- 𝘶-substitution intro
- 𝘶-substitution: multiplying by a constant
- 𝘶-substitution: defining 𝘶
- 𝘶-substitution: defining 𝘶 (more examples)
- 𝘶-substitution: defining 𝘶
- 𝘶-substitution: rational function
- 𝘶-substitution: logarithmic function
- 𝘶-substitution warmup
- 𝘶-substitution: indefinite integrals
- 𝘶-substitution: definite integrals
- 𝘶-substitution with definite integrals
- 𝘶-substitution: definite integrals
- 𝘶-substitution: definite integral of exponential function
- 𝘶-substitution: special application
- 𝘶-substitution: double substitution
- 𝘶-substitution: challenging application
Manipulating the expression to make u-substitution a little more obvious. Created by Sal Khan.
Want to join the conversation?
- Sorry if this seems silly, but I still do not quite understand where the 1/7 comes from at1:47instead of just 7?(36 votes)
- Unfortunately, we can't simply make up a constant to bring in. To remedy this, he effectively multiplies the primary equation by 1, which we can do, since 1 multiplied by any number is itself. 1/7*7=7/7=1, so by using this method, he is able to bring in a constant without changing the primary equation to something it is not.(89 votes)
- at4:35, when that 1/7 is distributed to the rest of the problem, wouldn't the result end up with a 1/7 C?(8 votes)
- Any constant multiplied by a scalar is still just some constant. Right after that Sal talks about how you could call them C_1 and C_2 to show that they are different constants. However, this precision is not normally required and is often left out entirely since it doesn't change the result. If you wanted to be thorough, you would call them C_1 and C_2 and also state that C_1 = 1/7 C_2.(33 votes)
- So is it typically a good strategy to choose "u" as something that either is (1) in the denominator, (2) in parenthesis, or (3) under a radical? I'm just looking for good strategies to choosing good "u" candidates. Thank you.(7 votes)
- Whatever you choose as
u, you MUST have the derivative of that thing you call
ubeing multiplied by whatever standard integral function you wish to use.
All of the standard integral forms you will learn amount to:
∫ f(u) duwhere
duis NOT just some notation, it needs to be treated as the derivative of whatever you call
So, an effective strategy is to break the function into two pieces that are multiplied by each other. The first piece will be something you can use u substitution for to get the integral into a form you know how to integrate. The second piece will be the derivative of whatever you called
u. If you cannot do that, you cannot use u substitution.
An example I recently gave to another post. Suppose you had:
∫ cos(x²) dxyou might think make
u=x²and this becomes
∫cos (u) duwhich is easy to integrate. But this is WRONG.
u=x²you must also have
du = 2x dx. There is no
2xin the integral, so you cannot use that substitution. You would need:
∫ 2x cos (x²) dxyou have
du = 2x dxand that gives you:
∫ cos (u) du = sin (u) + C = sin (x²) + C
It turns out, though it looks simpler,
∫ cos(x²) dxcannot be integrated by any means taught in introductory integral calculus courses, but is a very advanced level problem.
So, remember, the
duis not just some notation you can tack on.(14 votes)
- What happens to du when you take the integral of du? It just disappears? What happens when you take the derivative of dx? does it disappear too? WHY R THINGS DISAPPEARING?(13 votes)
- The derivative of 'dx' on its own is zero, because you're deriving 1 dx, and the derivative of a constant is zero. You can however derive f'(x)dx, and you get the second derivative of your function f(x), or f"(x)dx. 'dx' means "a little change in the x-direction", so when integrating, you're multiplying the function by these minute changes. Remember Riemann sums? If you write integrals in terms of the sums, it looks like this: sigma f(x) delta(x); in other words, you're summing the area of all of the little rectangles created by your function taken over small changes in x, or delta(x). If these changes are truly minuscule, then delta(x) is transformed into 'dx', and the sigma, or sum, into the integral. I hope this helps!(3 votes)
- Im not seeing how U-Sub is basically anti-chain rule. Could someone attempt to make this more clear? Thank you.(6 votes)
- Watch the first video on u-substitution (https://www.khanacademy.org/math/calculus/integral-calculus/u_substitution/v/u-substitution). The integral he has there is
∫ (3x² + 2x) e^(x³ + x²) dx.
This integral evaluates to e^(x³ + x²) + C.
You will notice that if you take the derivative of e^(x³ + x²) using the chain rule you get
d/dx e^(x³ + x²) = (x³ + x²)′ e^(x³ + x²) = (3x² + 2x) e^(x³ + x²),
which is in fact the expression whose antiderivative you were trying to find in the first place.
Therefore, u-substitution can in such cases be used to unravel the chain rule.(2 votes)
- If I don't use U-substitution will I get the same answer?(3 votes)
- U-substitution is never actually required. I often don't use it. However, not using it requires keeping track of a lot of different things. U-sub greatly simplifies keeping track of things so you don't overlook something important.
However, if you are able to keep track of everything and do it all without the u-sub, then go for it.(6 votes)
- Can you use indefinite integrals to find out definite integrals?
Sort of like how you can use derivatives to figure out the tangent line of a point?(3 votes)
- Yes look at the fundamental theorem of calculus. Definite and indefinite integrals are essentially the same thing only one of them you are applying the first part of the fundamental theorem of calculus.(4 votes)
- I've always learned to use U-subs when something and its derivative are in my integral. Here Sal just manipulates the function to add in that derivative. Is there any trick to knowing when to use a U-sub like this? Or can they be used almost anywhere?(3 votes)
- Well, when the "u" is a first degree polynomial ax+b you can just multiply and divide by the coefficient a so that you have the function and its derivative. As my calc teacher says, "If you want something, give it to yourself".
However, this is not possible with higher order polynomials or other functions. For example, ∫ e^sinx dx. I can't just multiply and divide by cosx because it won't help me to solve the integral. And if I have ∫ 1/(x^2+5) dx, I can't just multiply and divide by 2x because, again, it does not help me solve the integral.(5 votes)
- Okay I saw the comments but I am still confused on why he did not use the power rule. (7x+9)^1/2, add 1 to 1/2 and then put the new number on the bottom. So in the end it would be 2(7x-9)^3/2 over 3.(3 votes)
- He did, but you also have to apply the chain rule. The derivative of 7x+9 is 7, and you have to make that 7 vanish too. So, divide your solution by 7.(3 votes)
- why did he bring out 1/7 , i think he should bring 7 out? correct me if i am wrong(3 votes)
- Because the derivative of 7x+9 which is u is 7 so du= 7*dx then he will replace 7*dx in the original equation by du if he had taken 7 out he wouldnt have been able to replace du in the equation thus wont be able to use the u substitution(2 votes)
Let's take the indefinite integral of the square root of 7x plus 9 dx. So my first question to you is, is this going to be a good case for u-substitution? Well, when you look here, maybe the natural thing to set to be equal to u is 7x plus 9. But do I see its derivative anywhere over here? Well, let's see. If we set u to be equal to 7x plus 9, what is the derivative of u with respect to x going to be? Derivative of u with respect to x is just going to be equal to 7. Derivative of 7x is 7. Derivative of 9 is 0. So do we see a 7 lying around anywhere over here? Well, we don't. But what could we do in order to have a 7 lying around, but not change the value of the integral? Well, the neat thing-- and we've seen this multiple times-- is when you're evaluating integrals, scalars can go in and outside of the integral very easily. Just to remind ourselves, if I have the integral of let's say some scalar a times f of x dx, this is the same thing as a times the integral of f of x dx. The integral of the scalar times a function is equal to the scalar times the integral of the functions. So let me put this aside right over here. So with that in mind, can we multiply and divide by something that will have a 7 showing up? Well, we can multiply and divide by 7. So imagine doing this. Let's rewrite our original integral. So let me draw a little arrow here just to go around that aside. We could rewrite our original integral as being 9 to the integral of times 1/7 times 7 times the square root of 7x plus 9 dx. And if we want to, we could take the 1/7 outside of the integral. We don't have to, but we can rewrite this as 1/7 times the integral of 7, times the square root of 7x plus 9 dx. So now if we set u equal to 7x plus 9, do we have its derivative laying around? Well, sure. The 7 is right over here. We know that du-- if we want to write it in differential form-- du is equal to 7 times dx. So du is equal to 7 times dx. That part right over there is equal to du. And if we want to care about u, well, that's just going to be the 7x plus 9. That is are u. So let's rewrite this indefinite integral in terms of u. It's going to be equal to 1/7 times the integral of-- and I'll just take the 7 and put it in the back. So we could just write the square root of u du, 7 times dx is du. And we can rewrite this if we want as u to the 1/2 power. It makes it a little bit easier for us to kind of do the reverse power rule here. So we can rewrite this as equal to 1/7 times the integral of u to the 1/2 power du. And let me just make it clear. This u I could have written in white if I want it the same color. And this du is the same du right over here. So what is the antiderivative of u to the 1/2 power? Well, we increment u's power by 1. So this is going to be equal to-- let me not forget this 1/7 out front. So it's going to be 1/7 times-- if we increment the power here, it's going to be u to the 3/2, 1/2 plus 1 is 1 and 1/2 or 3/2. So it's going to be u to the 3/2. And then we're going to multiply this new thing times the reciprocal of 3/2, which is 2/3. And I encourage you to verify the derivative of 2/3 u to the 3/2 is indeed u to the 1/2. And so we have that. And since we're multiplying 1/7 times this entire indefinite integral, we could also throw in a plus c right over here. There might have been a constant. And if we want, we can distribute the 1/7. So it would get 1/7 times 2/3 is 2/21 u to the 3/2. And 1/7 times some constant, well, that's just going to be some constant. And so I could write a constant like that. I could call that c1 and then I could call this c2, but it's really just some arbitrary constant. And we're done. Oh, actually, no we aren't done. We still just have our entire thing in terms of u. So now let's unsubstitute it. So this is going to be equal to 2/21 times u to the 3/2. And we already know what u is equal to. u is equal to 7x plus 9. Let me put a new color here just to ease the monotony. So it's going to be 2/21 times 7x plus 9 to the 3/2 power plus c. And we are done. We were able to take a kind of hairy looking integral and realize that even though it wasn't completely obvious at first, that u-substitution is applicable.