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Current time:0:00Total duration:7:28

AP.CALC:

CHA‑5 (EU)

, CHA‑5.A (LO)

, CHA‑5.A.1 (EK)

what I've got here is the graph of y is equal to cosine of X and what I want to do is figure out the area under the curve Y is equal to f of X and above the x-axis I'm going to do it over various intervals so first let's think about the area under the curve between X is equal to zero and X is equal to PI over two x is equal to PI over two so we're talking about this area this area right over here well the way we denote it is the definite integral from 0 to PI over 2 of cosine of X DX and all this is is kind of it's reminiscent of taking a sum of a bunch of super thin rectangles with width DX and height f of X at each if for each of those rectangles and then you take an infinite number of those infinitely thin rectangles and that's kind of what this notation is trying to depict but we know how to do this already the second fundamental theorem of calculus helps us we just have to figure out what the antiderivative of cosine of X is or what an antiderivative of cosine of X is evaluated at PI over 2 and from that subtract it evaluated at 0 so what's the antiderivative of cosine of X or what's an antiderivative well we know if we take the derivative let me write this up here we know that if we take the derivative of sine of X sine of X we get cosine of X so the antiderivative of cosine of X is sine of X now why do I keep saying sine of X is an antiderivative it's not just the antiderivative well I can also take the antiderivative of sine of X sine of X plus any arbitrary constant and still get cosine of X because the derivative of a constant is 0 this could be PI this could be 5 this could be a million this could be Google this could be any crazy number but the derivative of this is still going to be cosine X so when I say that we just have to we have to find an antiderivative I'm just saying look we just have to find one of the derivatives sine of X is probably the simplest because in this case the constant is 0 so let's evaluate so one way that we can denote this the antiderivative of cosine of X where an antiderivative of cosine of X is sine of X and we're going to evaluate it at PI over 2 and from that subtract it evaluated at 0 so this is going to be equal to let me do it right over here this is going to be equal to sine of PI over 2 minus sine of 0 which is equal to sine of PI over 2 is 1 sine of 0 is 0 so it's 1 minus 0 is equal to 1 so the area of this region right over here this area is equal to 1 now let's do something interesting let's think about the area let's think about the area under the curve between let's think about the curve area under the curve between let's say PI over 2 PI over 2 and 3 PI over 2 3 3 PI over 2 3 PI over 2 3 PI over 2 so between here and here so we're talking about this area we're talking about that area right over here this is 3 PI over 2 so once again the way we denote the area is the definite integral from PI over 2 to 3 PI over 2 of cosine of X DX the antiderivative or n antiderivative of cosine of X is sine of X evaluated at 3 PI over 2 and PI over 2 so this is going to be equal to sine of 3 PI over 2 minus sine of PI over 2 what's sine of 3 PI over 2 if you visualize the unit circle if we visualize the unit circle real fast 3 PI over 2 is going all the way 3 fourths around the unit circle so it's right over there so sine is Z is the y-coordinate on that unit circle so it's negative 1 so this right over here is negative 1 this right over your sine of PI over 2 PI over 2 is just going straight up like that so sine of PI over 2 is 1 so this is interesting we get negative 1 minus 1 which equal to negative two we got a negative area here we got a negative area now how does that make sense we know in the real world areas are always positive but what is negative two really trying to depict well it's trying to sign it for the based on the idea that now our function is below the x-axis so we could kind of think that we have an area of two but it's all below the x-axis in this case and so it is signed as negative to the actual area is two but since it's below the x-axis we get a negative right over here now let's do one more interesting one let's find the definite integral from 0 to 3 PI over 2 0 to 3 PI over 2 of cosine of X DX now this is just denoting this entire area going from 0 all the way all the way to 3 PI over 2 and what do you think is going to happen well let's evaluate it it's going to be sine of 3 PI over 2 sine of 3 PI over 2 minus sine of 0 minus sine of 0 which is equal to negative 1 minus 0 which is equal to negative 1 so what just happened here the area under all this orange area that I just outlined is clearly not negative or any area isn't a negative number and there isn't even one but what just happened here well we saw in the first case that this first area that this first area was 1 the area of this first region right over here is 1 and then the area of the second region we got negative 2 we got negative 2 negative 2 and so one way to interpret it is is that your net area above your net area above the x-axis is negative 1 or you another way to say it your net area is negative 1 so it's taking the one might one air one region above the x-axis and then subtracting the two below it so what the definite integral is doing when we evaluate it using the second fundamental theorem of calculus it's essentially finding the net area above the x axes and if we get a negative number that means that the net area is that actually most of the area is below the x-axis if we get zero then that means it all Nets out and if you want to see a case that's zero take the integral from from zero all the way all the way to two pi and this will evaluate to zero because you have an area of one and in another area of one but it nets out with this area of negative two let's try that out so if we go from zero to two pi of cosine of X DX this is going to be equal to sine of two pi minus sine of zero which is equal to zero minus zero which is indeed equal to zero now clearly there was area here but all the area that was above the x-axis netted out with all the area that was below the x-axis

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