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This right over here, x to the twelfth is the same thing as x to the third to the fourth power. This right over here is the same thing as x to the third to the sixth power. So if we could replace each of these x's with x to the thirds, we would get this power series up here. Well, how do we do that? Well, we would just... | Worked example cosine function from power series Series AP Calculus BC Khan Academy.mp3 |
Well, how do we do that? Well, we would just say, well, it's the cosine of x to third. And actually, let me do that in a different color. So the cosine, and that's not a different color, so the cosine of x to the third is going to be equal to, and once again, everywhere we see an x, we replace it with x to the third. S... | Worked example cosine function from power series Series AP Calculus BC Khan Academy.mp3 |
So the cosine, and that's not a different color, so the cosine of x to the third is going to be equal to, and once again, everywhere we see an x, we replace it with x to the third. So it's one minus, and actually, I'm just going to put in parentheses squared, two factorial, which I wanted to do that in the green. Let m... | Worked example cosine function from power series Series AP Calculus BC Khan Academy.mp3 |
All right, so it's going to be equal to one minus parentheses squared over two factorial plus parentheses to the fourth power over four factorial minus parentheses to the sixth power over six factorial. And now let me get changed back to that mauve color. And since I'm taking the cosine of x to the third, well, this is... | Worked example cosine function from power series Series AP Calculus BC Khan Academy.mp3 |
This is going to be x to the third to the fourth power. This is going to be x to the third to the sixth power, which is exactly what I have right over here. So this right over here is the power series for cosine of x to the third. So evaluating this when x is equal to the cube root of pi over two is the same thing as e... | Worked example cosine function from power series Series AP Calculus BC Khan Academy.mp3 |
So evaluating this when x is equal to the cube root of pi over two is the same thing as evaluating this when x is equal to the cube root of pi over two. Let me write that down, because this is interesting. So this, so I'll just rewrite it, from n equals zero to infinity of negative one to the n, x to the sixth n over t... | Worked example cosine function from power series Series AP Calculus BC Khan Academy.mp3 |
And if you've been following some of the videos on differentiability implies continuity and what happens to a continuous function as our change in x, if x is our independent variable, as that approaches zero, how the change in our function approaches zero, then this proof is actually surprisingly straightforward. So le... | Chain rule proof Derivative rules AP Calculus AB Khan Academy.mp3 |
So the chain rule tells us that if y, y is a function of u, which is a function of x, and we want to figure out the derivative of this, so we want to differentiate this with respect to x. So we're gonna differentiate this with respect to x. We could write this as the derivative of y with respect to x, which is going to... | Chain rule proof Derivative rules AP Calculus AB Khan Academy.mp3 |
This is what the chain rule tells us. But how do we actually go about proving it? Well, we just have to remind ourselves that the derivative of y with respect to x, the derivative of y with respect to x, is equal to the limit as delta x approaches zero of change in y over change in x. Now we can do a little bit of alge... | Chain rule proof Derivative rules AP Calculus AB Khan Academy.mp3 |
Now we can do a little bit of algebraic manipulation here to introduce a change in u. So let's do that. So this is going to be the same thing as the limit as delta x approaches zero. And I'm gonna rewrite this part right over here. I'm gonna essentially divide and multiply by a change in u. So I can rewrite this as del... | Chain rule proof Derivative rules AP Calculus AB Khan Academy.mp3 |
And I'm gonna rewrite this part right over here. I'm gonna essentially divide and multiply by a change in u. So I can rewrite this as delta y over delta u times delta u, whoops, times delta u over delta x. Change in y over change in u times change in u over change in x. And you can see, these are just going to be numbe... | Chain rule proof Derivative rules AP Calculus AB Khan Academy.mp3 |
Change in y over change in u times change in u over change in x. And you can see, these are just going to be numbers here. So our change in u, this would cancel with that, and you'd be left with change in y over change in x, which is exactly what we had here. So nothing earth-shattering just yet. But what's this going ... | Chain rule proof Derivative rules AP Calculus AB Khan Academy.mp3 |
So nothing earth-shattering just yet. But what's this going to be equal to? What's this going to be equal to? Well, the limit of the product is the same thing as the product of the limits. So this is going to be the same thing as the limit as delta x approaches zero of, and I'll color-code it, of this stuff, of delta y... | Chain rule proof Derivative rules AP Calculus AB Khan Academy.mp3 |
Well, the limit of the product is the same thing as the product of the limits. So this is going to be the same thing as the limit as delta x approaches zero of, and I'll color-code it, of this stuff, of delta y over delta u times, maybe I'll put parentheses around it, times, times the limit, the limit as delta x approa... | Chain rule proof Derivative rules AP Calculus AB Khan Academy.mp3 |
Delta u over delta x. So what does this simplify to? Well, this right over here, this is the definition, and we're assuming, in order for this to even be true, we have to assume that u and y are differentiable at x. So we assume, in order for this to be true, we're assuming, we're assuming y, comma, u are differentiabl... | Chain rule proof Derivative rules AP Calculus AB Khan Academy.mp3 |
So we assume, in order for this to be true, we're assuming, we're assuming y, comma, u are differentiable, are differentiable, are differentiable at x. And remember, also, if they're differentiable at x, that means they're continuous at x. But if u is differentiable at x, then this limit exists, and this is the derivat... | Chain rule proof Derivative rules AP Calculus AB Khan Academy.mp3 |
So this right over here, we can rewrite as du dx. I think you see where this is going. Now, this right over here, just looking at it the way it's written right here, we can't quite yet call this dy du, because this is the limit as delta x approaches zero, not the limit as delta u approaches zero. But we just have to re... | Chain rule proof Derivative rules AP Calculus AB Khan Academy.mp3 |
But we just have to remind ourselves the results from probably the previous video, depending on how you're watching it, which is, if we have a function u that is continuous at a point that as delta x approaches zero, delta u approaches zero. So we can actually rewrite this. We can rewrite this right over here. Instead ... | Chain rule proof Derivative rules AP Calculus AB Khan Academy.mp3 |
Let's do a few more examples of finding the limit of functions as x approaches infinity or negative infinity. So here I have this crazy function, 9x to the 7th minus 17x to the 6th plus 15 square roots of x, all of that over 3x to the 7th plus 1,000x to the 5th minus log base 2 of x. So what's going to happen as x appr... | Limits at infinity of quotients (Part 2) Limits and continuity AP Calculus AB Khan Academy.mp3 |
And the key here, like we've seen in other examples, is just to realize which terms will dominate. So for example, in the numerator, out of these three terms, the 9x to the 7th is going to grow much faster than any of these other terms. So this is the dominating term in the numerator. And the denominator, 3x to the 7th... | Limits at infinity of quotients (Part 2) Limits and continuity AP Calculus AB Khan Academy.mp3 |
And the denominator, 3x to the 7th, is going to grow much faster than an x to the 5th term, definitely much faster than a log base 2 term. So at infinity, as we get closer and closer to infinity, this function is going to be roughly equal to 9x to the 7th over 3x to the 7th. And so we can say, especially as we get larg... | Limits at infinity of quotients (Part 2) Limits and continuity AP Calculus AB Khan Academy.mp3 |
We can say this limit is going to be the same thing as this limit, which is going to be equal to the limit as x approaches infinity. Well, we can just cancel out the x to the 7th, so it's going to be 9 thirds or just 3, which is just going to be 3. So that is our limit as x approaches infinity of all of this craziness.... | Limits at infinity of quotients (Part 2) Limits and continuity AP Calculus AB Khan Academy.mp3 |
Now let's do the same with this function over here. Once again, crazy function. We're going to negative infinity, but the same principles apply. Which terms dominate as the absolute value of x get larger and larger and larger, as x gets larger in magnitude? Well, in the numerator, it's the 3x to the 3rd term. In the de... | Limits at infinity of quotients (Part 2) Limits and continuity AP Calculus AB Khan Academy.mp3 |
Which terms dominate as the absolute value of x get larger and larger and larger, as x gets larger in magnitude? Well, in the numerator, it's the 3x to the 3rd term. In the denominator, it's the 6x to the 4th term. So this is going to be the same thing as the limit of 3x to the 3rd over 6x to the 4th as x approaches ne... | Limits at infinity of quotients (Part 2) Limits and continuity AP Calculus AB Khan Academy.mp3 |
So this is going to be the same thing as the limit of 3x to the 3rd over 6x to the 4th as x approaches negative infinity. And if we simplify this, this is going to be equal to the limit as x approaches negative infinity of 1 over 2x. And what's this going to be? Well, if the denominator, even though it's becoming a lar... | Limits at infinity of quotients (Part 2) Limits and continuity AP Calculus AB Khan Academy.mp3 |
Well, if the denominator, even though it's becoming a larger and larger and larger negative number, it becomes 1 over a very, very large negative number, which is going to get us pretty darn close to 0, just as 1 over x as x approaches negative infinity gets us close to 0. So this right over here, the horizontal asympt... | Limits at infinity of quotients (Part 2) Limits and continuity AP Calculus AB Khan Academy.mp3 |
The key realization here is to simplify the problem by just thinking about which terms are going to dominate the rest. Now let's think about this one. What is the limit of this crazy function as x approaches infinity? Well, once again, what are the dominating terms? In the numerator, it's 4x to the 4th. In the denomina... | Limits at infinity of quotients (Part 2) Limits and continuity AP Calculus AB Khan Academy.mp3 |
Well, once again, what are the dominating terms? In the numerator, it's 4x to the 4th. In the denominator, it's 250x to the 3rd. These are the highest degree terms. So this is going to be the same thing as the limit as x approaches infinity of 4x to the 4th over 250x to the 3rd, which is going to be the same thing as t... | Limits at infinity of quotients (Part 2) Limits and continuity AP Calculus AB Khan Academy.mp3 |
These are the highest degree terms. So this is going to be the same thing as the limit as x approaches infinity of 4x to the 4th over 250x to the 3rd, which is going to be the same thing as the limit of 4. Well, this is going to be the same thing as we could divide 200 and, well, I'll just leave it like this. It's goin... | Limits at infinity of quotients (Part 2) Limits and continuity AP Calculus AB Khan Academy.mp3 |
It's going to be the limit of 4 over 250x to the 4th divided by x to the 3rd is just x times x as x approaches infinity. Or we could even say this is going to be 4 250ths times the limit as x approaches infinity of x. Now what's this? What's the limit of x as x approaches infinity? Well, it's just going to keep growing... | Limits at infinity of quotients (Part 2) Limits and continuity AP Calculus AB Khan Academy.mp3 |
What's the limit of x as x approaches infinity? Well, it's just going to keep growing forever. So this right over here is just going to be infinity. Infinity times some number right over here is going to be infinity. So the limit as x approaches infinity of all of this, it's actually unbounded. It's infinity. And a kin... | Limits at infinity of quotients (Part 2) Limits and continuity AP Calculus AB Khan Academy.mp3 |
Infinity times some number right over here is going to be infinity. So the limit as x approaches infinity of all of this, it's actually unbounded. It's infinity. And a kind of obvious way of seeing that right from the get-go is to realize that the numerator has a 4th degree term, while the highest degree term in the de... | Limits at infinity of quotients (Part 2) Limits and continuity AP Calculus AB Khan Academy.mp3 |
I drop a rock in the middle of that pool of water, and a little while later, a ripple has, a little wave, a ripple has formed that is moving radially outward from where I dropped the rock. So let me see how well I can draw that. So it's moving radially outward. So that is the ripple that is formed from me dropping the ... | Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3 |
So that is the ripple that is formed from me dropping the rock into the water. So it's a circle centered at where the rock initially hit the water. And let's say right at this moment, the radius of this circle is equal to 3 centimeters. And we also know that the radius is increasing at a rate of 1 centimeter per second... | Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3 |
And we also know that the radius is increasing at a rate of 1 centimeter per second. So radius growing at rate of 1 centimeter per second. So given this, right now our circle, our ripple circle, has a radius of 3 centimeters. And we know that the radius is growing at 1 centimeter per second. Given that, at what rate is... | Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3 |
And we know that the radius is growing at 1 centimeter per second. Given that, at what rate is the area of circle growing? Area of circle growing. Interesting. So let's think about what we know and then what we don't know, what we're trying to figure out. So if we call this radius r, we know that right now r is equal t... | Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3 |
Interesting. So let's think about what we know and then what we don't know, what we're trying to figure out. So if we call this radius r, we know that right now r is equal to 3 centimeters. We also know the rate at which r is changing with respect to time. We also know this information right over here. dr dt, the rate ... | Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3 |
We also know the rate at which r is changing with respect to time. We also know this information right over here. dr dt, the rate at which the radius is changing with respect to time, is 1 centimeter per second. Now what do we need to figure out? Well, they said, what rate is the area of the circle growing? So we need ... | Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3 |
Now what do we need to figure out? Well, they said, what rate is the area of the circle growing? So we need to figure out at what rate is the area of the circle, where a is the area of the circle, at what rate is this growing? This is what we need to figure out. So what might be useful here is if we can come up with a ... | Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3 |
This is what we need to figure out. So what might be useful here is if we can come up with a relationship between the area of the circle and the radius of the circle, and maybe take the derivative with respect to time. And we'll have to use a little bit of the chain rule to do that. So what is the relationship at any g... | Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3 |
So what is the relationship at any given point in time between the area of the circle and the radius of the circle? Well, this is elementary geometry. The area of a circle is going to be equal to pi times the radius of the circle squared. Now, what we want to do is figure out the rate at which the area is changing with... | Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3 |
Now, what we want to do is figure out the rate at which the area is changing with respect to time. So why don't we take the derivative of both sides of this with respect to time? And let me give myself a little more real estate. Actually, let me just rewrite what I just had. So pi r squared. Area is equal to pi r squar... | Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3 |
Actually, let me just rewrite what I just had. So pi r squared. Area is equal to pi r squared. Now I'm going to take the derivative of both sides with respect to time. So the derivative with respect to time. I'm not taking the derivative with respect to r. I'm taking the derivative with respect to time. So on the left-... | Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3 |
Now I'm going to take the derivative of both sides with respect to time. So the derivative with respect to time. I'm not taking the derivative with respect to r. I'm taking the derivative with respect to time. So on the left-hand side, right over here, I'm going to have the derivative of our area. Actually, let me just... | Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3 |
So on the left-hand side, right over here, I'm going to have the derivative of our area. Actually, let me just write it in that green color. I'm going to have the derivative of our area with respect to time on the left-hand side. And on the right-hand side, what do I have? Well, if I'm taking the derivative of a consta... | Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3 |
And on the right-hand side, what do I have? Well, if I'm taking the derivative of a constant times something, I can take the constant out. So let me just do that. Pi times the derivative with respect to time of r squared. And to make it a little bit clearer what I'm about to do, why I'm using the chain rule, we're assu... | Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3 |
Pi times the derivative with respect to time of r squared. And to make it a little bit clearer what I'm about to do, why I'm using the chain rule, we're assuming that r is a function of time. If r wasn't a function of time, then area wouldn't be a function of time. So instead of just writing r, let me make it explicit ... | Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3 |
So instead of just writing r, let me make it explicit that it is a function of time. I'll write r of t. So it's r of t, which we're squaring. We want to find the derivative of this with respect to time. And here, we just have to apply the chain rule. We're taking the derivative of something squared with respect to that... | Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3 |
And here, we just have to apply the chain rule. We're taking the derivative of something squared with respect to that something. So the derivative of that something squared with respect to the something is going to be 2 times that something to the first power. And then we're going to have, let me make it clear, this is... | Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3 |
And then we're going to have, let me make it clear, this is the derivative of r of t squared with respect to r of t, the derivative of something squared with respect to that something. If it was the derivative of x squared with respect to x, we'd have 2x. If it's the derivative of r of t squared with respect to r of t,... | Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3 |
This is just the derivative with respect to r of t. The derivative which this changes with respect to time, we have to multiply this times the rate at which r of t changes with respect to time. So the rate at which r of t changes with respect to time, well, we could just write that as dr dt. These are equivalent expres... | Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3 |
And of course, we have our pi out front. And I just want to emphasize, this is just the chain rule right over here. The derivative of something squared with respect to time is going to be the derivative of the something squared with respect to the something. So that's 2 times the something times the derivative of that ... | Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3 |
So that's 2 times the something times the derivative of that something with respect to time. I can't emphasize enough what we did right over here. This is the chain rule. So we're left with pi times this is equal to the derivative of our area with respect to time. Now, let me rewrite all of this again, just so it clean... | Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3 |
So we're left with pi times this is equal to the derivative of our area with respect to time. Now, let me rewrite all of this again, just so it cleans up a little bit. So we have the derivative of our area with respect to time is equal to pi times, actually, let me put that 2 out front, is equal to 2 times pi times, I ... | Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3 |
Actually, let me make the r in blue. 2 pi r dr dt. Now, what do we know? We know what r is. We know that r, at this moment right in time, is 3 centimeters. Right now, r is 3 centimeters. We know dr dt right now is 1 centimeter per second. | Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3 |
We know what r is. We know that r, at this moment right in time, is 3 centimeters. Right now, r is 3 centimeters. We know dr dt right now is 1 centimeter per second. We know this is 1 centimeter per second. So what's da dt going to be equal to? What's going to be equal to, do that same green, 2 pi times 3 times 1 times... | Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3 |
We know dr dt right now is 1 centimeter per second. We know this is 1 centimeter per second. So what's da dt going to be equal to? What's going to be equal to, do that same green, 2 pi times 3 times 1 times, that's purple, times 1 centimeter per second. And let's make sure we get the units right. So we have a centimete... | Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3 |
What's going to be equal to, do that same green, 2 pi times 3 times 1 times, that's purple, times 1 centimeter per second. And let's make sure we get the units right. So we have a centimeter times a centimeter. So it's going to be centimeters. That's too dark of a color. It's going to be square centimeters, centimeters... | Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3 |
So it's going to be centimeters. That's too dark of a color. It's going to be square centimeters, centimeters times centimeters, square centimeter per second, which is the exact units we need for a change in area. So we have da dt is equal to this. The rate at which area is changing with respect to time is equal to 6 p... | Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3 |
A particle moves in the xy plane so that at any time t is greater than or equal to zero. Its position vector is, and they give us the x component and the y component of our position vectors, and they're both functions of time. What is the particle's acceleration vector at time t equals three? Alright, so our position, ... | Planar motion example acceleration vector Advanced derivatives AP Calculus BC Khan Academy.mp3 |
Alright, so our position, let's denote that. It's a vector-valued function. It's gonna be a function of time. It is a vector. And they already told us that the x component of our position is negative three t to the third power plus four t squared, and the y component is t to the third power plus two. And so you give me... | Planar motion example acceleration vector Advanced derivatives AP Calculus BC Khan Academy.mp3 |
It is a vector. And they already told us that the x component of our position is negative three t to the third power plus four t squared, and the y component is t to the third power plus two. And so you give me any time greater than or equal to zero, I put it in here, and I can give you the corresponding x and y compon... | Planar motion example acceleration vector Advanced derivatives AP Calculus BC Khan Academy.mp3 |
And this is one form of notation for a vector. Another way of writing this, you might be familiar with engineering notation. It might be written like, or sometimes people write this as unit vector notation. Negative three to the third plus four t squared times the unit vector in the horizontal direction plus t to the t... | Planar motion example acceleration vector Advanced derivatives AP Calculus BC Khan Academy.mp3 |
Negative three to the third plus four t squared times the unit vector in the horizontal direction plus t to the third plus two times the unit vector in the vertical direction. This is just denoting the same thing. This is the x component, this is the y component. This is the component in the horizontal direction, this ... | Planar motion example acceleration vector Advanced derivatives AP Calculus BC Khan Academy.mp3 |
This is the component in the horizontal direction, this is the component in the vertical direction, or the y component. Now the key realization is if you have the position vector, well, the velocity vector's just going to be the derivative of that. So v of t is just going to be equal to r prime of t, which is going to ... | Planar motion example acceleration vector Advanced derivatives AP Calculus BC Khan Academy.mp3 |
So let's do that. So if we want to take the derivative of the x component here with respect to time, we're just gonna use the power rule a bunch. So it's three times negative three, so it's negative nine t squared, and then plus two times four is eight, so plus eight t to the first, so plus eight t. And then over here ... | Planar motion example acceleration vector Advanced derivatives AP Calculus BC Khan Academy.mp3 |
So actually I have space to write that, three t squared even bigger. Three t squared. All right, and if we want to find the acceleration function or the vector-valued function that gives us acceleration as a function of time, well, that's just going to be the derivative of the velocity function with respect to time. So... | Planar motion example acceleration vector Advanced derivatives AP Calculus BC Khan Academy.mp3 |
So this is going to be equal to, this is going to be equal to, let me give myself some space, and so the x component, well, I just take the derivative of the x component again, and let me find a color I haven't used yet. I'll use this green. So let's see, two times negative nine, negative 18 times t to the first power ... | Planar motion example acceleration vector Advanced derivatives AP Calculus BC Khan Academy.mp3 |
Two times three is six t to the first power, just six t. So this is, we've just been able to, by taking the derivative of this position vector-valued function twice, I'm able to find the acceleration function. And now I just have to evaluate it at t equals three. So our acceleration at t is equal to three is equal to, ... | Planar motion example acceleration vector Advanced derivatives AP Calculus BC Khan Academy.mp3 |
And so what does this simplify to? Well, this is going to be equal to, let's see, negative 18 times three is negative 54, negative 54 plus eight is negative 46, negative 46, and then six times three is 18. Did I do that arithmetic right? So this is negative 54 plus eight. So negative 54 plus four would be negative 50, ... | Planar motion example acceleration vector Advanced derivatives AP Calculus BC Khan Academy.mp3 |
And like always, I encourage you to pause this video and see if you can 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 that my brain does, it says, well, I'm used to taking derivatives and antiderivatives of e to the x. N... | _-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3 |
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 could say that the antiderivative of e to the x is equal to e to the x plus c. So since I'm dealing with something raised to a, this particular situation, something raised to a function of x, it seems like ... | _-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3 |
Well, the way I would do that is re-express two in terms of e. So what would be two in terms of e? Well, two is equal to e, is equal to e raised to the power that you need to raise e to to get to two. Well, what's the power that you have to raise e to to get to two? Well, that's the natural log of two. Once again, the ... | _-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3 |
Well, that's the natural log of two. Once again, the natural log of two is the exponent that you have to raise e to to get to two. So if you actually raise e to it, you're going to get two. So this is what two is. Now what is two to the x to the third? Well, if we raise both sides of this to the x to the third power, i... | _-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3 |
So this is what two is. Now what is two to the x to the third? Well, if we raise both sides of this to the x to the third power, if we raise both sides to the x to the third power, two to the x to 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 ... | _-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3 |
So that already seems pretty interesting. So let's rewrite this. And actually, what I'm going to do is 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 evaluate the definite ones. So let's just think about this. Let's think about the indefinite int... | _-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3 |
So let's just think about this. Let's think about the indefinite integral of x squared times two to the x to the third power dx. I really want to find the antiderivative 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 two to the x to the third... | _-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3 |
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 two to the x to the third, I'm going to write all of this business. Let me just copy and paste that. We already established this is the same thing as two to the x to the third power. Copy and paste it jus... | _-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3 |
Copy and paste it just like that. And then let me just close it with a dx. 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 u substitution could be at play here because I have this kind o... | _-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3 |
That makes me a little bit more comfortable, but it still seems pretty complicated. But you might be saying, well, okay, look, maybe u substitution could be at play here because I have this kind of crazy expression, x to the third times the natural log of two, but what's the derivative of that? Well, that's going to be... | _-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3 |
Well, that's just a constant times x squared. We already have a 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 du g... | _-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3 |
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 du going to be? Well, du is going to be, it's 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... | _-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3 |
We could say that this is the same thing as x squared times three natural log of two, which is the same thing, just using logarithm properties, as x squared times the natural log of two to the third power. Three natural log of two is the same thing as the natural log of two to the third power. So this is equal to x squ... | _-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3 |
So let's see, if this is u, where is du? Oh, and of course, 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. | _-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3 |
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. | _-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3 |
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 eight. So if we, ideally, we would have a natural log of eight right over here. | _-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3 |
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 eight. So if we, ideally, we would have a natural log of eight right over here. And we could put it there, as long as we also, we can multiply by a natural log of eight, as long as we also divide by... | _-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3 |
So if we, ideally, we would have a natural log of eight right over here. And we could put it there, as long as we also, we can multiply by a natural log of eight, as long as we also divide by a natural log of eight. And so we could do it right over here. We could divide by natural log of eight. But we know that the ant... | _-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3 |
We could divide by natural log of eight. But we know that the antiderivative of some constant times a function is the same thing as the constant times the antiderivative 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 write this in terms of u and du. This... | _-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3 |
So it's one over the natural log of eight. So let's write this in terms of u and du. This simplifies to one over the natural log of eight times the antiderivative 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 ... | _-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3 |
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 one over natural log of eight out here. One over natural log of eight times e to the u. Times e to the u, e to the u. And of course, if we're thinking in terms of j... | _-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3 |
This is going to be equal to, so let me just write the one over natural log of eight out here. One over natural log of eight times e to the u. Times e to the u, e to the u. And of course, if we're thinking in terms of just antiderivative, there would be some constant out there. And then we would just reverse the substi... | _-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3 |
And of course, if we're thinking in terms of just antiderivative, there would be some constant out there. And then we would just reverse the substitution. We already know what u is. So this is going to be equal to the antiderivative of this expression is one over the natural log of eight times e to the, instead of u, w... | _-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3 |
So this is going to be equal to the antiderivative of this expression is one over the natural log of eight times e to the, instead of u, we know that u is x to the third times the natural log of two. And of course, we could put a plus c there. Now, going back to the original problem. We just need to evaluate the antide... | _-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3 |
We just need to evaluate the antiderivative of this at each of these points. So let's 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 antiderivative evaluated at one minus the antiderivative evaluated at zero. We don't have... | _-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3 |
This is just going to be equal to, it's going to be equal to the antiderivative evaluated at one minus the antiderivative evaluated at zero. We don't have to worry about the constants because those will cancel out. And so we are going to get, we are going to get one, let me evaluate it first at one. So you're gonna get... | _-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3 |
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 evaluated at one. And then we're gonna have minus it evaluated at zero. So it's going to be one over the natural log of eight times e to the, well, whe... | _-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3 |
So it's going to be one over the natural log of eight times e to the, well, when x is zero, this whole thing is going to be zero. Well, e to the zero is just one. And e to the natural log of two, well, that's just going to be two. We already established that early on. This is just going to be equal to two. So we are le... | _-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3 |
And also, the diameter of the top of the cup is also 4 centimeters. And I'm pouring water into this cup right now. And I'm pouring the water at a rate of 1 cubic centimeter per second. And right at this moment, there is a height of 2 centimeters of water in the cup right now. So the height right now from the bottom of ... | Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3 |
And right at this moment, there is a height of 2 centimeters of water in the cup right now. So the height right now from the bottom of the cup to this point right over here is 2 centimeters. So my question to you is, at what rate? We know the rate at which the water is flowing into the cup. We're being given a volume p... | Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3 |
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