Riemann Sums Upper and Lower limits

**nyhc_is_br00tal** · February 7th 2008, 03:32 PM

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

**galactus** · February 7th 2008, 03:47 PM

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.

**nyhc_is_br00tal** · February 7th 2008, 03:50 PM

Originally Posted by galactus

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 $(i-1)$ ?

**galactus** · February 7th 2008, 03:57 PM

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.

**nyhc_is_br00tal** · February 7th 2008, 04:23 PM

Originally Posted by galactus

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.

**galactus** · February 7th 2008, 04:34 PM

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?.

**nyhc_is_br00tal** · February 7th 2008, 05:40 PM

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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