yego.me
💡 Stop wasting time. Read Youtube instead of watch. Download Chrome Extension

_-substitution: definite integrals | AP Calculus AB | Khan Academy


4m read
·Nov 11, 2024

What we're going to do in this video is get some practice applying u-substitution to definite integrals. So let's say we have the integral. So we're going to go from x equals 1 to x equals 2, and the integral is (2x \times (x^2 + 1)^3 , dx).

So, I already told you that we're going to apply u-substitution, but it's interesting to be able to recognize when to use it. The key giveaway here is, well, I have this (x^2 + 1) business of 3rd power, but then I also have the derivative of (x^2 + 1) which is (2x) right over here.

So I could do the substitution. I could say (u) is equal to (x^2 + 1), in which case the derivative of (u) with respect to (x) is just (2x + 0) or just (2x). I could write that in differential form. A mathematical hand-wavy way of thinking about it is multiplying both sides by (dx), and so you get (du = 2x , dx).

Now, at least this part of the integral I can rewrite. So let me at least write this. This is going to be, I'll write the integral—we're going to think about the bounds in a second—so we have (u^3).

You do the same orange color: (u^3), that's this stuff right over here. And then (2x , dx), remember you could just view this as (2x \times (x^2 + 1)^3 \times dx). So (2x , dx), well (2x , dx) that is (du).

So that and that together, that is (du).

Now, an interesting question because this isn't an indefinite integral—we're not just trying to find the antiderivative, this is a definite integral—so what happens to our bounds of integration? Well, there's two ways that you can approach this.

You can change your bounds of integration because this one is (x = 1) to (x = 2), but now we're integrating with respect to (u). So one way, if you want to keep your bounds of integration—or if you want to keep this a definite integral, I guess you could say—you would change your bounds from (u =) something to (u =) something else.

So your bounds of integration, let's see, when (x = 1), what is (u)? Well, when (x = 1), you have (1^2 + 1), so you have (2): (u = 2) in that situation. When (x = 2), what is (u)? Well, you have (2^2) which is (4) plus (1) which is (5), so (u = 5).

And you won't typically see someone writing (u =) or (u = 5); it's often just from (2) to (5) because we're integrating with respect to (u). You assume it's (u = 2) to (u = 5).

So we could just rewrite this as being equal to the integral from (2) to (5) of (u^3 , du). But it's really important to realize why we changed our bounds. We are now integrating with respect to (u) and the way we did it is we used our substitution right over here.

What (x = 1), (u = 2); when (x = 2): (2^2 + 1) (u = 5). Then we can just evaluate this right over here. Let’s see: this is going to be equal to the antiderivative of (u^3) which is (\frac{u^4}{4}), and we're going to evaluate that at (5) and (2).

So this is going to be (\frac{5^4}{4} - \frac{2^4}{4}), and then we could simplify this if we like, but we've just evaluated this definite integral.

Now, another way to do it is to think about the— is to try to solve the indefinite integral in terms of (x) and use (u) substitution as an intermediate. So one way to think about this is to say, well, let's just try to evaluate what the indefinite integral of (2x \times (x^2 + 1)^3 , dx) is.

And then whatever this expression ends up being algebraically, I'm going to evaluate it at (x = 2) and at (x = 1). So then you use the (u) substitution right over here, and you would get this would simplify using the same substitution as the integral of (u^3 , du).

Once again, you're going to evaluate this whole thing from (x = 2) and then subtract from that it evaluated at (x = 1).

And so this is going to be equal to, well let's see, this is (\frac{u^4}{4}) and once again evaluating it at (x = 2) and then subtracting from that at (x = 1).

And then now we can just back substitute. We could say, "Hey look, (u = x^2 + 1)". So this is the same thing as (\frac{(x^2 + 1)^4}{4}) and we're now going to evaluate that at (x = 2) and (x = 1).

And you will notice that you'll get the exact same thing. When you put (x = 2) here, you get (2^2 + 1) which is (5) to the (4)th power over (4), that right over there. And then minus (1^2 + 1) is (2) to the (4)th power over (4), that right over there.

So, either way, you'll get the same result. You can either keep it a definite integral and then change your bounds of integration and express them in terms of (u). That's one way to do it. The other way is to try to evaluate the indefinite integral, use (u) substitution as an intermediary step, then back substitute back, and then evaluate at your bounds.

More Articles

View All
Musical Fire Table!
Just press play, you mean? [Voiceover] Yeah, go for it. Whoa! [Music] Now, you may have seen a Ruben’s tube before. That’s basically a pipe with a bunch of holes in it, and you pump in a flammable gas and light it on fire, so you basically create a row …
The EPA Talks Climate Change | StarTalk
So, climate change, is that real? Presumably, the EPA is ready to do something about it. I went straight to Gina McCarthy, the administrator of the EPA, to find out what are they doing about climate change. Let’s check it out. I’m moving forward to devel…
How to Improve Your Life in 24 HOURS
[Music] If you’ve ever browsed self-improvement forums like on Reddit, then you’ll often come across some pretty good advice, some pretty questionable stuff—no doubt. But every once in a while, there’s a little nugget of wisdom that sticks with you, and …
2015 AP Physics 1 free response 3b
The spring is now compressed twice as much to Δx = 2D. A student is asked to predict whether the final position of the block will be twice as far at x = 6D. The student reasons that since the spring will be compressed twice as much as before, the block wi…
I Made A Solenoid Engine!
I built a solenoid engine. Unlike most motors out there that hide how they work, this beauty bears all. A solenoid is a kind of electromagnet. When electricity flows through this coil, a magnetic field pulls the magnet-topped piston inside up. But when th…
Grand Canyon Adventure: The 750-Mile Hike That Nearly Killed Us (Part 1) | Nat Geo Live
What we’re gonna do tonight, Kevin and I are gonna take you on an unusual and somewhat remarkable journey through a remarkable place, the Grand Canyon. But before we do that, we felt it’s important to get a little bit of an idea of how we know each other,…