anonymous
  • anonymous
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
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
\[\int\limits_{3}^{0} (2x^2 + x +3) dx\] \[\Delta x = -3/n\] \[x_{k} = 3 + -3k/n\] \[f(x_k) = (3+ -3k/n)^2 - (3 + -3k/n) +1\]
anonymous
  • anonymous
im not sure where to go from there
anonymous
  • anonymous
*

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.