Math Help - [SOLVED] Help simplifying this def. integral by limit definition

1. [SOLVED] Help simplifying this def. integral by limit definition

http://i34.tinypic.com/23r747b.jpg

I think I get the setup, but I'm having trouble simplifying it. It's getting so long with the cube factored out ...is there an easier way?

I have: E (as in sigma) [2(-5+3i/n)^3(3/n)]

I simplified it further by factoring out the cube but it seems like it's way too complicated now. Any ideas how to avoid this, or am I way off?

Thanks :]

2. Originally Posted by janedoe
http://i34.tinypic.com/23r747b.jpg

I think I get the setup, but I'm having trouble simplifying it. It's getting so long with the cube factored out ...is there an easier way?

I have: E (as in sigma) [2(-5+3i/n)^3(3/n)]

I simplified it further by factoring out the cube but it seems like it's way too complicated now. Any ideas how to avoid this, or am I way off?

I hope you're not saying that you in any way factored out the index i?

Thanks :]
In any case....you were right in your setup.

$\sum_{i=1}^n\left[2\left(-5+3i/n\right)^3(3/n)\right]=\frac{6}{n}\sum_{i=1}^n(-5+3i/n)^3$

So, now multiply out the binomial. Then arranging sums in the forms...

$\sum{i},\sum{i^2},\sum{i^3}$

and then make use of the well know formulas for these sums.

Simplify and take the limit.

Note: Everyone knows that the limit definition is highly tedious, this is why they have been defined in terms of definite integrals, so that we may make the process easier. But, a knowledge of why definite integrals work (because of the riemann integral) is essential in an in depth understandind of them.

3. Originally Posted by janedoe
http://i34.tinypic.com/23r747b.jpg

I think I get the setup, but I'm having trouble simplifying it. It's getting so long with the cube factored out ...is there an easier way?

I have: E (as in sigma) [2(-5+3i/n)^3(3/n)]

I simplified it further by factoring out the cube but it seems like it's way too complicated now. Any ideas how to avoid this, or am I way off?

Thanks :]
Limit definition?

Seeing as the answers are multiple choice, why not just use the standard method?

$\int_{-5}^{-2}{2u^3\,du} = \left[\frac{1}{2}u^4\right]_{-5}^{-2}$

$= \frac{1}{2}\left[(-2)^4 - (-5)^4\right]$

$= \frac{1}{2}\left[16 - 625\right]$

$= \frac{1}{2}(-609)$

$= -\frac{609}{2}$.

4. Originally Posted by Prove It
Limit definition? Seeing as the answers are multiple choice, why not just use the standard method?
That's a good point.

5. As for using the 'standard method', I guess because I wanted to practice doing it the other way (this quiz is only for practice). Also, I never learned to do it that way. I mean, I get how to do general antiderivatives but I didn't know you could plug in the upper and lower limits and subtract like that.

Originally Posted by VonNemo19

So, now multiply out the binomial.
That's what I meant that I did...but it got very messy and I just couldn't seem to get it down to the 3 i forms. Would you mind showing me how you would finish simplifying?

6. Originally Posted by janedoe
As for using the 'standard method', I guess because I wanted to practice doing it the other way (this quiz is only for practice). Also, I never learned to do it that way. I mean, I get how to do general antiderivatives but I didn't know you could plug in the upper and lower limits and subtract like that.

That's what I meant that I did...but it got very messy and I just couldn't seem to get it down to the 3 i forms.
$(-5+\frac{3i}{n})^3=\left(-5+\frac{3i}{n}\right)\left[25-(10)\frac{3i}{n}+\frac{9i^2}{n^2}\right]$ $=-125+50\frac{3i}{n}-5\frac{9i^2}{n^2}+25\frac{3i}{n}-10\frac{9i^2}{n^2}+\frac{27i^3}{n^3}$

OK, there's your multiplication. Can you continue?

7. I'm sorry, I just can't seem to get this lol...I worked at it for so long last night, but I got such a huge summation and still didn't get the right answer. I'm very frustrated with this problem, but I'm going to try to get in touch with my teacher (I'm in a distance learning program) and see what he says, I feel bad to keep bothering you.

8. Originally Posted by janedoe
I feel bad to keep bothering you.

$=\frac{6}{n}\sum_{i=1}^{n}\left(-125+\frac{225i}{n}-\frac{135i^2}{n^2}+\frac{27i^3}{n^3}\right)$

Now, from here you want to get all of the i's alone. So...

Since the sum of a sum is the sum of the sums, we can kinda pass the sigma throughout
$\frac{6}{n}\left[-\sum_{i=1}^{n}(125)+\sum_{i=1}^{n}\left(\frac{225i }{n}\right)-\sum_{i=1}^{n}\left(\frac{135i^2}{n^2}\right)+\sum _{i=1}^{n}\left(\frac{27i^3}{n^3}\right)\right]$

$\frac{6}{n}\left[-125n+\frac{225}{n}\sum_{i=1}^{n}i-\frac{135}{n^2}\sum_{i=1}^{n}i^2+\frac{27}{n^3}\su m_{i=1}^{n}i^3\right]$

Now, from the formulas for these special sums...

9. Originally Posted by VonNemo19

$=\frac{6}{n}\sum_{i=1}^{n}\left(-125+\frac{225i}{n}-\frac{135i^2}{n^2}+\frac{27i^3}{n^3}\right)$

Now, from here you want to get all of the i's alone. So...

Since the sum of a sum is the sum of the sums, we can kinda pass the sigma throughout
$\frac{6}{n}\left[-\sum_{i=1}^{n}(125)+\sum_{i=1}^{n}\left(\frac{225i }{n}\right)-\sum_{i=1}^{n}\left(\frac{135i^2}{n^2}\right)+\sum _{i=1}^{n}\left(\frac{27i^3}{n^3}\right)\right]$

$\frac{6}{n}\left[-125n+\frac{225}{n}\sum_{i=1}^{n}i-\frac{135}{n^2}\sum_{i=1}^{n}i^2+\frac{27}{n^3}\su m_{i=1}^{n}i^3\right]$

Now, from the formulas for these special sums...
Thank you SO much! I figured it out from there, finally lol. I appreciate the help.

10. Originally Posted by janedoe
http://i34.tinypic.com/23r747b.jpg

I think I get the setup, but I'm having trouble simplifying it. It's getting so long with the cube factored out ...is there an easier way?

I have: E (as in sigma) [2(-5+3i/n)^3(3/n)]

I simplified it further by factoring out the cube but it seems like it's way too complicated now. Any ideas how to avoid this, or am I way off?

Thanks :]