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Speedy232 · Mar 7, 2009, 11:30 PM

How would you find the area under a curve of F(x)=X^2 between x=2 and x=10.

I'm puzzled. Ive done as much as I can. Here's what I come up with, but I can't seem to work it out. Lim┬(x→∞)⁡〖((1024n^3+1536n^2+512n)/(6n^3 )〗)+8

Any help would be greatly appreciated.

Perito · Mar 8, 2009, 04:09 AM

Is this a calculus problem? Evaluate the definite integral between X=2 and X=10.

galactus · Mar 8, 2009, 06:34 AM

It appears you must use a Riemann sum to evaluate?

The algebra is the killer with these things.

${\Delta}x=\frac{10-2}{n}=\frac{8}{n}$

Using the right endpoint method: $x_{k}=2+\frac{8k}{n}$

$f(x_{k}){\Delta}x=(2+\frac{8k}{n})^{2}\cdot\frac{8 }{n}=\frac{512k^{2}}{n^{3}}+\frac{256k}{n^{2}}+32/n$ ... [1]

Now, remember the identities for the sums of the squares of integers and integers.

$\sum_{k=1}^{n}k^{2}=\frac{n(n+1)(2n+1)}{6}$

$\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$

Sub these in [1] and get:

$\frac{384}{n}+\frac{256}{3n^{2}}+\frac{992}{3}$

Take the limit of this:

$\lim_{n\to {\infty}}\left[\frac{384}{n}+\frac{256}{3n^{2}}+992/3\right]$

Note, all terms head toward 0 except 992/3 and that is all that remains. Thus, that must be the area.

Compare this to the actual area which can be found by integrating, as Perito mentioned.

Speedy232 · Mar 8, 2009, 10:47 PM

Wow, No wonder it was like a week homework question. This is hard as anything on earth. Why would the teacher make us do this lol? Thank you for all your help. But where you say " Sub these into [1]" what do you actually mean there. Like do you go back to where you said [1] in the 2nd line of the equation? But yeah, that part confuses me. But yeah the area is the definite integral. Thank you so much.

Speedy232 · Mar 8, 2009, 11:44 PM

I got another question. How did you get all the numerical values. Obviously my numbers were wrong because I just tried copying off the sheet and multiplying it by whatever it was. But you somehow got the right figures... How..!. Can you explain please

galactus · Mar 9, 2009, 11:10 AM

Yes, sub in the identities for k and k^2 into the formula I provided. It simplifies down and is all in terms of n. Then, if you take the limit as $n\to {\infty}$ , you get your area under the curve.

What you need to do is see what is going on here. What we are doing is counting up the area of the infinite number of rectangles under the curve.

The area of each rectangle is $f(x_{k}){\Delta}x$ .

As the number of rectangles, n, becomes unbounded we get the area under the curve.

It's the idea behind integration.

Look it up in any calc book.

When we sub the identities, we get:

$\frac{512}{n^{3}}\sum_{k=1}^{n}k^{2}+\frac{256}{n^ {2}}\sum_{k=1}^{n}k+\frac{32}{n}\sum_{k=1}^{n}1$

Remember the identities I showed you for k and k^2, sub them in:

$\frac{512}{n^{3}}\cdot\frac{n(n+1)(2n+1)}{6}+\frac {256}{n^{2}}\cdot\frac{n(n+1)}{2}+32$

This all whittles down with algebra to:

$\frac{384}{n}+\frac{256}{3n^{2}}+\frac{992}{3}$

Now, it is easy to see as $n\to {\infty}$ , all we are left with is 992/3 and that is the area under the curve.

Speedy232 · Mar 9, 2009, 04:47 PM

Wow... Wow... Wow.. Thank you soooo much. I understand it now and it is actually really simple once you look at it. Can't believe I didn't get that. Thanks for explaining it. I am in your Debt...

Sincere Speedy232

Speedy232 · Mar 9, 2009, 10:48 PM

Sorry to bother again. But see how you have $\frac{512}{n^{3}}\sum_{k=1}^{n}k^{2}+\frac{256}{n^ {2}}\sum_{k=1}^{n}k+\frac{32}{n}\sum_{k=1}^{n}1$

How does that whittle down... As you can tell Integration and algebra arnt really my strength when mixed together. I honestly don't know how you can cancel it down...

galactus · Mar 10, 2009, 10:25 AM

I explained it all in my previous posts.

I got to go now, but I will try to get back with you later.

Speedy232 · Mar 12, 2009, 04:22 AM

$=512/n^3 [n(n+1)(2n+1)/6]+256/n^2 [n(n+1)/2]+32<br /> <br /> =384/n+256/(3n^2 )+992/3$

Ok its my last day and i know I'm missing a step, can anybody elaborate this for me please?