The limit as cats approaches infinity is...

The answer is 8.

You turn the infinity sign veritical then it becomes 8 :D

Tricky. :)

Hint, hint.... RobDog888. I'm triple infinity. :D

Quote:

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Originally Posted by eyeRmonkey

but when you use the power rule thingy:

∫[xⁿ]
= (xⁿ⁺¹)/(n+1)

Then

∫[ax*(1/bx)]
= ∫[ax*(bx)^-1]

That makes n=-1 in this case and n+1 = 0 so that part would go away... right? Thats the part that was confusing me before. Or am I missing something involving log or ln?

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ax * 1/bx = a/b since x crosses out hence the answer is
(a/b)x + c

and lim cats → ∞ is 3.1415926535897932384626433832795

I didn't intend that :blush:

I meant,

∫[(ax+c)/(bx+d)]

which according to your integrator link is

[(ax+c)x]/(bx+d)

Quote:

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Originally Posted by eyeRmonkey

I wish I was out of school this early.

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I'm out of school forever :D

I think the integrator is stuffed, it doesn't seem to recognise the x hence it calculates thinking that everything is a number.

Heres an example

*Image Removed*

Whoa!

Andrew, you lost me on the second step. Why are there 2 terms all of the sudden? Where did 15/2 come from?

I'll stare at it some more later. Maybe I'll come up with something.

The top is the same as it was before - that's algebra for you.

3/2 * (2x+1) + (15/2) = 3x + 3/2 + 15/2 = 3x + 18/2 = 3x+9.

Which is what you had originally.

Okay I see it now, and I figured it was something like that. I am just curious how you separated it out like that. I mean I see how. But why? Is there a pattern to the madness or is it just something you figure out after working with intergrals for a while?

I guess the point was to get (2x+1)/(2x+1)? But I never would have figured out which fractions to factor out.

Thanks for the help guys.

If anyone is interested, I took my calculus final today. I got a 94%. :)

good job. now get programming.

Congratulations :)
As for why i split them up, you'll notice that one of the fractions becomes 1 and the other is simple to integrate, hence by spliting a hard integral into 2, 3 or even more easier integrals, you are able to get the answer easier.

Okay. I just never would have picked those fractions. I see it now, but I never would have thought of it. I was hoping there was some rule or rule-of-thumb that helped. I guess I will just have to get used to it. :)

The only problem I missed on the calc final was about maximizing the area of 2 adjecent pig pens with only 18 feet of fencing. I don't even know what I did wrong, but I will check it on Monday (we don't have school on Friday because today was the last day of finals).

Quote:

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Originally Posted by eyeRmonkey

The only problem I missed on the calc final was about maximizing the area of 2 adjecent pig pens with only 18 feet of fencing. I don't even know what I did wrong, but I will check it on Monday (we don't have school on Friday because today was the last day of finals).

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Reminds me of maths when i was still as school (1 month ago) ... Oh to pick the fractions, make the first fraction the same as the bottom (only if top degree = bottom degree, like example) then add any extras to make sure you don't change the question, like i added the 3/2. Its really just rearranging the integral to something simplier without modifying the actual question.