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# Negative definite integrals

AP.CALC:
FUN‑6 (EU)
,
FUN‑6.A (LO)
,
FUN‑6.A.1 (EK)

## Video transcript

we've already thought about what a definite integral means if I'm taking the definite integral from A to B of f of X DX I can just view that as the area below my function f so if this is my y axis this is my x axis and y is equal to f of X - something like that y is equal to f of X and if this is a and if this is B I could just view this expression as being equal to this area but what if my function was not above the x-axis what if it was below the x-axis so these are going to be equivalent let's say let me just draw that scenario so let me draw a scenario where it's my x-axis that is my y-axis and let's say I have let's say I have a function that looks like that so that is y is equal to G of X and let's say that this right over here is a and this right over here is B and let's say that this area right over here is equal to 5 well if I were to ask you what is the definite integral from A to B of G of X DX what do you think it is going to be well you might be tempted to say hey well it's just the area again between my curve and the x-axis you might you might be time to say hey this is just going to be equal to 5 but you have to be very careful because if you're looking at the area above your curve and below your x-axis vs. below your curve and above the x-axis this definite integral is actually going to be the negative of the area now we'll see later on why this will work out nicely with a whole set of integration properties but if you want to get some intuition for it let's just think about velocity versus time graphs so if I in my horizontal axis that is time my vertical axis this is velocity and velocity is going to be measured in meters per second time is going to be measured in seconds time is measured in seconds and actually I'm going to do two scenarios here so let's say that I have a first velocity time graph let's just call it V 1 of T which is equal to 3 and it would be 3 meters per second so 1 2 3 so it would look like that that is V 1 of T and if I were to look at the definite integral going from time equals 1 to time equals 5 of V sub 1 of T DT what would this be equal to well here my function is both above my T axis so I'll just go from 1 to 5 which will be around there and I could just think about the area here and this area is pretty easy to calculate it's going to be 3 meters per second times 4 seconds that's my change in time and so this is going to be 12 meters and so this is going to be equal to 12 and one way to conceptualize this is this gives us our change in position if my velocity is 3 meters per second and since it's positive you can conceptualize that is it's going to the right at 3 meters per second what is my change of in position well I would have gone 12 meters to the right and you don't need calculus to figure that out 3 meters per second times 4 seconds would be 12 meters but what if it were the other way around what if I had another velocity function let's call that V sub 2 of T that is equal to negative 2 meters per second and it's just a constant negative 2 meters per second so this is V sub 2 of T right over here what would or what should the definite integral from 1 to 5 of V sub 2 of T V DT be equal to well it should be equal to my change in position but if my velocity is negative that means I'm moving to the left that means my change in position should be to the left as opposed to to the right and so we can just look at this area right over here when if you just look at as the rectangle is going to be 2 times 4 which is equal to eight but you have to be very careful since it is below my horizontal axis and above my function this is going to be negative and this should make a lot of sense if I'm going to m/s to the left for four seconds or another way to think about it if I'm going negative two meters per second for four seconds then my change in position is going to be negative eight meters I would have moved eight meters to the left if we say the convention is negative means to the left so the big takeaway is if it's below your function and above the horizontal axis the definite integral and if your a is less than B then your definite integral is going to be positive if your a is less than B but your function over that interval is below the horizontal axis then your definite integral is going to be negative and in the future we'll also look at definite integrals that are a mix of both but that's a little bit more complicated
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