anonymous
  • anonymous
Desperately need help with this calculus question!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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anonymous
  • anonymous
The hint was that dx equals 4/n and x_i equals 4i/n
jim_thompson5910
  • jim_thompson5910
Do you have this so far? \[\Large f(x) = \frac{x^2}{4} + 6\] \[\Large f\left(x_i\right) = \frac{\left(x_i\right)^2}{4} + 6\] \[\Large f\left(x_i\right) = \frac{\left(\frac{4i}{n}\right)^2}{4} + 6\]

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anonymous
  • anonymous
Yes!
jim_thompson5910
  • jim_thompson5910
ok what do you get when you simplify that?
anonymous
  • anonymous
\[\left( 4*i \right)^{2}/n ^{2}+6\]
jim_thompson5910
  • jim_thompson5910
you forgot about the 4 down below
jim_thompson5910
  • jim_thompson5910
Here is what I get \[\Large f(x) = \frac{x^2}{4} + 6\] \[\Large f\left(x_i\right) = \frac{\left(x_i\right)^2}{4} + 6\] \[\Large f\left(x_i\right) = \frac{\left(\frac{4i}{n}\right)^2}{4} + 6\] \[\Large f\left(x_i\right) = \frac{\frac{16i^2}{n^2}}{4} + 6\] \[\Large f\left(x_i\right) = \frac{16i^2}{4n^2} + 6\] Hopefully you agree?
anonymous
  • anonymous
Yup, I cancelled out the 4 in the denominator
jim_thompson5910
  • jim_thompson5910
oh right, then you'd have \[\Large f\left(x_i\right) = \frac{4i^2}{n^2} + 6\]
jim_thompson5910
  • jim_thompson5910
\[\Large R_n = \sum_{i=1}^{n}f\left(x_i\right)\Delta x\] \[\Large R_n = \sum_{i=1}^{n}\left(\frac{4i^2}{n^2} + 6\right)*\frac{4}{n}\] \[\Large R_n = \sum_{i=1}^{n}\left(\frac{4i^2}{n^2}*\frac{4}{n} + 6*\frac{4}{n}\right)\] \[\Large R_n = \sum_{i=1}^{n}\left(\frac{4i^2}{n^2}*\frac{4}{n}\right) + \sum_{i=1}^{n}\left(6*\frac{4}{n}\right)\] \[\Large R_n = \sum_{i=1}^{n}\left(\frac{16i^2}{n^3}\right) + \sum_{i=1}^{n}\left(\frac{24}{n}\right)\] \[\Large R_n = \frac{16}{n^3}\sum_{i=1}^{n}\left(i^2\right) + \frac{24}{n}*\sum_{i=1}^{n}\left(1\right)\]
jim_thompson5910
  • jim_thompson5910
now you'll use the identities \[\Large \sum_{i = 1}^{n}(i^2) = \frac{n(n+1)(2n+1)}{6}\] \[\Large \sum_{i = 1}^{n}(1) = n\]
anonymous
  • anonymous
So then you just plug the identities into your last equation for R_n?
jim_thompson5910
  • jim_thompson5910
yes then you simplify
anonymous
  • anonymous
I got this \[\left( 32*n ^{2}+48*n+16 \right)/(6*n^2)\]
jim_thompson5910
  • jim_thompson5910
this is what I'm getting (see the attached PDF)
1 Attachment
jim_thompson5910
  • jim_thompson5910
if it's too small, zoom in to at least 150%
anonymous
  • anonymous
What the bah-jesus. You're right. Sorry, we didn't do anything like this in class and I'm incredibly confused
jim_thompson5910
  • jim_thompson5910
so you're probably a lesson or two ahead?
jim_thompson5910
  • jim_thompson5910
where are you stuck at?
anonymous
  • anonymous
what I actually put for Rn and how I know when the work is done haha
jim_thompson5910
  • jim_thompson5910
there is a number of different possible ways to enter Rn but one thing that I would type in, if I were doing the HW is to type in what you see on the last line (since it's the most simplified)
anonymous
  • anonymous
Yes, that's correct! Thank you so much! How would you go about the limit as n goes to infinity?
anonymous
  • anonymous
Actually, I figured it out! Thanks so much for your help!

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