Integration technique: Dummy Variables & Proof: Odd function over symmetric integral is always 0

A new integration technique AND a proof in one video?! What the fck are those?!

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[cookie]

During the stress of my online exam this was the most precious and helpful thing I could have come across

[dox]

It doesnt matter if theres a dog a tree a circle a triangle or YOUR MOM lol <3

[smartsport]

I think the cooler part of this video is when you proved that the integral of f(x) from -a to a is always equal to the integral of f(-x) from -a to a. I guess it's kind of obvious but it's interesting to think about.

[jeromesnail]

6:03 THAT'S THE SOUND OF DA POLICE!!

[cooldawg]

This might sound knit picky but it’s crucial for someone first learning this concept and trying to follow along:

When you first substituted you (-a) into the integral function you do it in the order that made it seem like your “b” was the (-a) when in fact it was the original “a” term that became the (-a). So I think next time you can write:

Intg. from -a to a of f((-a) + a - x) dx

That way your a’s and b’s stay consistent and rearrange order if you need

[kevinnaidoo]

The community of signals and systems thanks you.

[trevr]

This has saved me more time over the years than almost any other technique

[wuyqrbt]

Beautiful ❤️
Your definition for "King rule" as a dummy variable are great.
Just 🔥Flammable 🔥 .
Thank you

[CloudSkywalker]

If x and t are just names, then ∫f(x)dx = ∫f(t)dt. BUT if they are different variables with an explicit relationship between them (as when you do a variable substitution), that equality doesn't hold generally, does it?

It holds in definite integrals because you also adjust the limits of integration. And in the example you put, it just happens that you can arrange the integral in a way that the limits remain the same. But this is just specific case.

Also, in indefinite integrals the results migh or might not be the same. They could be off by a constant term.

Let's say f(x) = x+1. Then, ∫f(x)dx = ∫f(t)dt if x and t are "just" names.
But it t = x+3 (and dt = dx), then:
∫f(t)dt = ∫(t+1)dt = t^2/2+t+C1
∫f(t)dt = ∫f(x+3)dx = ∫(x+3+1)dx = ∫(x+4)dx = x^2/2 + 4x + C2 and you could not change t -> x as though they were "just" names.

In the video about ∫sinx/(sinx+cosx)dx you could change u to x as though they were "just" names because they really were just names and the variabl substitution wasn't really needed since sin and cos are periodic functions with the property that sin(pi/2-x) = cos(x) and cos(pi/2-x) = sin(x). The u wasn't needd and it was just a name. But those are specific cases not a generality.

[MrAssassins]

Four houndred visits within 10 hours it's underrated i guess

[pyrotas]

Almost correct. If f(x)=-f(-x) then the integral over the [-a, a] domain or (-a, a) domain is zero provided f(x) is everywhere integrable within the domain. Otherwise, one ends in all kinds of paradoxes with functions like f(x)=1/x.

[oxman]

1:30 I'm busy trying to actually learn and this boi's just busy roasting me

[unknownknown]

Sadly, that is already proven, by EW Weisstein(Unless it is you!) But, since I believe you independently discovered it, hats off!

[pulanemathe]

Wow that was amazing explanation. It is the first i understood this part of maths. And btw you have an amazing voice and very cute.

[Anthony-dbou]

It’s crazy how much basic properties of functions can help in integration. The things I wish I had known in calc 1, 2, and 3 😂

[johndoe]

I think you should put a set of parentheses like this (f(x) + f(-x))dx

[alimghazzawi]

Ok but sometimes the graph of the function change and originally the integral give us an area under a graph and the only way to mange and get the same area in all region of -♾ to♾ is if the graph was exactly the same

[TonnyF]

this is D:


by the way quite nice intro

[cooldawg]

I’m still confused after you do your substitutions, how does int[f(x)dx] = int[f(a+b-x)dx] accross the same bounds

How can x = a+b-x? Does that mean that x is always (a+b)/2

[thepistac]

"I plus I is indeed just 2I"
*takes off glasses*
"My god"