Calculus, all content (2017 edition)
- Sal interviews the AP Calculus Lead at College Board
- 2015 AP Calculus AB/BC 1ab
- 2015 AP Calculus AB/BC 1c
- 2015 AP Calculus AB/BC 1d
- 2015 AP Calculus AB 2a
- 2015 AP Calculus AB 2b
- 2015 AP Calculus 2c
- 2015 AP Calculus AB/BC 3a
- 2015 AP Calculus AB/BC 3b
- 2015 AP Calculus AB/BC 3cd
- 2015 AP Calculus AB/BC 4ab
- 2015 AP Calculus AB/BC 4cd
- 2015 AP Calculus AB 5a
- 2015 AP Calculus AB 5b
- 2015 AP Calculus AB 5c
- 2015 AP Calculus AB 5d
- 2015 AP Calculus AB 6a
- 2015 AP Calculus AB 6b
- 2015 AP Calculus AB 6c
- 2011 Calculus AB free response #1a
- 2011 Calculus AB free response #2 (a & b)
- 2011 Calculus AB free response #2 (c & d)
- 2011 Calculus AB free response #3 (a & b)
- 2011 Calculus AB free response #3 (c)
- 2011 Calculus AB free response #4a
- 2011 Calculus AB free response #4b
- 2011 Calculus AB free response #4c
- 2011 Calculus AB free response #4d
- 2011 Calculus AB free response #5a
- 2011 Calculus AB free response #5b
- 2011 Calculus AB free response #5c
- 2011 Calculus AB free response #6a
- 2011 Calculus AB free response #6b
- 2011 Calculus AB free response #6c
Equation of a tangent line and area between curves. Created by Sal Khan.
Want to join the conversation?
- Don't you have to write units squared right after the -1/8 + 1/pi because you are finding the area?(3 votes)
- Unless units are given, you don't have to add a generic 'units^2' to your answer on the AP exam. That said it doesn't count against you if you do add it.(3 votes)
- 4:55Where do these trig function properties of derivatives/antiderivatives come from? I'm having trouble grasping how to take the derivative of functions in which x is being multiplied by a constant, such as sin(2x), cos(3x/4) etc.
I perused the obvious vids on here, but didn't see any mention of these types of derivatives/antiderivatives.
thanks in advance(4 votes)
- I found this on wolfram alpha but I'm still not sure what is going on in the u substitution step.
Possible intermediate steps:
integral sin(pi x) dx
For the integrand sin(pi x), substitute u = pi x and du = pi dx:
= 1/pi integral sin(u) du
The integral of sin(u) is -cos(u):
Substitute back for u = pi x:
= -(cos(pi x))/pi+constant
Is this saying to multiply the entire integral by pi/pi? So just using a little trick of multiplying it by 1 and hence not changing anything?
- @3:46Why are we evaluating/integrating from 0 to 1/2 instead of 0 to 1?(1 vote)
- Because we are integrating with respect to x, and the x values in consideration range from 0 to 1/2. We could change the integration to be with respect to y, in which case you would integrate from 0 to 1.(3 votes)
- at3:10how do you know which function is above/below?(1 vote)
- Usually the best way is to make a graph of the region and determine the answer by looking at the graph. In this case they've given you the graph, so you just have to recognize which line looks like part of the sine function (sin(πx) looks like sin(x) except for greater frequency) and which line looks like x^3 (again, there's a coefficient, but that doesn't change the general shape of the function).
If you're unsure, you can figure the slope of each function at the left boundary (x = 0), and the one with the higher slope would be the one that's above the other. In a pinch you can try to find a point within the range where you can calculate the value of both functions. By far the most valuable skill here, though, is to understand the graphs.(2 votes)
- I see that the derivative of -1/pi(cos-pi-x) is sin-pi-x, but going the other way is not overtly obvious. What is the operation for taking the integral of sin-pi-x(1 vote)
- If you know how to do u-subs, then make u=pi*x, and du=pi*dx. If you know what to do from here, great. Otherwise, let me know.(1 vote)
- Where did Sal get that he couldn't use the calculator to evaluate the integral?(1 vote)
Let R be the region in the first quadrant enclosed by the graphs of f of x is equal to 8x to the third and g of x is equal to sine of pi x, as shown in the figure above. And they drew the figure right over here. Part a, write the equation for the line tangent to the graph of f at x is equal to 1/2. So let me just redraw it here just so that-- I like drawing on the black background, I guess, is the main reason why I'm redrawing it. So the function f of x is equal to 8x to the third, looks like this. It looks like that. This is our f of x-axis, and this is our x-axis. And we want the equation for the line tangent at x is equal to 1/2. So this is x is equal to 1/2. If you go up here, if you evaluate f of 1/2, you get 8 times 1/2 to the third, which is 8 times 1/8, which is 1. And they actually gave us that already on this point. This is the point 1/2 comma 1. And we need to find the equation for the tangent line. So the tangent line will look something like that. And to figure out this equation, we just need to figure out its slope, and then we know a point that it's on-- and you could use point-slope or you could just use your kind of standard slope-intercept form to give an equation for that line. So the first part, let's figure out its slope. And the slope of the tangent line is going to be the same slope as the slope of our function at that point. Or another way to think about it, it is going to be f prime of 1/2, or the derivative evaluated at 1/2. The derivative gives us the slope of that line at any point. So what is f prime of x? f prime of x is just the derivative of this. So 3 times 8 is 24 times x squared. 24 x squared. f prime of 1/2 is equal to 24 times 1/2 squared, which is equal to 24 times 1/4, which is equal to 6. So the slope of this line is equal to 6. I'll use m for slope. That's the convention that we used when we first learned it in algebra. So the slope is going to be 6. So the general equation for this line is y is equal to mx plus b. This is the slope. This is the y-intercept. We already know that the slope is 6, and then we can use the fact that the line goes through the point 1/2, 1 to figure out what b is. So when y is 1, 1 is equal to our slope times x. x is 1/2. Or another way to say it, when x is 1/2, y is 1 plus some y-intercept. Or if I take x is 1/2, I multiply it times the slope plus the y-intercept, I should get 1. And so I get 1 is equal to 3 plus b. I can subtract 3 from both sides, and I get negative 2 is equal to b. So the equation of the line is going to be y is equal to 6x minus 2. That is the equation of the tangent line. Now part b, find the area of R. So R is this region right over here. It's bounded above by g of x, which they've defined as sine of pi x. It's bounded below by f of x, or 8x to the third. So the area is going to be-- actually, let me just do it-- I'll scroll down a little bit. I still want to be able to see this graph right over here. Part b. The area of R is going to be equal to the integral from 0, that's this point of intersection right over here, to 1/2. So let me make it clear. This is 0 to 1/2. And then the function on the top-- so we could just take the area of that, but then we're going to have to subtract from that the area underneath the function below. Or one way to think about it is the integral from 0 to 1/2 of the top function is g of x, which is sine of pi x. But if we just evaluated this integral-- let me just put a dx over here. If we just evaluated this, we would get the area of this entire region. But what we need to do is subtract out the area underneath the second, underneath f of x. We just subtract out the area under that. So we just subtract from that f of x. And f of x, we already saw, is 8x to the third. Is 8x to the third power. And now we can just evaluate this. So let me draw a little line here. It's getting a little bit messy. I'll just do it down here. So we need to take the antiderivative of sine of pi x. Well, the derivative of cosine of x is negative sine x. The derivative of cosine of pi x is negative pi cosine of pi x. So the antiderivative of sine of pi x is going to be negative 1 over pi cosine of pi x. And you can verify it for yourself. And you might say, wait. Sal, how did you know it was a negative. Well, I put the negative there so that when I take the derivative of the cosine of pi x, I would get a negative sign, but that negative will cancel out the negative to give me a positive here. And you say, why do you put a 1 over pi here. Well, when you take the derivative of this thing using the chain rule, you take the derivative of the pi x, you'll get pi, that you would multiply everything by, and then you would get negative sine of pi x. And that pi doesn't show up here. So I need something for it to cancel out with. And that's what this 1 over pi is going to cancel out with. And you could do use substitution and all the rest, if you found something like that useful. But it's in general, a good habit or I guess it's good to be able to do this almost by sight. So, and you can verify that this derivative is equal to sine of pi x. So the antiderivative of sine of pi x is this. The antiderivative of negative 8x to the third power is negative 8-- I'm going to divide it by 4-- so negative 2x to the fourth power. And all I did is I incremented the 3 to a 4, and then I divided by 8 by the 4. And you could take the derivative of this to verify this it is the same thing as negative 8x to the third power. And we're going to have to evaluate that from 0 to 1/2. When you evaluated 1/2, I'm going to get negative 1 over pi cosine of pi over 2 minus 2 times 1/2 to the fourth power is 1/16. So that's evaluated at 1/2. And then from that, I'm going to subtract negative 1 over pi cosine of 0. Let me just write it out. So minus negative 1 over pi cosine of pi times 0. Let me just write cosine of 0 pi, I could write, or pi times 0, minus 2 times 0 to the fourth. So that's just going to be minus 0. So let's evaluate this. So to simplify it, we have a cosine of pi over 2. Cosine of pi over 2 is just going to be 0, so this whole thing just becomes zero. And then you have a negative 2 divided by 16. That's negative 1/8. And then from that, I'm going to subtract this business here. Cosine of 0, this is 1. So this is just a negative 1 over pi. And then I have a 0 there. so I can ignore that. So this is equal to negative 1 over 8 plus 1 over pi. And we are done. This part of it, you're not allowed to use a calculator. So this is about as far as I would expect them to expect you to get. And so I'll leave you there. In the next video, we will do part c.