[ SECRET POST #2317 ]

[case]

⌈ Secret Post #2317 ⌋

Warning: Some secrets are NOT worksafe and may contain SPOILERS.

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Notes:

Secrets Left to Post: 02 pages, 048 secrets from Secret Submission Post #331.
Secrets Not Posted: [ 0 - broken links ], [ 1 - not!secrets ], [ 0 - not!fandom ], [ 0 - too big ], [ 0 - repeat ].
Current Secret Submissions Post: here.
Suggestions, comments, and concerns should go here.

Comments

Re: I need help

[dreemyweird]

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.

Re: I need help

[(Anonymous)]

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?

Re: I need help

[(Anonymous)]

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?

Re: I need help

[(Anonymous)]

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?

Re: I need help

[dreemyweird]

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.

Re: I need help

[(Anonymous)]

Oh. Okay, that makes sense!

Re: I need help

[dreemyweird]

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)

Re: I need help

[(Anonymous)]

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!

Re: I need help

[dreemyweird]

You're welcome!