Question

I have the Calculus BC final tomorrow. Any last minute study tips? I'm pretty much good to go, curriculum wise, but I'm scared out of my mind!

Answer 1

Did you just delete something that you posted? I didn't quite catch that.

Answer 2

Sample problem:
\[\int\limits \frac{d \beta}{\left( \beta^2 + \xi^2\right)^{\frac{n}{2}}}\]
:p

Answer 3

0_0 Ok. Well. In that case I guess I should prepare myself to fail the test xD The beta thing throws me off... Are you certain that this is High school level Calc BC? Because I dont believe I've ever integrated something with so many variables in it! Hahaha

Answer 4

Well you've done like:
\[\int\limits \frac{dx}{(x^2+49)^{\frac{3}{2}}}\] right?
Its the same thing but with beta exchanged for x and xi is a constant :P. The n/2 makes it more general and significantly harder, I was just messing really :P

I'm in quantum mechanics right now and check out my hw:

Answer 5

http://openstudy.com/study#/updates/5184549de4b09587cd396e42

Answer 6

Sure, I've integrated that before. But I figured you were just messing. I genuinely attempted the problem though XD Way to pull at my nerves! haha. And quantum mechanics is pretty impressive. Good job. I wish I had the mental capacity to follow something like that.

Answer 7

If you're interested I could probably work that problem out for you :P

Answer 8

I would love that. But only if you wish. I would probably make the 3/2 a negative, so that f(x) could be in the numerator, then derive from there. Unless there is some sort of trig function formula that would work better. I don't think u substitution will work.

Answer 9

Since 49 is a perfect square, im assuming there is some sort of formula that has slipped my mind.

Answer 10

Well no need for formulas :P Rewrite it as:
\[\int\limits \frac{dx}{(x^2+7^2)^{\frac{3}{2}}}\]
And let:
\[x = 7 \tan(\phi) \implies dx = 7 \sec^2(\phi)d \phi\]
This gives us:
\[\int\limits \frac{7 \sec^2(\phi) d \phi}{7^3(\tan^2(\phi)+1)^{\frac{3}{2}}}\]
But:
\[\cos^2(\phi)+\sin^2(\phi) = 1 \implies \frac{\cos^2(\phi)}{\cos^2(\phi)} + \frac{\sin^2(\phi)}{\cos^2(\phi)} = \frac{1}{\cos^2(\phi)} \]
\[\implies 1 + \tan^2(\phi) = \sec^2(\phi)\]
So we now have:
\[\frac{1}{7^2} \int\limits \frac{\sec^2(\phi)d \phi}{(\sec^2(\phi))^{\frac{3}{2}}} = \frac{1}{49}\int\limits \frac{d \phi}{\sec(\phi)} = \frac{1}{49} \int\limits \cos(\phi) d \phi = \frac{1}{49}\sin(\phi)+C\]

Answer 11

That problem>My understanding. xD But thank you! It is very impressive. I wish you could take my final for me tomorrow, you seem very well prepared. ;)

Answer 12

However:

So we have that:
\[\sin(\phi) = \frac{\textrm{opposite}}{\textrm{hypotenuse}} = \frac{x}{\sqrt{x^2+49}}\]
So our final answer is:
\[\frac{x}{49} \frac{1}{\sqrt{x^2+49}} + C\]


You haven't done trig subs? :O

Answer 13

And to confirm:
https://www.wolframalpha.com/input/?i=integral+of+1%2F%28x^2%2B49%29^%283%2F2%29

Answer 14

Not at such a magnitude, no. We have, but more simplistically, I believe. Simple, precalc-ish trig subs. Nothing like that.

Answer 15

Oh :x my bad then lol

However, if you have any other problems you don't know fire away!

Answer 16

Thanks! I wont hesitate to ask. As you can see, I'm pretty ill-prepared. Lots of little mistakes and slip ups that need refining. Thank you for your help.

Answer 17

Not a problem, I get to keep my calc skills sharp as well :P