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Current time:0:00Total duration:8:13

let R be the region in the first quadrant enclosed by the graphs of f of X is equal to 8 X to the 3rd 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 8 is equal to 8 X to the 3rd looks like this it looks like that this is our f of x axis and this is our x axis and we want to the equation for the line tangent at X is equal to 1/2 so this is X is equal to a 1/2 if you go up here if you evaluate F of 1/2 you get F 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 the equation we just really figure out its slope and then we know a point that it's on and we could use there we could you could use point-slope or you can 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 F prime of 1/2 or the derivative evaluated 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 what we the convention that we use an out 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 a 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 and plus some plus some y-intercept 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 is this region right over here it's bounded above by G of X which they've defined a sine of pi sine of PI X it's bounded below it's bounded below by f of X or 8x to the third so the area 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 Part B the area of R is going to be equal to the integral from 0 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 0 to 1/2 and then the the function on the top so we could just take the area of that but then we're going to subtract from that the area underneath the function below or not one way to think about it is the integral from 0 to 1/2 of the the top function is G of X which is sine of PI X so a sine of PI X 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 area of this entire region but what we need to do is subtract out the area underneath the second underneath f of we need to subtract out the area under that so we just subtract from that f of X and f of X we already saw is 8 X to the third is 8 X to the third power and now we could just evaluate this so let me draw a little line here it's getting a little bit messy 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 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 derivative the antiderivative of sine of PI X is going to be negative 1 over PI cosine cosine of PI X and you can verify it for yourself and you might say wait so 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'm going to I would get a negative sign but that negative will cancel out the negative to give me a positive here and you said why did 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 then 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 1 over PI is going to cancel out with and you could do u substitution and all the rest if if if you found that if you found something like that useful but it's it's it's in general a good a good habit or I guess it's a good it's it's it's good to be able to do this almost 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 8 X to the third power is negative 8 I'm going to divide it by 4 so negative 2x to the fourth power and I just all I did is I incremented the 3 to a 4 and then I divided the 8 by the 4 and you could take the derivative of this to verify that it is the same thing as negative 8 X to the 3rd power and we're gonna have to evaluate that from 0 to 1/2 when you evaluate it 1/2 so I'm gonna get I'm going to get negative 1 over pi cosine of pi over 2 cosine of PI over 2 minus 2 times 1/2 to the 4th power is 1/16 so that's it 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 negative 1 over PI cosine of PI times 0 cosine of PI let me just write cosine of 0 PI I can 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 0 and then you have a negative 2 divided by 16 that's negative 1/8 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 a negative 1 over PI and then I have a zero there so I can ignore that so this is equal to negative 1 over 8 plus 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 get to to for 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 we will do Part C

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