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Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | We have 20 times, let's see, let me subtract 15 dx dt from both sides of this equation. And we get 5 dx dts. 5 dx dts, I just subtracted this from both sides of the equation. This is 15 dx dts, this is 20. So we have 5 dx dts is equal to, this part right over here is negative 600. And this part right over here is negative 200. So it's equal to negative 800 feet per second. |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | This is 15 dx dts, this is 20. So we have 5 dx dts is equal to, this part right over here is negative 600. And this part right over here is negative 200. So it's equal to negative 800 feet per second. Or negative 800, and actually this will actually be in feet per second. And so dx dt is equal to, dividing both sides by 5, 5 times 16 is 80. So this is negative 160 feet per second. |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | So it's equal to negative 800 feet per second. Or negative 800, and actually this will actually be in feet per second. And so dx dt is equal to, dividing both sides by 5, 5 times 16 is 80. So this is negative 160 feet per second. And we're done. And we see the shadow is moving very, very, very, very fast to the left. x is decreasing. |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | So this is negative 160 feet per second. And we're done. And we see the shadow is moving very, very, very, very fast to the left. x is decreasing. And we see that. That's why we have this negative sign here. The value of x is decreasing. |
Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | In the last video, we attempted to approximate the area under a curve by constructing four rectangles of equal width and using the left boundary of each rectangle, the function evaluated at the left boundary, to determine the height. And we came up with an approximation. What I want to do in this video is generalize things a bit, using the exact same method, but doing it for an arbitrary function with arbitrary boundaries and an arbitrary number of rectangles. So let's do it. I'm going to draw the diagram as large as I can to make things clear, to make things as clear as possible. So that's my y-axis. And this right over here is my x-axis. |
Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | So let's do it. I'm going to draw the diagram as large as I can to make things clear, to make things as clear as possible. So that's my y-axis. And this right over here is my x-axis. Let me draw an arbitrary function. So let's say my function looks something like that. So that is y is equal to f of x. |
Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | And this right over here is my x-axis. Let me draw an arbitrary function. So let's say my function looks something like that. So that is y is equal to f of x. And let me define my boundaries. So let's say this right over here is x equals a. This is x equals a. |
Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | So that is y is equal to f of x. And let me define my boundaries. So let's say this right over here is x equals a. This is x equals a. And this right over here is x. This is x equals b. So this is b. |
Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | This is x equals a. And this right over here is x. This is x equals b. So this is b. And I'm going to use n rectangles. n rectangles. And I'm going to use the function evaluated at the left boundary of the rectangle to determine its height. |
Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | So this is b. And I'm going to use n rectangles. n rectangles. And I'm going to use the function evaluated at the left boundary of the rectangle to determine its height. So, for example, this will be rectangle 1. I'm going to evaluate what f of a is. I'm going to evaluate what f of a is. |
Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | And I'm going to use the function evaluated at the left boundary of the rectangle to determine its height. So, for example, this will be rectangle 1. I'm going to evaluate what f of a is. I'm going to evaluate what f of a is. So this right over here is f of a. And then I'm going to use that as the height of my first rectangle. The height of my first rectangle. |
Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | I'm going to evaluate what f of a is. So this right over here is f of a. And then I'm going to use that as the height of my first rectangle. The height of my first rectangle. So just like that. So rectangle number 1 looks like this. And I'll even number it. |
Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | The height of my first rectangle. So just like that. So rectangle number 1 looks like this. And I'll even number it. Rectangle 1 looks just like that. And just to have a convention here, because I'm going to want to label each of the boundaries of the left boundary. Each of the x values of the left boundary. |
Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | And I'll even number it. Rectangle 1 looks just like that. And just to have a convention here, because I'm going to want to label each of the boundaries of the left boundary. Each of the x values of the left boundary. So we'll say a is equal to x naught. a is equal to x naught. So we can also call this point right over here x naught. |
Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | Each of the x values of the left boundary. So we'll say a is equal to x naught. a is equal to x naught. So we can also call this point right over here x naught. That x value. And then we go to the next rectangle. And we can call this one right over here. |
Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | So we can also call this point right over here x naught. That x value. And then we go to the next rectangle. And we can call this one right over here. This x value, we'll call it x1. It's the left boundary of the next rectangle. If we evaluate f of x1, we get this value right over here. |
Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | And we can call this one right over here. This x value, we'll call it x1. It's the left boundary of the next rectangle. If we evaluate f of x1, we get this value right over here. This right over here is f of x1. So it tells us our height. And we want an equal width to the previous one. |
Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | If we evaluate f of x1, we get this value right over here. This right over here is f of x1. So it tells us our height. And we want an equal width to the previous one. And we'll think about what the width is going to be in a second. So this right over here is our second rectangle. This right over here is our second rectangle that we're going to use to approximate the area under the curve. |
Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | And we want an equal width to the previous one. And we'll think about what the width is going to be in a second. So this right over here is our second rectangle. This right over here is our second rectangle that we're going to use to approximate the area under the curve. That's rectangle number 2. Let's do rectangle number 3. Well, rectangle number 3, the left boundary, we're just going to call it x sub 2. |
Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | This right over here is our second rectangle that we're going to use to approximate the area under the curve. That's rectangle number 2. Let's do rectangle number 3. Well, rectangle number 3, the left boundary, we're just going to call it x sub 2. And its height is going to be f of x sub 2. And its width is going to be the same width as the other ones. I'm just eyeballing it right over here. |
Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | Well, rectangle number 3, the left boundary, we're just going to call it x sub 2. And its height is going to be f of x sub 2. And its width is going to be the same width as the other ones. I'm just eyeballing it right over here. So this is rectangle number 3. And we're going to continue this process all the way until we get to rectangle number n. So this is the nth rectangle. The nth rectangle right over here. |
Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | I'm just eyeballing it right over here. So this is rectangle number 3. And we're going to continue this process all the way until we get to rectangle number n. So this is the nth rectangle. The nth rectangle right over here. And what am I going to label this point right over here? Well, we already see a pattern. The left boundary of the first rectangle is x sub 0. |
Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | The nth rectangle right over here. And what am I going to label this point right over here? Well, we already see a pattern. The left boundary of the first rectangle is x sub 0. The left boundary of the second rectangle is x sub 1. The left boundary of the third rectangle is x sub 2. So the left boundary of the nth rectangle is going to be x sub n minus 1. |
Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | The left boundary of the first rectangle is x sub 0. The left boundary of the second rectangle is x sub 1. The left boundary of the third rectangle is x sub 2. So the left boundary of the nth rectangle is going to be x sub n minus 1. Whatever the rectangle number is, the left boundary is x sub that number minus 1. And this is just based on the convention that we've defined. Now the next thing that we need to do in order to actually calculate this area is think about what is the width. |
Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | So the left boundary of the nth rectangle is going to be x sub n minus 1. Whatever the rectangle number is, the left boundary is x sub that number minus 1. And this is just based on the convention that we've defined. Now the next thing that we need to do in order to actually calculate this area is think about what is the width. So let's call the width of any of these rectangles. And for these purposes, or the purpose of this example, I'm going to assume that it's constant. Although you can do these sums where you actually vary the width of the rectangle, but then it gets a little bit fancier. |
Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | Now the next thing that we need to do in order to actually calculate this area is think about what is the width. So let's call the width of any of these rectangles. And for these purposes, or the purpose of this example, I'm going to assume that it's constant. Although you can do these sums where you actually vary the width of the rectangle, but then it gets a little bit fancier. So I want an equal width. So I want delta x to be equal width. And to think about what that has to be, we just have to think, what's the total width that we're covering? |
Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | Although you can do these sums where you actually vary the width of the rectangle, but then it gets a little bit fancier. So I want an equal width. So I want delta x to be equal width. And to think about what that has to be, we just have to think, what's the total width that we're covering? Well, the total distance here is going to be b minus a. And we're just going to divide by the number of rectangles that we want, the number of sections that we want. So we want to divide by n. So if we assume this is true, and then we assume that a is equal to x0, and then x1 is equal to x0 plus delta x, x2 is equal to x1 plus delta x, and we go all the way to xn is equal to xn minus 1 plus delta x, then we've essentially set up this diagram right over here. |
Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | And to think about what that has to be, we just have to think, what's the total width that we're covering? Well, the total distance here is going to be b minus a. And we're just going to divide by the number of rectangles that we want, the number of sections that we want. So we want to divide by n. So if we assume this is true, and then we assume that a is equal to x0, and then x1 is equal to x0 plus delta x, x2 is equal to x1 plus delta x, and we go all the way to xn is equal to xn minus 1 plus delta x, then we've essentially set up this diagram right over here. b is actually going to be equal to xn. So this is xn. It's equal to xn minus 1 plus delta x. |
Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | So we want to divide by n. So if we assume this is true, and then we assume that a is equal to x0, and then x1 is equal to x0 plus delta x, x2 is equal to x1 plus delta x, and we go all the way to xn is equal to xn minus 1 plus delta x, then we've essentially set up this diagram right over here. b is actually going to be equal to xn. So this is xn. It's equal to xn minus 1 plus delta x. So now I think we've set up all of the notation and all of the conventions in order to actually calculate the area or our approximation of the area. So our approximation, approximate area, is going to be equal to what? Well, it's going to be the area of the first rectangle. |
Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | It's equal to xn minus 1 plus delta x. So now I think we've set up all of the notation and all of the conventions in order to actually calculate the area or our approximation of the area. So our approximation, approximate area, is going to be equal to what? Well, it's going to be the area of the first rectangle. So let me write this down. So it's going to be rectangle 1, so the area of rectangle 1, so rectangle 1 plus the area of rectangle 2, plus the area of rectangle 2, plus the area of rectangle 3. I think you get the point here. |
Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | Well, it's going to be the area of the first rectangle. So let me write this down. So it's going to be rectangle 1, so the area of rectangle 1, so rectangle 1 plus the area of rectangle 2, plus the area of rectangle 2, plus the area of rectangle 3. I think you get the point here. Area of rectangle 3 all the way, plus all the way to the area of rectangle n. And so what are these going to be? Rectangle 1 is going to be its height, which is f of x0 or f of a. Either way, x0 and a are the same thing. |
Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | I think you get the point here. Area of rectangle 3 all the way, plus all the way to the area of rectangle n. And so what are these going to be? Rectangle 1 is going to be its height, which is f of x0 or f of a. Either way, x0 and a are the same thing. So it's f of a times our delta x, times our width, our height times our width. So times delta, actually, let me write f of x0. I wanted to write f of x0 times delta x. |
Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | Either way, x0 and a are the same thing. So it's f of a times our delta x, times our width, our height times our width. So times delta, actually, let me write f of x0. I wanted to write f of x0 times delta x. What is our height of rectangle 2? It's f of x1 times delta x. f of x1 times delta x. What's our area of rectangle 3? |
Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | I wanted to write f of x0 times delta x. What is our height of rectangle 2? It's f of x1 times delta x. f of x1 times delta x. What's our area of rectangle 3? It's f of x2 times delta x. And then we go all the way to our area. We're taking all the sums all the way to rectangle n. What's its area? |
Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | What's our area of rectangle 3? It's f of x2 times delta x. And then we go all the way to our area. We're taking all the sums all the way to rectangle n. What's its area? It's f of x sub n minus 1. Actually, that's a different shade of orange. We'll use that same shade. |
Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | We're taking all the sums all the way to rectangle n. What's its area? It's f of x sub n minus 1. Actually, that's a different shade of orange. We'll use that same shade. It is f of x sub n minus 1 times delta x. And we're done. We've written it in a very general way. |
Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | We'll use that same shade. It is f of x sub n minus 1 times delta x. And we're done. We've written it in a very general way. But to really make us comfortable with the various forms of notation, especially the types of notation you might see when people are talking about approximating areas or sums in general, I'm going to use the traditional sigma notation. So another way we could write this as the sum. This is equal to the sum from, and remember, this is just based on the conventions that I set up. |
Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | We've written it in a very general way. But to really make us comfortable with the various forms of notation, especially the types of notation you might see when people are talking about approximating areas or sums in general, I'm going to use the traditional sigma notation. So another way we could write this as the sum. This is equal to the sum from, and remember, this is just based on the conventions that I set up. I'll let i count which rectangle we're in, from i equals 1 to n. And then we're going to look at each rectangle. So the first rectangle, that's rectangle 1. So it's going to be f of, well, if we're in the i-th rectangle, then we're going to take the left boundary is going to be x sub i minus 1 times delta x. |
Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | This is equal to the sum from, and remember, this is just based on the conventions that I set up. I'll let i count which rectangle we're in, from i equals 1 to n. And then we're going to look at each rectangle. So the first rectangle, that's rectangle 1. So it's going to be f of, well, if we're in the i-th rectangle, then we're going to take the left boundary is going to be x sub i minus 1 times delta x. And so right over here is a general way of considering, of thinking about approximating the area under a curve using rectangles where the height of the rectangles are defined by the left boundary. And this tells us it's the left boundary. And we see for each, if this is the i-th rectangle right over here, if this is rectangle i, then this right over here is x sub i minus 1, and this height right over here is f of x sub i minus 1. |
Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3 | So it's going to be f of, well, if we're in the i-th rectangle, then we're going to take the left boundary is going to be x sub i minus 1 times delta x. And so right over here is a general way of considering, of thinking about approximating the area under a curve using rectangles where the height of the rectangles are defined by the left boundary. And this tells us it's the left boundary. And we see for each, if this is the i-th rectangle right over here, if this is rectangle i, then this right over here is x sub i minus 1, and this height right over here is f of x sub i minus 1. So that's all we did right over there, times delta x. And then you sum all of these from the first rectangle all the way to the n-th. So hopefully that makes you a little bit more comfortable with this notation. |
Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3 | Let's say that f of x is equal to x squared plus x minus 6 over x minus 2. And we're curious about what the limit of f of x as x approaches 2 is equal to. Now, the first attempt that you might want to do right when you see something like this is to see what happens. What is f of 2? Now, this won't always be the limit even if it's defined, but it's a good place to start just to see if something reasonable could pop out. So looking at it this way, if we just evaluate f of 2, on our numerator we're going to get 2 squared plus 2 minus 6. So that's going to be 4 plus 2, which is 6, minus 6. |
Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3 | What is f of 2? Now, this won't always be the limit even if it's defined, but it's a good place to start just to see if something reasonable could pop out. So looking at it this way, if we just evaluate f of 2, on our numerator we're going to get 2 squared plus 2 minus 6. So that's going to be 4 plus 2, which is 6, minus 6. You're going to get 0 in the numerator, and you're going to get 0 in the denominator. So we don't have the function is not defined. So not defined at x is equal to 2. f not defined. |
Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3 | So that's going to be 4 plus 2, which is 6, minus 6. You're going to get 0 in the numerator, and you're going to get 0 in the denominator. So we don't have the function is not defined. So not defined at x is equal to 2. f not defined. So there's no simple thing there. Even if this did evaluate, if it was a continuous function, then actually the limit would be whatever the function is, but that doesn't necessarily mean the case. But we see very clearly that the function is not defined here. |
Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3 | So not defined at x is equal to 2. f not defined. So there's no simple thing there. Even if this did evaluate, if it was a continuous function, then actually the limit would be whatever the function is, but that doesn't necessarily mean the case. But we see very clearly that the function is not defined here. So let's see if we can simplify this, and we'll also try to graph it in some way. So one thing that might have jumped out at your head is you might want to factor this expression on top. So if we want to rewrite this, we could rewrite the top expression, and this just goes back to your algebra 1, two numbers whose product is negative 6, whose sum is positive 3. |
Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3 | But we see very clearly that the function is not defined here. So let's see if we can simplify this, and we'll also try to graph it in some way. So one thing that might have jumped out at your head is you might want to factor this expression on top. So if we want to rewrite this, we could rewrite the top expression, and this just goes back to your algebra 1, two numbers whose product is negative 6, whose sum is positive 3. Well, that could be positive 3 and negative 2. So this could be x plus 3 times x minus 2, all of that, over x minus 2. So as long as x does not equal 2, these two things will cancel out. |
Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3 | So if we want to rewrite this, we could rewrite the top expression, and this just goes back to your algebra 1, two numbers whose product is negative 6, whose sum is positive 3. Well, that could be positive 3 and negative 2. So this could be x plus 3 times x minus 2, all of that, over x minus 2. So as long as x does not equal 2, these two things will cancel out. So we could say this is equal to x plus 3 for all x's, except for x is equal to 2, as long as x does not equal 2. So that's another way of looking at it. Another way we could rewrite our f of x. I'm doing it in blue just to change the colors. |
Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3 | So as long as x does not equal 2, these two things will cancel out. So we could say this is equal to x plus 3 for all x's, except for x is equal to 2, as long as x does not equal 2. So that's another way of looking at it. Another way we could rewrite our f of x. I'm doing it in blue just to change the colors. We could rewrite f of x. This is the exact same function. f of x is equal to x plus 3 when x does not equal 2. |
Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3 | Another way we could rewrite our f of x. I'm doing it in blue just to change the colors. We could rewrite f of x. This is the exact same function. f of x is equal to x plus 3 when x does not equal 2. And we could even say it's undefined when x is equal to 2. So given this definition, it becomes much clearer to us of how we can actually graph f of x. So let's try to do it. |
Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3 | f of x is equal to x plus 3 when x does not equal 2. And we could even say it's undefined when x is equal to 2. So given this definition, it becomes much clearer to us of how we can actually graph f of x. So let's try to do it. So that is not anywhere near being a straight line. That is much better. So let's call this the y-axis. |
Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3 | So let's try to do it. So that is not anywhere near being a straight line. That is much better. So let's call this the y-axis. I'll call it y equals f of x. And then let's, over here, let me make a horizontal line. That is my x-axis. |
Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3 | So let's call this the y-axis. I'll call it y equals f of x. And then let's, over here, let me make a horizontal line. That is my x-axis. So defined this way, f of x is equal to x plus 3. So if this is 1, 2, 3, we have a y-intercept at 3, and then the slope is 1. The slope is 1. |
Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3 | That is my x-axis. So defined this way, f of x is equal to x plus 3. So if this is 1, 2, 3, we have a y-intercept at 3, and then the slope is 1. The slope is 1. But it's defined for all x's, except for x is equal to 2. So this is x is equal to 1, x is equal to 2. So when x is equal to 2, it is undefined. |
Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3 | The slope is 1. But it's defined for all x's, except for x is equal to 2. So this is x is equal to 1, x is equal to 2. So when x is equal to 2, it is undefined. So let me make sure I can. So it's undefined right over there. It's undefined right over there. |
Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3 | So when x is equal to 2, it is undefined. So let me make sure I can. So it's undefined right over there. It's undefined right over there. So this is what f of x looks like. Now, given this, let's try to answer our question. What is the limit of f of x as x approaches 2? |
Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3 | It's undefined right over there. So this is what f of x looks like. Now, given this, let's try to answer our question. What is the limit of f of x as x approaches 2? Well, we can look at this graphically. As x approaches 2 from lower values than 2, so this right over here is x is equal to 2. If we get to maybe, let's say this is 1.7, we see that our f of x is right over there. |
Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3 | What is the limit of f of x as x approaches 2? Well, we can look at this graphically. As x approaches 2 from lower values than 2, so this right over here is x is equal to 2. If we get to maybe, let's say this is 1.7, we see that our f of x is right over there. If we get to 1.9, our f of x is right over there. So it seems to be approaching this value right over there. Similarly, as we approach 2 from values greater than it, if we're at, I don't know, this could be like 2.5, 2.5, our f of x is right over there. |
Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3 | If we get to maybe, let's say this is 1.7, we see that our f of x is right over there. If we get to 1.9, our f of x is right over there. So it seems to be approaching this value right over there. Similarly, as we approach 2 from values greater than it, if we're at, I don't know, this could be like 2.5, 2.5, our f of x is right over there. If we get even closer to 2, our f of x is right over there. Once again, we look like we are approaching this value. Or another way of thinking about it, if we ride this line from the positive direction, we seem to be approaching this value for f of x. |
Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3 | Similarly, as we approach 2 from values greater than it, if we're at, I don't know, this could be like 2.5, 2.5, our f of x is right over there. If we get even closer to 2, our f of x is right over there. Once again, we look like we are approaching this value. Or another way of thinking about it, if we ride this line from the positive direction, we seem to be approaching this value for f of x. If we ride this line from the negative direction, from values less than 2, we seem to be approaching this value right over here. This is essentially the value of x plus 3 if we set x is equal to 2. So this is essentially going to be this value right over here is equal to 5. |
Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3 | Or another way of thinking about it, if we ride this line from the positive direction, we seem to be approaching this value for f of x. If we ride this line from the negative direction, from values less than 2, we seem to be approaching this value right over here. This is essentially the value of x plus 3 if we set x is equal to 2. So this is essentially going to be this value right over here is equal to 5. If we just look at it visually, if we just graphed a line with slope 1 with a y-intercept of 3, this value right over here is 5. Now, we can also try to do this numerically. So let's try to do that. |
Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3 | So this is essentially going to be this value right over here is equal to 5. If we just look at it visually, if we just graphed a line with slope 1 with a y-intercept of 3, this value right over here is 5. Now, we can also try to do this numerically. So let's try to do that. So if this is our function definition, completely identical to our original definition, then we just try values as x gets closer and closer to 2. So let's try values less than 2. So 1.9999, this is almost obvious, 1.9999 plus 3, well, that gets you pretty darn close to 5. |
Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3 | So let's try to do that. So if this is our function definition, completely identical to our original definition, then we just try values as x gets closer and closer to 2. So let's try values less than 2. So 1.9999, this is almost obvious, 1.9999 plus 3, well, that gets you pretty darn close to 5. If I put even more 9s here, got even closer to 2, we'd get even closer to 5 here. If we approach 2 from the positive direction, and then we once again, we're getting closer and closer to 5 from the positive direction. If we were even closer to 2, we'd be even closer to 5. |
Connecting limits and graphical behavior Limits and continuity AP Calculus AB Khan Academy.mp3 | Well, we've done this multiple times. Let's think about what g of x approaches as x approaches five from the left. G of x is approaching negative six. As x approaches five from the right, g of x looks like it's approaching negative six. So a reasonable estimate based on looking at this graph is that as x approaches five, g of x is approaching negative six. And it's worth noting that that's not what g of five is. G of five is a different value. |
Connecting limits and graphical behavior Limits and continuity AP Calculus AB Khan Academy.mp3 | As x approaches five from the right, g of x looks like it's approaching negative six. So a reasonable estimate based on looking at this graph is that as x approaches five, g of x is approaching negative six. And it's worth noting that that's not what g of five is. G of five is a different value. But the whole point of this video is to appreciate all that a limit does. A limit only describes the behavior of a function as it approaches a point. It doesn't tell us exactly what's happening at that point, what g of five is, and it doesn't tell us much about the rest of the function, about the rest of the graph. |
Connecting limits and graphical behavior Limits and continuity AP Calculus AB Khan Academy.mp3 | G of five is a different value. But the whole point of this video is to appreciate all that a limit does. A limit only describes the behavior of a function as it approaches a point. It doesn't tell us exactly what's happening at that point, what g of five is, and it doesn't tell us much about the rest of the function, about the rest of the graph. For example, I could construct many different functions for which the limit as x approaches five is equal to negative six, and they would look very different from g of x. For example, I could say the limit of f of x as x approaches five is equal to negative six, and I can construct an f of x that does this that looks very different than g of x. And in fact, if you're up for it, pause this video and see if you could do the same. |
Connecting limits and graphical behavior Limits and continuity AP Calculus AB Khan Academy.mp3 | It doesn't tell us exactly what's happening at that point, what g of five is, and it doesn't tell us much about the rest of the function, about the rest of the graph. For example, I could construct many different functions for which the limit as x approaches five is equal to negative six, and they would look very different from g of x. For example, I could say the limit of f of x as x approaches five is equal to negative six, and I can construct an f of x that does this that looks very different than g of x. And in fact, if you're up for it, pause this video and see if you could do the same. If you have some graph paper, even just sketch it. Well, the key thing is that the behavior of the function as x approaches five from both sides, from the left and the right, it has to be approaching negative six. So for example, a function that looks like this, so let me draw f of x, an f of x that looks like this and is even defined right over there and then does something like this, that would work. |
Connecting limits and graphical behavior Limits and continuity AP Calculus AB Khan Academy.mp3 | And in fact, if you're up for it, pause this video and see if you could do the same. If you have some graph paper, even just sketch it. Well, the key thing is that the behavior of the function as x approaches five from both sides, from the left and the right, it has to be approaching negative six. So for example, a function that looks like this, so let me draw f of x, an f of x that looks like this and is even defined right over there and then does something like this, that would work. As we approach from the left, we're approaching negative six as we approach from the right, we are approaching negative six You could have a function like this. Let's say the limit, let's call it h of x as x approaches five is equal to negative six. You could have a function like this. |
Connecting limits and graphical behavior Limits and continuity AP Calculus AB Khan Academy.mp3 | So for example, a function that looks like this, so let me draw f of x, an f of x that looks like this and is even defined right over there and then does something like this, that would work. As we approach from the left, we're approaching negative six as we approach from the right, we are approaching negative six You could have a function like this. Let's say the limit, let's call it h of x as x approaches five is equal to negative six. You could have a function like this. Maybe it's defined up to there, then it's, you have a circle there, then it keeps going. Maybe it's not defined at all for any of these values. And then maybe down here, it is defined for all x values greater than or equal to four and it just goes right through negative six. |
Connecting limits and graphical behavior Limits and continuity AP Calculus AB Khan Academy.mp3 | You could have a function like this. Maybe it's defined up to there, then it's, you have a circle there, then it keeps going. Maybe it's not defined at all for any of these values. And then maybe down here, it is defined for all x values greater than or equal to four and it just goes right through negative six. So notice, all of these functions as x approaches five, they all have the limit defined and it's equal to negative six, but these functions all look very, very, very different. Now another thing to appreciate is, for a given function, let me delete these, oftentimes we're asked to find the limits as x approaches some type of an interesting value. So for example, x approaches five, five is interesting right over here because we have this point discontinuity. |
Connecting limits and graphical behavior Limits and continuity AP Calculus AB Khan Academy.mp3 | And then maybe down here, it is defined for all x values greater than or equal to four and it just goes right through negative six. So notice, all of these functions as x approaches five, they all have the limit defined and it's equal to negative six, but these functions all look very, very, very different. Now another thing to appreciate is, for a given function, let me delete these, oftentimes we're asked to find the limits as x approaches some type of an interesting value. So for example, x approaches five, five is interesting right over here because we have this point discontinuity. But you could take the limit on an infinite number of points for this function right over here. You could say the limit of g of x as x approaches, not x equals, as x approaches one. What would that be? |
Connecting limits and graphical behavior Limits and continuity AP Calculus AB Khan Academy.mp3 | So for example, x approaches five, five is interesting right over here because we have this point discontinuity. But you could take the limit on an infinite number of points for this function right over here. You could say the limit of g of x as x approaches, not x equals, as x approaches one. What would that be? Pause the video and try to figure it out. Let's see. As x approaches one from the left-hand side, it looks like we are approaching this value here. |
Connecting limits and graphical behavior Limits and continuity AP Calculus AB Khan Academy.mp3 | What would that be? Pause the video and try to figure it out. Let's see. As x approaches one from the left-hand side, it looks like we are approaching this value here. And as x approaches one from the right-hand side, it looks like we are approaching that value there. So that would be equal to g of one. That is equal to g of one based on, that would be a reasonable, that's a reasonable conclusion to make looking at this graph. |
Connecting limits and graphical behavior Limits and continuity AP Calculus AB Khan Academy.mp3 | As x approaches one from the left-hand side, it looks like we are approaching this value here. And as x approaches one from the right-hand side, it looks like we are approaching that value there. So that would be equal to g of one. That is equal to g of one based on, that would be a reasonable, that's a reasonable conclusion to make looking at this graph. And if we were to estimate that, g of one looks like it's approximately negative 5.1 or 5.2, negative 5.1. We could find the limit of g of x as x approaches pi. So pi is right around there. |
Connecting limits and graphical behavior Limits and continuity AP Calculus AB Khan Academy.mp3 | That is equal to g of one based on, that would be a reasonable, that's a reasonable conclusion to make looking at this graph. And if we were to estimate that, g of one looks like it's approximately negative 5.1 or 5.2, negative 5.1. We could find the limit of g of x as x approaches pi. So pi is right around there. As x approaches pi from the left, we're approaching that value, which looks actually pretty close to the one we just thought about. As we approach from the right, we're approaching that value. And once again, in this case, this is gonna be equal to g of pi. |
Connecting limits and graphical behavior Limits and continuity AP Calculus AB Khan Academy.mp3 | So pi is right around there. As x approaches pi from the left, we're approaching that value, which looks actually pretty close to the one we just thought about. As we approach from the right, we're approaching that value. And once again, in this case, this is gonna be equal to g of pi. We don't have any interesting discontinuities there or anything like that. So there's two big takeaways here. You can construct many different functions that would have the same limit at a point. |
Connecting limits and graphical behavior Limits and continuity AP Calculus AB Khan Academy.mp3 | And once again, in this case, this is gonna be equal to g of pi. We don't have any interesting discontinuities there or anything like that. So there's two big takeaways here. You can construct many different functions that would have the same limit at a point. And for a given function, you can take the limit at many different points. In fact, an infinite number of different points. And it's important to point that out, no pun intended, because oftentimes we get used to seeing limits only at points where something strange seems to be happening. |
Functions defined by integrals switched interval AP Calculus AB Khan Academy.mp3 | Now at first when you see this, you're like, wow, this is strange. I have a function that is being defined by an integral, a definite integral, but one of its bounds are x. And you should just say, well, this is okay. A function can be defined any which way. And as we'll see, it's actually quite straightforward to evaluate this. So g of negative two, g of negative two, and I'll do the negative two in a different color, g of negative two, well, what we do is we take this expression right over here, this definite integral, and everywhere we see an x, we replace it with a negative two. So this is going to be equal to the integral from zero to x of, and I'll write x in a second, f of t dt. |
Functions defined by integrals switched interval AP Calculus AB Khan Academy.mp3 | A function can be defined any which way. And as we'll see, it's actually quite straightforward to evaluate this. So g of negative two, g of negative two, and I'll do the negative two in a different color, g of negative two, well, what we do is we take this expression right over here, this definite integral, and everywhere we see an x, we replace it with a negative two. So this is going to be equal to the integral from zero to x of, and I'll write x in a second, f of t dt. Well, x is now negative two. This is now negative two. And so how do we figure out what this is? |
Functions defined by integrals switched interval AP Calculus AB Khan Academy.mp3 | So this is going to be equal to the integral from zero to x of, and I'll write x in a second, f of t dt. Well, x is now negative two. This is now negative two. And so how do we figure out what this is? Now before we even look at this graph, you might say, okay, this is the region under, the area of the region under the graph y equals f of t between negative two and zero, but you have to be careful. Notice our upper bound here is actually a lower number than our lower bound right over here. So it will be nice to swap those bounds so we can truly view it as the area of the region under f of t above the t-axis between those two bounds. |
Functions defined by integrals switched interval AP Calculus AB Khan Academy.mp3 | And so how do we figure out what this is? Now before we even look at this graph, you might say, okay, this is the region under, the area of the region under the graph y equals f of t between negative two and zero, but you have to be careful. Notice our upper bound here is actually a lower number than our lower bound right over here. So it will be nice to swap those bounds so we can truly view it as the area of the region under f of t above the t-axis between those two bounds. And so when you swap the bounds, this is going to be equal to negative definite integral from negative two, negative two to zero of f of t dt. And now what we have right over here, what I'm squaring off in magenta, this is the area under the curve f between negative two and zero. So between negative two and zero. |
Functions defined by integrals switched interval AP Calculus AB Khan Academy.mp3 | So it will be nice to swap those bounds so we can truly view it as the area of the region under f of t above the t-axis between those two bounds. And so when you swap the bounds, this is going to be equal to negative definite integral from negative two, negative two to zero of f of t dt. And now what we have right over here, what I'm squaring off in magenta, this is the area under the curve f between negative two and zero. So between negative two and zero. So that is this area right over here that we care about. Now what is that going to be? Well, you could, there's a bunch of different ways that you could do this. |
Functions defined by integrals switched interval AP Calculus AB Khan Academy.mp3 | So between negative two and zero. So that is this area right over here that we care about. Now what is that going to be? Well, you could, there's a bunch of different ways that you could do this. You could split it off into a square and a triangle. The area of this square right over here is four. It's two by two. |
Functions defined by integrals switched interval AP Calculus AB Khan Academy.mp3 | Well, you could, there's a bunch of different ways that you could do this. You could split it off into a square and a triangle. The area of this square right over here is four. It's two by two. And just make sure to, make sure you look at the units. Sometimes each square doesn't represent one square unit, but in this case it does. So that's four. |
Functions defined by integrals switched interval AP Calculus AB Khan Academy.mp3 | It's two by two. And just make sure to, make sure you look at the units. Sometimes each square doesn't represent one square unit, but in this case it does. So that's four. And then up here, this is half of four, right? If it was all of this, that would be four. This triangle is half of four. |
Functions defined by integrals switched interval AP Calculus AB Khan Academy.mp3 | So that's four. And then up here, this is half of four, right? If it was all of this, that would be four. This triangle is half of four. So this is two right over there. Or you could view this as base times height times 1 1 2, which is gonna be two times two times 1 1 2. And so this area right over here is six. |
Functions defined by integrals switched interval AP Calculus AB Khan Academy.mp3 | This triangle is half of four. So this is two right over there. Or you could view this as base times height times 1 1 2, which is gonna be two times two times 1 1 2. And so this area right over here is six. So this part is six, but we can't forget that negative sign. So this is going to be equal to negative, negative six. And so g of negative two is negative six. |
2015 AP Calculus AB BC 4cd AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | Justify your answer. Well, to think about whether we have a relative minimum or a relative maximum, we can see, well, what's the derivative at that point? If it's zero, then it's a good candidate that we're dealing with a, that it could be a relative minimum or a maximum. If it's not zero, then it's neither. And then if it is zero, if we want to figure out relative minimum or relative maximum, we can evaluate the sign of the second derivative. So let's just think about this. So we want to evaluate f prime, we want to figure out what f prime of, f prime of two is equal to. |
2015 AP Calculus AB BC 4cd AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | If it's not zero, then it's neither. And then if it is zero, if we want to figure out relative minimum or relative maximum, we can evaluate the sign of the second derivative. So let's just think about this. So we want to evaluate f prime, we want to figure out what f prime of, f prime of two is equal to. So we know that f prime of x, f prime of x, which is the same thing as dy, dx, is equal to two times x minus y. We saw that in the last problem. And so f prime of two, right this way, f prime of two is going to be equal to two times two, two times two, minus whatever y is when x is equal to two. |
2015 AP Calculus AB BC 4cd AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | So we want to evaluate f prime, we want to figure out what f prime of, f prime of two is equal to. So we know that f prime of x, f prime of x, which is the same thing as dy, dx, is equal to two times x minus y. We saw that in the last problem. And so f prime of two, right this way, f prime of two is going to be equal to two times two, two times two, minus whatever y is when x is equal to two. Well, do we know what y is when x equals to two? Sure, they tell us right over here. Y is equal to f of x. |
2015 AP Calculus AB BC 4cd AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | And so f prime of two, right this way, f prime of two is going to be equal to two times two, two times two, minus whatever y is when x is equal to two. Well, do we know what y is when x equals to two? Sure, they tell us right over here. Y is equal to f of x. So when x is equal to two, when x is equal to two, y is equal to three. So two times two minus three, and so this is going to be equal to four minus three is equal to one. And so since the derivative at two is not zero, this is not going to be a minimum, a relative minimum, or a relative maximum. |
2015 AP Calculus AB BC 4cd AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | Y is equal to f of x. So when x is equal to two, when x is equal to two, y is equal to three. So two times two minus three, and so this is going to be equal to four minus three is equal to one. And so since the derivative at two is not zero, this is not going to be a minimum, a relative minimum, or a relative maximum. So we could say since f prime of two, f prime of two does not equal zero, this, we have a, f has, right this way, f has neither, neither minimum, or relative minimum I guess I could say, relative min, or relative max at x equals two. Alright, let's do the next one. Find the values of the constants m and b for which y equals mx plus b is a solution to the differential equation. |
2015 AP Calculus AB BC 4cd AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | And so since the derivative at two is not zero, this is not going to be a minimum, a relative minimum, or a relative maximum. So we could say since f prime of two, f prime of two does not equal zero, this, we have a, f has, right this way, f has neither, neither minimum, or relative minimum I guess I could say, relative min, or relative max at x equals two. Alright, let's do the next one. Find the values of the constants m and b for which y equals mx plus b is a solution to the differential equation. Alright, this one is interesting. So let's actually, let's just write down everything we know before we even think that y equals mx plus b could be a solution to the differential equation. So we know that, we know that dy over dx is equal to two x minus y. |
2015 AP Calculus AB BC 4cd AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | Find the values of the constants m and b for which y equals mx plus b is a solution to the differential equation. Alright, this one is interesting. So let's actually, let's just write down everything we know before we even think that y equals mx plus b could be a solution to the differential equation. So we know that, we know that dy over dx is equal to two x minus y. They told us that. We also know that the second derivative, the second derivative of y with respect to x is equal to two minus dy dx. We figured that out in part b of this problem. |
2015 AP Calculus AB BC 4cd AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | So we know that, we know that dy over dx is equal to two x minus y. They told us that. We also know that the second derivative, the second derivative of y with respect to x is equal to two minus dy dx. We figured that out in part b of this problem. And then we could also express this, we saw that we could also write that as two minus two x plus y if you just substitute, if you substitute this in for that. So it's two minus two x plus y. So let me write it that way. |
2015 AP Calculus AB BC 4cd AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | We figured that out in part b of this problem. And then we could also express this, we saw that we could also write that as two minus two x plus y if you just substitute, if you substitute this in for that. So it's two minus two x plus y. So let me write it that way. So this is also equal to two minus two x plus y. So that's everything we know before we even thought that maybe there's a solution of y equals mx plus b. So now let's start with y equals mx plus b. |
2015 AP Calculus AB BC 4cd AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | So let me write it that way. So this is also equal to two minus two x plus y. So that's everything we know before we even thought that maybe there's a solution of y equals mx plus b. So now let's start with y equals mx plus b. So if y is equal to mx plus b, y equals mx plus b, so this is the equation of a line, then dy dx is going to be equal to, well the derivative of this with respect to x is just m, derivative of this with respect to x, it's a constant, it's not going to change with respect to x, it's just a zero. And that makes sense. The rate of change of y with respect to x is the slope of our line. |
2015 AP Calculus AB BC 4cd AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | So now let's start with y equals mx plus b. So if y is equal to mx plus b, y equals mx plus b, so this is the equation of a line, then dy dx is going to be equal to, well the derivative of this with respect to x is just m, derivative of this with respect to x, it's a constant, it's not going to change with respect to x, it's just a zero. And that makes sense. The rate of change of y with respect to x is the slope of our line. So can we use, and this is really all that we know, we could keep, actually we could go even further. We could take the second derivative here. The second derivative of y with respect to x, well that's going to be zero. |
2015 AP Calculus AB BC 4cd AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | The rate of change of y with respect to x is the slope of our line. So can we use, and this is really all that we know, we could keep, actually we could go even further. We could take the second derivative here. The second derivative of y with respect to x, well that's going to be zero. The second derivative of a linear function, well it's going to be zero, we see that here. So this is all of the information that we have. We get this from the previous parts of the problem and we get this just taking the first and second derivatives of y equals mx plus b. |
2015 AP Calculus AB BC 4cd AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | The second derivative of y with respect to x, well that's going to be zero. The second derivative of a linear function, well it's going to be zero, we see that here. So this is all of the information that we have. We get this from the previous parts of the problem and we get this just taking the first and second derivatives of y equals mx plus b. So given this, can we figure out, can we figure out what m and b are? Alright, so we could, if we said m is equal to two x minus y, that doesn't seem right. This one is a tricky one. |
2015 AP Calculus AB BC 4cd AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | We get this from the previous parts of the problem and we get this just taking the first and second derivatives of y equals mx plus b. So given this, can we figure out, can we figure out what m and b are? Alright, so we could, if we said m is equal to two x minus y, that doesn't seem right. This one is a tricky one. Let's see, we know that the second derivative is going to be equal to zero. We know that this is going to be equal to zero for this particular solution. And we know dy dx is equal to m. We know this is m. And so there you have it, we have enough information to solve for m. We know that zero is equal to two minus m. So zero is equal to two minus m. And so we can add m to both sides and we get m is equal to two. |
2015 AP Calculus AB BC 4cd AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | This one is a tricky one. Let's see, we know that the second derivative is going to be equal to zero. We know that this is going to be equal to zero for this particular solution. And we know dy dx is equal to m. We know this is m. And so there you have it, we have enough information to solve for m. We know that zero is equal to two minus m. So zero is equal to two minus m. And so we can add m to both sides and we get m is equal to two. So that by itself was quite useful. And then what we could say, let's see, can we solve this further? Well we know that this right over here, dy dx, this is m. This is m. And it's equal to two. |
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