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
Article V of the Constitution | US government and civics | Khan Academy
Hey, this is Kim from Khan Academy, and today I’m learning about Article 5 of the U.S. Constitution, which describes the Constitution’s amendment process. To learn more about Article 5, I talked to two experts: Professor Michael Rappaport, who is the Darl…
Doing donuts in $150k+ cars…on the front lawn
[Music] Let me show my hair first. What’s up you guys? Brendan. So, I’m so excited this morning! I am on my way to Frank Out, he’s OC, a private car. If it’s working, backyard. He has an insanely cool house in the middle of Los Angeles, and the inside ya…
Refugees Welcomed in New York | Explorer
[music playing] HOST: Of approximately 61,000 residents in Utica, New York, nearly 11,000 are immigrants and refugees. And 450 or more arrive here each year. Utica was a manufacturing town in the 1970s and 1980s. Some of our factories began to leave, and…
Vincent Kartheiser: Playing William Bradford | Saints & Strangers
[Music] William Bradford was a man who was born in England, and at a very young age, was exposed to church and religion. There were some people on the outskirts of their religion that were beginning to be arrested for their beliefs. In his early adulthoo…
Rounding to the nearest tenth and hundredth
Joey used 0.432 lbs of cheese to make mac and cheese for dinner. We could also call this 432,000 pounds of cheese to make mac and cheese for dinner. Round the amount of cheese to the nearest tenth. So, the amount of cheese, once again, is 0.432 lbs. Just…
Revealing My Entire $6 Million Investment Portfolio | 29 Years Old
What’s up you guys, it’s Graham here. So, a few months ago, I made a video breaking down in graphic detail each of my seven income sources: how I built them up, what’s involved in running them, and then most importantly, the question everyone wants to kno…