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u-substitution with exponential function

You can never get too much u-substitution practice. Created by Sal Khan.
Video transcript
Let's see if we can calculate the definated roll from zero to one of x squared, times two, times two to the x to the third power dx. And, like always I encourage you to pause this video and see if you can figure this out, figure this out on your own. So I'm assuming you've had a go at it and so there's a couple of interesting things here. The first thing, at least in my brain does, it says, well I'm used to taking derivatives and antiderivatives of e to the x, not some other base to the x. So we know that the derivative with respect to x of e to the x. Is e to the x or we can say the ante derivirate of e to the x is equal to e to the x plus c. So, since I'm doing to the, something raise to this particular situation from the raise to the function of x, it seems like I might wanna put some, I might wanna change the base here. But how do I do that? Well the way I would do that is re-express 2 in terms of e, so what would be 2 in terms of e. Well 2 is equal to e is equal to e raised to the power that you had to raise e to to get to 2. Well what's the power that you have to raise e to to get to 2? Well that's the natural log of 2. Once again, the natural log of 2 is the exponent that you have to raise e to, to get to 2. So if you actually raise e to it, you're going to get 2. So this is what 2 is, now what is 2 to the x to the 3rd? Well, if we raise both sides of this to the x to the 3rd power. If we raise both side of the x to the third power, two to the x of the third is equal to, if I raise something to an exponent and then raise that to an exponent it's going to be equal to e to the x to the third times the natural log of two, times the natural log of two. So that already seems pretty interesting. So let's, let's rewrite this. And actually, what I'm gonna do, let's just focus on the indefinite integral first. See if we can figure that out, and then we can apply. And then we can take, we can evaluate the definite ones. So let's just, let's just think about this. Let's think about the indefinite integral of x squared. Times 2 to the x to the 3rd power, dx. So I really wanna find the anti-derivative of this. Well, this is going to be the exact same thing as the integral of. So I'll write my x squared, still. But instead of 2 to the x to the 3rd, I'm gonna write all of this business. Let me just copy and paste that. We already established, this is the same thing, as two to the ex to the third power. Copy, and paste, just like that. And then, let me just, close it with a dee ex. So, I was able to get it, in terms of e as a base. That makes me a little bit more comfortable, but it still seems pretty complicated. But, you might be saying, well okay look, maybe use substitution could be at play here because I had this, this kind of crazy expression, x to the third times the natural log of 2. But what's the derivative of that? Well that's going to be 3x squared times the natural log of 2. Or 3 times the natural log of two times x squared. Well that's just a constant times x squared. We already have an x squared here. And so maybe we can engineer this a little bit to have the constant there as well. So let's think about that. So if we made this, if we defined this as u, so if we said u is equal to x to the third times the natural log of two, what is d,u going to be? Well do you is going to be, is going to be, well natural log of two is just a constant. So it's going to be three x squared, times the natural log of two. And we can actually just change the order or multiplying a little bit. We could say that this is the same thing as x squared. Time three natural log of two. Which is the same thing, just using logarithm properties, as x squared times the natural log of 2 to the third power. So this is equal to x squared time the natural log of eight. So let's see. If this is u, where's du? And of course, we can't we can't forget the dx. This is a dx right over here. dx, dx, dx, so where is the du? Well we have a dx. Let me circle things. So you have a dx here, you have a dx there. You have an x squared here, you have an x squared here. So really, all we need is. All we need here is the natural log of 8. So if we, ideally we would have a natural log if 8 right over here. And we could put it there as long as we also, we can multiply by a natural log of 8 as long as we also divide by a natural log of 8. And so, we could, we could do it by, right over here. We could write, we could divide by a natural log of 8. But we know that the anti-derivative of some constant times a function is the same thing as the constant times the anti-derivative of that function. So we could just take that on the outside. So it's one over the natural log of eight. So let's writes this in terms of UNDU. This simplifies to one over. The natural log of eight times the anti-derivative of e, e to the u, e to the u, that's the u, du. This times this times that is du, du. And this is straightforward, we know what this is going to be. This is going to be equal to, so let me just write the 1 over natural log of 8 out here, 1 over natural log of 8, times, times e to the u times e to the u. e to the u of course, if we're thinking int terms of anti-derivative there'd be some constant out there and then we would just reverse the substitution. We already know what us is so this is going to be equal to the anti-derivative of this expression is 1 over the natural log of 8 times e to the, instead of u, we know that u is. x to the third times a natural log of 2 and of course we can put a plus C there. Now going back to the original problem, we just need to evaluate the anti derivative of this at each of these points. So lets rewrite this. So given what we just figured out, so let me copy and paste that. This is just going to be equal to, it's going to be equal to the anti-derivative evaluated at one, minus the anti-derivative evaluated at zero. We don't have to worry about the constants, because those will cancel out. So we are going to get, we are going to get one. Let me evaluate it first at one. So you're gonna get one over the natural log of eight, times E to the one to the third power. Which is just one times the natural log of two. Natural log of two, that's it at evaluated at one. And then we're gonna minus it evaluated at zero. So it's going to be one over the natural log of 8 times e to the well, when x is 0 this whole thing is going to be 0. Well e to the 0 is just 1 and e to the natural log of 2, well that's just going to be 2. We already established that early on, that this is just going to be equal to 2.