FOR PEOPLE ASKING WHERE THE DU WENT: Derivatives and integrals are inverse functions, so they cancel each other out. When you INTEGRATE a DERIVATIVE (which is du) it cancels du out.
Huh? The norm on my university is 30. The department of technical physics rewards their students 30% less credits or so in general to circumvent the law of how many credits you can take per term in order for them to learn more, as well as their content being harder.
Personally, AP Calculus AB was easy for me this year except for this topic. But like usual, Sal makes it so much more easier. Thanks man, years from now I'm going to be thinking about my high school years and appreciating how much you've helped me in chemistry, physics, and maths. You're the GOAT
Professors spend 2-3 class periods talking about one concept and people still don't get it (I'm people.) Then I watch a 5 minute video from you and it's easy.
+Jem Celespara Think of 'du' as 'dx' ok? When we have a standard integral with respect to x, dx will be at the end to say 'with respect to x'. When we solve the integral the dx doesn't matter because that's all it was saying. In U-substitution when we are dealing with u instead of x we write 'du' to say 'with respect to u.
BCrafty121 thats the "power rule" which only works if the base of the exponent is a variable. E is not a variable but a number, the derivative of e^x is ln(e)e^x= e^x because the ln(e)=1
I don't understand, I initially tried it the conventional way and got (x^3+x^2)e^(x^3+x^2)+C by treating e as a constant. I don't fully understand how e works, I would imagine that int of e^u du=e^u * u + C. Sorry, if this question sounds a bit stupid, but most of my calculus is self taught and I am in geometry.
+NexDev Nop. e^x is basically defined in a way such that its derivative (I prefer differentiation) is e^x. Remember when you try to find the differentiation (or derivative) for e^x and you got some limit that cannot be evaluate, so that's where we defined that limit as = 1.
When you have e^u * du and you take the anti derivative, what happened to the du? I guess I'm not understanding why it can disappear instead of becoming "u" when you take the antiderivative.
The integral and the du both cancel. thus you see the end result as it is. Its just like when you see a square root of a number, like Square root of 3, to a power. The power cancels out the square root and you are left with a single value.
it would change it, but he does this to show how you would substitute for beginners. If it helps any, what we are essentially doing here is replacing dx with du so because the dx is operating on the first function it becomes e^u du. where du = f'(x) dx. So he moved it to show that it wont effect e^f(x) except for when we substitute u.
I assume you mean at 4:30 - du just means when taking the integral it's in respect to "u" - so when you take the integral of e^u (which is still e^u) the du goes away because you took the integral of it and that's the answer. Finally you just put a +C at the end to account for your constant.
If the derivative is a constant, multiply the inside of the integral by that constant, and divide by that same constant on the outside. This will keep the value of the integral the same, and now you can do the substitution. Otherwise, tough luck. Look for another method of integration.
At 1:58 saying that really isn't du divided by dx.... could someone just explain that to me? I mean I kind of know its a notation trick. But calculating the derivative of a function actually means dividing a tiny change in u by a tiny change in x. I know I mess up something really badly, but please, could someone explain that to me?
It isn't actually any particular ratio between a tiny change in u and a tiny change in x. It's what that ratio approaches as the tiny change in x approaches 0.
not trying to offend you but a teacher has the right to teach however they want to teach. A good teacher teaches how they presume the most students will understand and the fact that most teachers choose this method speaks volumes. You come across as someone trying to be pompous.
The problem is that this way is wrong. It works, yes, but isn't the right way to do it and it confuses students. dx and dy are not numbers. To do this in this way you need to study a lot of other things before.
Anacleta Ludovica I agree and I am aware of why the dx "dissapears". But some people can learn it later as it isn't important to know at the stage when most people learn it at.
You can teach this at that stage, without multiplying anything. This is the way I prefer: Let f'(x)=e^x and g(x)=x^3+x^2 Then f(x)=e^x and g'(x)=3x^2+2x So by the chain rule: Integral [f'(g(x)) * g'(x)] = f(g(x)) +c = e^(x^3+x^2) + c
I agree because i know this I can visually undo most usubstition problems. But its not necessary for someone to understand the way you learn and those wwho would benefit from it will simply understand it or will learn it on their own.
FOR PEOPLE ASKING WHERE THE DU WENT:
Derivatives and integrals are inverse functions, so they cancel each other out. When you INTEGRATE a DERIVATIVE (which is du) it cancels du out.
omgflyingbanana THANK YOU OMG.
derivatives and integrals are inverse operations, not inverse functions.
@@victorserras 100% of people knew what he means, don't be that guy
Honestly, I didn't XD
Bro i thought I would never understand. thank you
can i do a you-substitution to replace my calc prof?
@Kathan Jani are u still alive
@Kathan Jani no i diedededlyded
Are you still alive ??? Pls say yes 😭
@@tonyhaddad1394 not meeeee
XDDDDD
you just explained in 5 minutes what took my professor 2 class periods( 5 hours). I thank you.
By far the best explanation of this concept on youtube
You are literally the only reason why I'm surviving 32 credit hours per semester.
HOLY SHIT!
Wow... that's like... a lot of credits...... such schooling... many night... much coffee... so smart... numerous applause :D
***** That's twice the amount that my college even allows. Props.
Huh? The norm on my university is 30. The department of technical physics rewards their students 30% less credits or so in general to circumvent the law of how many credits you can take per term in order for them to learn more, as well as their content being harder.
What the hell?! All I have to say is this...
WOW
4:20 Nice rapping
HAHAHAHAHA
Top 10 Rappers Eminem Was Too Afraid To Diss
@@AscendedOnOsu hahaha
2x = speed bomb
Personally, AP Calculus AB was easy for me this year except for this topic. But like usual, Sal makes it so much more easier. Thanks man, years from now I'm going to be thinking about my high school years and appreciating how much you've helped me in chemistry, physics, and maths. You're the GOAT
Your absolutely correct.
Should've liked this video a year ago when I actually watched it. This is a great channel btw ty guys.
11 Years later, this is still one of the best math tutorial videos out there
People forever asking me how I study... THIS...
Thank you Khan Academy :)
Great review, thanks
Great video! Super helpful!
Brilliant. This shows this concept so well.
Khan why are you such a genius?!
Thanks Khan!
Thank you! This is super helpful!
These videos rock!!!! Merely stating the obvious :)
Thank you so much sir, it cleared up a lot of my confusions, God bless you
Professors spend 2-3 class periods talking about one concept and people still don't get it (I'm people.) Then I watch a 5 minute video from you and it's easy.
loved the explanation
Thanks ! that was very useful 😊
...this is equal to u.
That pun made me feel so integrated...
omg... Pun-ception
Thank you so much!
Khan is the best!
u r an amazing teacher
For me it's simple look
The primitive of ue^u ' is e^(u) so you just have to replace
very helpful, thanks :)
Nice example Khan!!!
Khan Academy 4 life!
The derivative of an exponent is the exponent times it's power's derivative so the integral will be e^(x^3 + x^2)
go ahead and logout for me omar
+ C
@@reallife2103What's your problem?
Integrate Sin[x]/sqrt(Cos[x]) dx
so does du just go away because it is part of undoing the chain rule?
when the integral sign goes away so does the du, its inverse
A bit of a late reply I'm afraid, that guy's probably graduated college from now XD
However, I've had the same doubt. So thanks!
derivitive of e^x is equal to e^x * dx and normally dx is just 1 so the anti dirivitive cancels out that du
the du would cancel out bc of the chain rule (inversed)
{f'(x)e^f(x) = e^f(x) + c uses this and you get the answer
Why DU vanishes? It's simply because you are integrating the given equation already.
+Jem Celespara Think of 'du' as 'dx' ok? When we have a standard integral with respect to x, dx will be at the end to say 'with respect to x'. When we solve the integral the dx doesn't matter because that's all it was saying. In U-substitution when we are dealing with u instead of x we write 'du' to say 'with respect to u.
+Ruben Marquez Thank you!
best explanation (Y)
Here's a hard problem.
Integrate (4x^3*e^x^4) dx
We need the integration in the form px+q by ax^2+bx+c
Thanks...
Where did the du go....?
Thanks@!!!!!!!!!!!!!!!!!!!!!!!!!!!
Wait, derrivative of e^u should be ue^(u-1). So integral of e^u should be ((u+1)e)^(u+1).
BCrafty121 thats the "power rule" which only works if the base of the exponent is a variable. E is not a variable but a number, the derivative of e^x is ln(e)e^x= e^x because the ln(e)=1
Emperor Khan..!
are u-substitution and integration by parts interchangeable?
Nope, they're totally different things. u-substitution undoes the chain rule; integration by parts undoes the product rule.
I don't understand, I initially tried it the conventional way and got (x^3+x^2)e^(x^3+x^2)+C by treating e as a constant. I don't fully understand how e works, I would imagine that int of e^u du=e^u * u + C. Sorry, if this question sounds a bit stupid, but most of my calculus is self taught and I am in geometry.
+Jason Zhao you can't treat e as a constant because it's to the power of x^3 + x^2
Huaidong Tang What I dont understand is why e is so special. e itself is a constant, yet it is not treated as one,
e IS a constant. However e^x is not a constant. Just like how 2 is a constant but 2^x isn't a constant
Huaidong Tang But we treat 2^x differently than with e^x
+Jason Zhao no we don't
awesome
why can’t my teacher explain it this well
guys, isn't that integration of e^u du is
e^u
------
u
right?
so, the answer should be
e^(x^3 + x^2)
------
x^3 + x^2
amirite?
+NexDev
Nop. e^x is basically defined in a way such that its derivative (I prefer differentiation) is e^x. Remember when you try to find the differentiation (or derivative) for e^x and you got some limit that cannot be evaluate, so that's where we defined that limit as = 1.
Marco Antonio Graziano de Castro thanks for the explanation m8 :D
Sal, where are you a professor so I can just take all my calculus courses through you? I'm paying $475/hr and would rather pay it to you.
3:25
Khan:
Me: Wait, That's illegal
but what if the integrand lacks a factor x needed for du??
If it's a constant, you can multiply the inside and divide the outside by that constant. Otherwise, try finding another way to do the integral.
When you have e^u * du and you take the anti derivative, what happened to the du? I guess I'm not understanding why it can disappear instead of becoming "u" when you take the antiderivative.
The integral and the du both cancel. thus you see the end result as it is. Its just like when you see a square root of a number, like Square root of 3, to a power. The power cancels out the square root and you are left with a single value.
Why didn't this video exist in 2009 :'(
magenda!
lmao I'd give an50331 a positive vote if I could. :D
Totallyyy... that was a good one :P
no wonder he got into MIT
At about 3:15, why are we able to "move" e^(x^3 + x^2) "behind" the dx. Doesn't that change our integral?
it would change it, but he does this to show how you would substitute for beginners. If it helps any, what we are essentially doing here is replacing dx with du so because the dx is operating on the first function it becomes e^u du. where du = f'(x) dx. So he moved it to show that it wont effect e^f(x) except for when we substitute u.
lol-to baad this vid was made 15 days after my final :Q
F in chat for you
It's unfortunate that Sal picked an "e" as part of this video. I think as part of the an introduction to u-substitutuion it confuses things.
hey is this high school or college topic?
both
First! :D
anyone explain how come we just dropped the du?
I assume you mean at 4:30 - du just means when taking the integral it's in respect to "u" - so when you take the integral of e^u (which is still e^u) the du goes away because you took the integral of it and that's the answer. Finally you just put a +C at the end to account for your constant.
You drop the du when you integrate it already, you know, tagging a + C on the given equation
what is the program?
Ok but what do you do if the derivative is not outside??
If the derivative is a constant, multiply the inside of the integral by that constant, and divide by that same constant on the outside. This will keep the value of the integral the same, and now you can do the substitution.
Otherwise, tough luck. Look for another method of integration.
where did du go?! i solved this problem on my own first and i got 1/6e^(x^3+x^2)+c ? :(
At 1:58 saying that really isn't du divided by dx.... could someone just explain that to me? I mean I kind of know its a notation trick. But calculating the derivative of a function actually means dividing a tiny change in u by a tiny change in x. I know I mess up something really badly, but please, could someone explain that to me?
It isn't actually any particular ratio between a tiny change in u and a tiny change in x. It's what that ratio approaches as the tiny change in x approaches 0.
Where does du go? It seems to just disappear when you sub your f(x) back in for u???
Isn’t the integral of e^u = (e^u)/u’
That is the derivative of u. Im a couple years late. I hope u doing good homie
makes sense!! lol
"oh lucky me the u' happens to be the ^ of e"...try showing harder examples.
I love you Sal. Please come to my house, I will make you food
Samantha davson late
Times Eeee
It's really sad to see how everybody teachs this by "multiplying by dx each side" and nobody teachs this the right way.
not trying to offend you but a teacher has the right to teach however they want to teach. A good teacher teaches how they presume the most students will understand and the fact that most teachers choose this method speaks volumes. You come across as someone trying to be pompous.
The problem is that this way is wrong. It works, yes, but isn't the right way to do it and it confuses students. dx and dy are not numbers. To do this in this way you need to study a lot of other things before.
Anacleta Ludovica I agree and I am aware of why the dx "dissapears". But some people can learn it later as it isn't important to know at the stage when most people learn it at.
You can teach this at that stage, without multiplying anything. This is the way I prefer:
Let f'(x)=e^x and g(x)=x^3+x^2
Then f(x)=e^x and g'(x)=3x^2+2x
So by the chain rule:
Integral [f'(g(x)) * g'(x)] = f(g(x)) +c = e^(x^3+x^2) + c
I agree because i know this I can visually undo most usubstition problems. But its not necessary for someone to understand the way you learn and those wwho would benefit from it will simply understand it or will learn it on their own.
....I failed to understand where the french come into this.
nerd
one person failed calculus
I am not happy with the explaination. It was purely mechanical without any intuition.
Okay. Teach us the method. You're acting as if we are mathematicians. Why use that as the U? Do better!
Lol magenta
Eeeeeeeeeeeeee
Thank you so much!