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Math Help - Riemann Sums Upper and Lower limits

  1. #1
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    Riemann Sums Upper and Lower limits

    Im having a hard time calculating Ln and Un of the function y=3(x+2)^2+1 bounded by x=0 and x=6. I know that Delta Xi should be 6/n and therefore xi-1 is (i-1)(6/n), but no matter what i do to work the problem out, i cant seem to find the final formula for Ln or Un.
    Please help me, someone.
    Thank You
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  2. #2
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    Riemann sums can be tedious. Lotsa algebra to contend with.

    Youa re correct, {\Delta}x=\frac{6}{n}

    Therefore, with the right endpoint method:

    x_{k}=a+k{\Delta}x=\frac{6k}{n}

    So, f(x_{k}){\Delta}x=\left[3(\frac{6k}{n}+2)^{2}+1\right](\frac{6}{n})

    Expand this out and get:

    \frac{648k^{2}}{n^{3}}+\frac{432k}{n^{2}}+\frac{78  }{n}

    Now, remember the sums of the squares and sums of the integers formulas?.

    Sub those in and you'll be entirely in terms of n. Then, take your limit and you should get the solution. The same as if you did it the quick integration way.
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  3. #3
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    Quote Originally Posted by galactus View Post
    Riemann sums can be tedious. Lotsa algebra to contend with.

    Youa re correct, {\Delta}x=\frac{6}{n}

    Therefore, with the right endpoint method:

    x_{k}=a+k{\Delta}x=\frac{6k}{n}

    So, f(x_{k}){\Delta}x=\left[3(\frac{6k}{n}+2)^{2}+1\right](\frac{6}{n})

    Expand this out and get:

    \frac{648k^{2}}{n^{3}}+\frac{432k}{n^{2}}+\frac{78  }{n}

    Now, remember the sums of the squares and sums of the integers formulas?.

    Sub those in and you'll be entirely in terms of n. Then, take your limit and you should get the solution. The same as if you did it the quick integration way.
    so your saying that k= i right? but what happened to (i-1)?
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  4. #4
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    I'm sorry, I am used to using k instead of i.Same thing. The i-1 is from the left endpoint method. I used the right. You should get the same thing, except I think the right method is a little easier to use.
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  5. #5
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    Quote Originally Posted by galactus View Post
    I'm sorry, I am used to using k instead of i.Same thing. The i-1 is from the left endpoint method. I used the right. You should get the same thing, except I think the right method is a little easier to use.
    i got up to this point...
    =\frac{648n(n+1)(2n+1)}{n^3} + \frac{432n(n+1)}{n^2} + \frac{78}{n}
    now, im not sure how i properly cancel out the n's, or if i should just foil it all and cancel out that way.
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  6. #6
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    You're missing a few things. Looks like your summation formulas are incomplete. Should be:

    \frac{648}{n^{3}}\cdot\frac{n(n+1)(2n+1)}{6}+\frac  {432}{n^{2}}\cdot\frac{n(n+1)}{2}+78

    Expand out and simplify and get:

    \frac{540}{n}+\frac{108}{n^{2}}+510

    Now, take the limit as \rightarrow{\infty}. What do you get?.
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  7. #7
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    510 units right?

    apparently im not as good in math as i thought i was because im still having a difficult time simplifying it to what you found the answer to be. hah
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