x' to the right height is 2*(2*delta x)^2 + 2*delta x +3 width is delta x area is (2*(2*delta x)^2 + 2*delta x +3)*delta x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so the sum of areas is going to look like: (2*(k*delta x)^2 + k*delta x +3)*delta x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.02*k^2*(delta x)^3 + k*(delta x)^2 +3*delta x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0summing from k=1 to n

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that's the sum I wrote above

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh alright, thanks i appreciate the help!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0does that make sense?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah it does, im trying to work through the problem. ill have an answer soon enough though

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I didn't write the last term on the summation properly

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it should be:\[\Delta x \sum_{k=1}^{n} 3\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so\[S _{n} = 2(\Delta x)^3\sum_{k=1}^{n} k^2 + (\Delta x)^2 \sum_{k=1}^{n} k +\Delta x \sum_{k=1}^{n} 3\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh ok, thats makes more sense now

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Its either 0 or 6 i think 0 because n = 3, am i correct?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0n is the number of divisions in the partition... you find the sum above (in terms of n) ... you sub.s in b/n for delta x and then you take the limit as n>infinity

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then use b=3 (the interval length) and finally, make the whole result negative ( due to the simplification we did at the very beginning)
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