fandomsecrets | [ SECRET POST #2317 ]

[ SECRET POST #2317 ]

⌈ Secret Post #2317 ⌋

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Notes:

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Math help

specifically, integral calculus help

specifically^specifically u-substitution.

I just don't get it. At all. My teacher's have tried to explain it, but I still don't get it.

So, do any of you guys get it? Do you have a tool or trick for doing it, or was it just memorization? Do you have a fandom reference?

I just really need to understand this concept, and for some reason it doesn't make any sense to me so far. =_=;

What do you understand so far? Show us an example question involving u-substitution and describe the parts you do understand?

Integrate the equation SQR(1+X) from 0 to 1.

So you substitute U = 1+X. Then you have to find the dU and the dX? I think? And then you have to solve for U or for dU, and that's about where I'm stuck.

I do know, though, that you substitute the values you get when you plug in 0 and 1 into some equation, and rework it so that it's the integral from those two points.

Hi! I'm logged in now. Take a look at this page (http://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/usubdirectory/) and start reading the section from the words "Now the method of u-substitution will be illustrated on this same example. Begin with"

The sentence "Now "pretend" that the differentiation notation" (the 4th sentence down from "Now the method of u-substitution...") is the part you're getting stuck at? Hm. Maybe if I wrote your example this way it'll help you understand.

Let Y = 1+X
dY/dx = 1
(dY/dx)*(dx) = (1)*(dx)
(dY)*(dx/dx) = (1)*(dx)
dY = 1 dx (because dx/dx = 1 )

Do you see how I got from the 2nd equation (dY/dx = 1) to the last equation (dY = 1 dx)?

Now instead of Y, we say U

Let U = 1+X
dU/dx = 1
(dU/dx)*(dx) = (1)*(dx)
(dU)*(dx/dx) = (1)*(dx)
dU = 1 dx

Are you okay with this so far?

Question, how did you get (dx) to (dx/dx)? Wouldn't (dx/dx)=1 and =/= (dx)?

The first (dx) you are talking about, which equation and which side of the equation are you talking about?

If you're talking about the (dx) I multiplied in, the *(dx), I multipled (dx) to the left side and the right side of the equation. It wasn't a random addition.

Let U = 1+X
dU/dx = 1
(dU/dx)*(dx) = (1)*(dx)
(dU)*(dx/dx) = (1)*(dx)
dU = 1 dx

That one. Is it because you had (^dU/_{dX_{)(dX), and you pulled the dx from ^dU/_{dX_over?}}}

Wow, htmfail, sorry. :(

Yep! Do you remember the associative property of multiplication? It's the one that says a multiplication equation has the same product no matter what order its factors are in? A*(B*C)=(A*C)*B=B*(C*A)=etc.?

Division is essentially multiplication written in the form of (1/something). Let's take a look at just the left side of equation 3, (dU/dx)*(dx)

(dU/dx)*(dx)
=(dU)*(1/dx)*(dx)
=(dU)*(dx)*(1/dx)
=(dU)*(dx/dx)

:D?

I won't lie, I had forgotten that. |D

Yes, that makes sense now, thank you~! And I understood everything in your post with that clarification. :3

You're welcome! I see dreemyweird and another nonnie have been helping you out too. Is there anything else about u-substitution you want clarified?

(Since you're working with definite integrals where x has upper and lower bound limits, don't forget to change the limits to upper and lower bounds of U when you're integrating f(u) instead of f(x) ;P)

So, I change the bounds by plugging the upper and lower bounds into the U-formula, right?

Re: I need help

agentcthulhu - 2013-05-08 00:18 (UTC) - Expand

Re: I need help

(Anonymous) - 2013-05-08 00:26 (UTC) - Expand

Re: I need help

agentcthulhu - 2013-05-08 00:40 (UTC) - Expand

Re: I need help

(Anonymous) - 2013-05-08 00:46 (UTC) - Expand

Re: I need help

agentcthulhu - 2013-05-08 01:11 (UTC) - Expand

Re: I need help

(Anonymous) - 2013-05-08 01:17 (UTC) - Expand

Re: I need help

agentcthulhu - 2013-05-08 01:30 (UTC) - Expand

Re: I need help

(Anonymous) - 2013-05-08 01:34 (UTC) - Expand

Re: I need help

agentcthulhu - 2013-05-08 01:46 (UTC) - Expand

Not OP

This isn't a homework help site, don't be douchy. Didn't she just say the concept itself was unclear to him/her?

Nayrt

That's... not douchey at all? That's asking for very relevant information in terms of how to help OP, not being a douche.

OP

I didn't think they were being douchey at all. :(

Are you nuts?

Well, IDK if I'll be able to explain it better than your teacher did, but it is essentially a way to make the integration process faster by replacing a part of the function with u, differentiating this part and subsequently writing both dx and the function in terms of u (and du). You just make it look simpler in order to understand what integration rule is applicable. Say, if you have (2x+3)^3, then you let 2x+3 be u; hence du/dx is 2; dx=du/2; 1/2 int(u^3du)= you integrate it in the regular way and get shit done. It can be more complicated with more complex functions, but that's the basic mechanism.

Edited 2013-05-07 23:30 (UTC)

This might be a stupid question, but how did you go from du/dx=2 to dx=du/2?

Like, how did you decide that it was ^du/_dx?

Integration by parts and u-substitution are the times you're allowed to treat du/dx the derivative as a fraction. Most of the time, you're not supposed to do that, but in this case you just go ahead and pretend it is. :)

Does that help?

So there's no reason, it just is? And it's like that?

THAT MAKES SO MUCH MORE SENSE OMG.

I THOUGHT the teachers were just assuming everyone knew why and not understanding when I was asking (I can't words sometimes), but that's just one of the rules of integral calculus?

Not really; it's because they're rates of change. Hence here they behave as parts of a fraction. Say, if u changes twice as fast as x, we can safely assume the relationship above is true.

Edited 2013-05-07 23:54 (UTC)

Oh. Okay, that makes sense!

Because that's the way it works when one differentiates a function with two unknowns. You kind of need to acknowledge the fact that you're differentiating the function with respect to x, not to u. So, when you have u, you can't just write "du", you need to add this /dx. Technically speaking, this is not a fraction, but it does behave as one.
Nope, I'm glad if I'm of any help)

I feel really stupid now, because I didn't get that part, but now I've gotten it explained to me, and I feel so much better omg.

Thank you!

You're welcome!

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