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## Integral Calculus (2017 edition)

### Course: Integral Calculus (2017 edition) > Unit 6

Lesson 2: u-substitution- 𝘶-substitution intro
- 𝘶-substitution: rational function
- 𝘶-substitution: multiplying by a constant
- 𝘶-substitution: logarithmic function
- 𝘶-substitution: challenging application
- 𝘶-substitution: special application
- 𝘶-substitution: indefinite integrals
- 𝘶-substitution warmup
- 𝘶-substitution: definite integral of exponential function
- 𝘶-substitution: double substitution
- 𝘶-substitution: definite integrals
- u-substitution challenge

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# 𝘶-substitution intro

Using u-substitution to find the anti-derivative of a function. Seeing that u-substitution is the inverse of the chain rule. Created by Sal Khan.

## Want to join the conversation?

- In these series of videos (U-substitution) you introduce the treatment of the derivative operators (dx, du, etc) as fractions. You specify that they really are not, but treat them like that anyway. Have you made a video where you prove, or at least explain why they can be treated like that? Are there also other nice ways you can treat the operators? Maybe there is a video that introduces all the notations in differential form?(192 votes)
- the dxs and dus can actually be thought of as an infinitesimal quantity (like Leibniz did) as well as a marker for the variable you are dealing with. This is because the d ultimately derives from the little
*delta*meaning difference (in differential calculus) that eventually becomes negligibly small. This d is actually subject to multiple interchangeable definitions that can be very useful like a very small area of value one.(26 votes)

- I am still confused about what happens to the du. I understand the dx portion but, how exactly is the du taken out of the equation.(47 votes)
- Imagine they are fractions. dy/du has du on the numerator and du/dx has du on the numerator. Multiplying them together, you get (du/du)*(dy/dx), and since du/du is 1, you are just left with dy/dx.(5 votes)

- Sal said that you
*could*treat du/dx as a fraction and in the later videos, he does that. But, isn't du/dx not really a fraction? Is there a proof for what Sal does in the video (du/dx=3x^2+2x --> du=3x^2+2x dx)?(12 votes)- Actually, du/dx
*is*a fraction! It's a bit different, though.

Derivatives come from the formula of slope:

(change in y)/(change in x)

When we take the derivative of y, we get dy/dx, or:

(infinitely small change in y)/(infinitely small change in x)

If we use technical terms, this is:

(differential of y)/(differential of x)

NOTE: dy/dx can change from x to x with the same function because while y changes its speed, x doesn't because it's the independent variable, meaning that dy changes but dx doesn't, changing dy/dx.

Back on topic, though, d(variable) simply means a differential of that variable. It is a quantity, but a very small one (an*infinitesimal*one).

Since differentials are technically quantities, so then dy/dx is just a fraction of quantities! Just like any fraction, you can multiply it by its denominator to leave just the numerator!

I kind of talk about this more here:

https://www.khanacademy.org/cs/anti-derivatives-anti-differentials-and/2047217722

I hope this helps!(32 votes)

- At1:01Sal says that he will go more in depth in an other video. Does anybody know where this video is?(11 votes)
- If anyone it's still wondering, the video is this

https://www.khanacademy.org/math/integral-calculus/integration-techniques/reverse-chain-rule/v/reverse-chain-rule-introduction(16 votes)

- why do we don't consider dx or du in the final antiderivitive answer? if they are concerned in definite integral, then they should also be considered in antiderivites also, but it's not like that,,,,Why??(10 votes)
- When we integrate with respect to a variable such as x or u, the corresponding differential, dx or du, seems to disappear. But it doesn't truly disappear: it gets
*integrated into*the final answer.

This is easiest to see with the definite integral. When you go back to the way Riemann sums work, you see that dx, an increment in the x direction, gets multiplied by a quantity in the y dimension to produce the result. Note that dx does not appear in the result, which is simply a constant, but dx was used in producing that result.

The fundamental theorem of calculus tells us that we can produce the same result that we get from taking the limit of Riemann sums much more easily by finding the antiderivative of a function. The dx has been incorporated into the result, so we don't show it separately.(7 votes)

- If you were subbing for the most imbedded part of the function but what you subbed also existed somewhere else in the function, sometimes it doesn't work to substitute for both. Does it break algebraic rules to selectively substitute for one and leave the other as is?(7 votes)
- Sorry, that is ordinarily not a valid means of substitution. A valid substitution, generally speaking, requires that
**ALL**references to the original variable be replaced ESPECIALLY including its`dx`

(or whatever the variable is). You cannot have ANY stray bits leftover.

The purpose of u substitution is to wind up with`∫ f(u) du`

Where`f(u) du`

is something you know how to integrate. And remember`du`

is the derivative of whatever you called`u`

, it is NOT just some notation.

So, the answer is, no, you cannot do u-substitution that way. With integration, being close to a standard form is not good enough: you must have an exact match. For example,`∫ (x)∙cos(x²) dx`

is very easy to integrate but the very similar looking`∫ cos (x²) dx`

is nightmarishly difficult (getting into something called Fresnel integrals).

It isn't really an exception, but you can sort of have leftover bits when you do integration by parts. In that case, you do two u-substitutions (but you call them by different variables). However, even with integration by parts taken together the two substitutions must completely replace the original variable, with no leftovers allowed.(8 votes)

- ʃ f(x).g(x) dx we can`t say that this expression is

ʃ f(x)dx . g(x) so how we could say that

ʃ (3x3 + 2x) . e(x3 + x2 ) dx is equal to the expression

ʃ (3x3 + 2x) dx . e(x3 + x2 )(7 votes)- You're right, ( ʃ f(x)*g(x)*dx ) is different from ( ʃ f(x)*dx )*g(x), but what Sal meant was ( ʃ f(x)*dx*g(x) ). It's a product, so you can change the order of the terms multiplied together.(5 votes)

- Why do we use u? Is it just tradition?

Also, if you see something like ∫(dx)∙e(x), why don't you just cancel out ∫(dx)? The sum of the little parts of x (Calculus Made Easy by Martin Gardner and Sylvanus B. Thompson say that's how you can think of it) must equal x, after all...(2 votes)- This is the u-substitution introduction:

"U-substitution is a must-have tool for any integrating arsenal (tools aren't normally put in arsenals, but that sounds better than toolkit). It is essentially the reverise chain rule. U-substitution is very useful for any integral where an expression is of the form g(f(x))f'(x)(and a few other cases). Over time, you'll be able to do these in your head without necessarily even explicitly substituting. Why the letter "u"? Well, it could have been anything, but this is the convention. I guess why not the letter "u" :)"

Do you agree that this is what an indefinite integral does?

int(y'dx)=y

If so, do you agree that y' equals dy/dx?

int((dy/dx)dx)=y

If so, do you agree that the (dx)s cancel out it the integral?

int(dy)=y

The integral sums up all the differentials of a variable and makes it that variable. I think that's Martin Gardner and Sylvanus B. Thompson were talking about. You can't just get rid of the dx because it needs to cancel out the the dx in the derivative of the function.

If you were referring to something different than what I just showed, please tell me by commenting on this answer.(11 votes)

- I have a question. How do you know on what does the integral symbol apply? I always assumed that if I saw an integral sign I had to integrate the part that was between the integral symbol and the dX. But in this example at3:24the e^(bla) part also has to be integrated, while it's behind the dx.

For example in this video; https://www.khanacademy.org/math/calculus/integral-calculus/indefinite_integrals/v/antiderivative-of-hairier-expression everything in between the integral sign and the dx has to be integrated. But if dx is a factor of the expression behind the integral sign then this doesn't apply.

Can somebody clear up for me, how I know what is part of an integrand and what not?

Many thanks.(5 votes)- While it is customary to put the dx at the end, it is not required.

So, you integrate everything after the ∫ symbol. If there is anything that you need to treat separately you would typically have the problem written with some clear distinction indicated such as setting the integrand off with { }. But, usually if there is something not to be integrated, you would put that before the ∫ symbol.(5 votes)

- I'm kinda loss on this integral thing, I understand that the integration by parts is for multiplying functions and understand completely the antiderivative process , however on this u substitution I don¡t quite get the intuition behind this. When should I use u substitution ? Also, on the following exercises they use something called reverse chain rule, and on the hints they provide the formula f'(g(x))*g'(t)dx = f(g'(t)) + C , and yet I can't figure out where are they getting that from, we were never given an specific formula for u-substitution... Thanks to all willing to help ! (:(4 votes)
- Anti-derivatives are harder than derivatives. While derivatives tend to be evaluated exactly, using a far more limited set of rules. Anti-derivatives require many more rules and methods to evaluate if it can be evaluated in terms of an exact function. There are a bunch of different methods and approaches to solve integrals, sometimes multiple methods can be used, sometimes only one can solve it exactly, and sometimes exact equations don't exist leading to numerical approximations. u substitution is another method of evaluating an integral in an attempt to transform an integral that doesn't match a known integral rule into one that does. I'll give an alternative formulation of the rule.

u = g(x), if the range is some interval and f(x) is continuous over that interval then,

integral(f(g(x))g'(x)dx) = integral(f(u) du), say you have some function x * sin(x^2+5) that you want to evaluate the integral of using this method.

Re-arrange that function to sin(x^2+5) * x, now the derivative of the function in the sine function looks remarkably like x, constant values are just fine in the formulation so.

x dx = 1/2 * du giving a straightforward way to define this, u is x^2+5, since there isn't a 2x value it is known there will be a 1/2 factor. the substituted integral would be 1/2 * integral(sin(u)*du) = - 1/2 cos(u) + C = -1/2 cos(x^2+5) + C.

The basic idea is try to try to evaluate a complicated integral using a more simple one, determining u depends a lot on intuition, experience, and trial and error, than an exact equation.(7 votes)

## Video transcript

Let's say that we have
the indefinite integral, and the function is 3x
squared plus 2x times e to x to the third
plus x squared dx. So how would we go
about solving this? So first when you
look at it, it seems like a really
complicated integral. We have this polynomial right
over here being multiplied by this exponential expression,
and over here in the exponent, we essentially have
another polynomial. It seems kind of crazy. And the key intuition
here, the key insight is that you might want
to use a technique here called u-substitution. And I'll tell you
in a second how I would recognize that we
have to use u-substitution. And then over time,
you might even be able to do this type
of thing in your head. u-substitution is essentially
unwinding the chain rule. And the chain rule-- I'll go
in more depth in another video, where I really talk
about that intuition. But the way I would
think about it is, well, I have this crazy
exponent right over here. I have the x to the
third plus x squared, and this thing right
over here happens to be the derivative of x
to the third plus x squared. The derivative of x to
the third is 3x squared, derivative of x squared is
2x, which is a huge clue to me that I could use u-substitution. So what I do here is this
thing, or this little expression here, where I also see its
derivative being multiplied, I can set that equal to u. So I can say u is equal to x
to the third plus x squared. Now, what is going to
be the derivative of u with respect to x? du dx. Well, we've done
this multiple times. It's going to be
3x squared plus 2x. And now we can write this
in differential form. And du dx, this isn't really
a fraction of the differential of du divided by
differential of dx. It really is a form
of notation, but it is often useful to kind of
pretend that it is a fraction, and you could kind of view this
if you wanted to just get a du, if you just wanted to get a
differential form over here, how much does u change
for a given change in x? You could multiply
both sides times a dx. So both sides times a dx. And so if we were to pretend
that they were fractions, and it will give you the
correct differential form, you're going to
be left with du is equal to 3x squared plus 2x dx. Now why is this over here? Why did I go to the
trouble of doing that? Well we see we have
a 3x squared plus 2x, and then it's being multiplied
by a dx right over here. I could rewrite this
original integral. I could rewrite this
as the integral of-- and let me do it in that color--
of 3x squared plus 2x times dx times e-- let me do that
in that other color-- times e to the x to the
third plus x squared. Now what's interesting
about this? Well the stuff that
I have in magenta here is exactly equal to du. This is exactly equal to du. And then this stuff I have up
here, x to the third plus x squared, that is what
I set u equal to. That is going to be equal to u. So I can rewrite
my entire integral, and now you might
recognize why this is going to simplify
things a good bit, it's going to be equal to--
and what I'm going to do is I'm going to
change the order. I'm going to put the
du, this entire du, I'm gonna stick it on
the other side here, so it looks like more
of the standard form that we're used to seeing
our indefinite integrals in. So we're going to have
our du times e to the u. And so what would the
antiderivative of this be in terms of u? Well, the derivative of
e to the u is e to the u. The antiderivative of e
to the u is e to the u. So it's going to be
equal to e to the u. Now, there is a
possibility that there was some type of a
constant factor here, so let me write that. So plus c. And now, to get
it in terms of x, we just have to
unsubstitute the u. We know what u is
equal to, so we could say that this is
going to be equal to e. Instead of writing u, we
could say u is x to the third plus x squared. And then we have our plus c. And we are done. We have found the
antiderivative. And I encourage you to take
the derivative of this, and I think you will find
yourself using the chain rule, and getting right back
to what we had over here.