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

Calculus1
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\[\int\limits_{3}^{0} (2x^2 + x +3) dx\] \[\Delta x = -3/n\] \[x_{k} = 3 + -3k/n\] \[f(x_k) = (3+ -3k/n)^2 - (3 + -3k/n) +1\]
im not sure where to go from there
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I 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
then \[f(c _{k}) = 2*k^2 *\Delta x ^3 + k*\Delta x^2 + 3 \Delta x\]
and \[S _{n} = 2 \Delta x ^3 \sum_{}^{} k^2 + \Delta x^2 \sum_{}^{} k +3 \Delta x\]
so if be better of making the original equation negative then using that formula to find the equation
a 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
you got it from here?
yea i think so i think what confuses me the most is 3+(-3k/n)
do what I did before you sub.s in 'delta x = b/n'
\[2 * (3+ \frac{ -3k }{ n }^2) * \frac{ -3 }{ n }^3 + (3+ \frac{ -3k }{ n }) * \frac{ -3 }{ n } + 3\frac{ -3 }{ n }\] is this what im looking for or do does k = 1
start at x=0... move delta x to the right... what's the height of the rectangle you'd draw there?
|dw:1354238714258:dw|
height 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
|dw:1354238854668:dw|
next 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
so the sum of areas is going to look like: (2*(k*delta x)^2 + k*delta x +3)*delta x
2*k^2*(delta x)^3 + k*(delta x)^2 +3*delta x
summing from k=1 to n
that's the sum I wrote above
oh alright, thanks i appreciate the help!
does that make sense?
Yeah it does, im trying to work through the problem. ill have an answer soon enough though
still there?
I didn't write the last term on the summation properly
it should be:\[\Delta x \sum_{k=1}^{n} 3\]
so\[S _{n} = 2(\Delta x)^3\sum_{k=1}^{n} k^2 + (\Delta x)^2 \sum_{k=1}^{n} k +\Delta x \sum_{k=1}^{n} 3\]
oh ok, thats makes more sense now
Its either 0 or -6 i think 0 because n = 3, am i correct?
it's -63/2
n 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
then 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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