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# Functions defined by definite integrals (accumulation functions)

AP.CALC:
FUN‑5 (EU)
,
FUN‑5.A (LO)
,
FUN‑5.A.1 (EK)
,
FUN‑5.A.2 (EK)

## Video transcript

you've already spent a lot of your mathematical lives talking about functions the basic idea is give a valid input into a function so a member of that functions domain and then the function is going to tell you for that input what is going to be the corresponding output and we call that corresponding output f of X so for example there's many ways of defining functions you could say something like f of X is equal to x squared so that means that whatever X whatever you input into the function the output is going to be that input squared you could have something defined like this f of X is equal to x squared if if X odd and you could say it's equal to X to the third otherwise other wise so if it's an odd integer it's an odd integer you square it but otherwise for any other real number you take it to the third power this is a valid way of defining a function what we're going to do in this video is explore a new way or potentially new way for you of defining a function and that's by using a definite integral but it's the same general idea so what we have graphed here this is the T axis this is the y axis and we have the graph of the function f or you could view this as the graph of y is equal to f of T now what I want and this is another way of representing what outputs you might get for a given input here if T is one f of T is five if T is four F of T is three but I'm now going to define a new function based on a definite integral of f of T let's define our new function let's say G let's call it G of X let's make it equal to the definite integral from negative two to X of F of T DT now pause this video to really take a look at it this might look really fancy but what's happening here is given an input X G of X is going to be based on what the definite integral here would be for that X and so we can set up a little table here to think about some potential values so let's say X and let's say G of X right over here so if X is 1 what is G of X going to be equal to all right so G of 1 is going to be equal to the definite integral going from negative 2 now X is going to be equal to 1 in this situation that's what we're inputting into the function so 1 is our upper bound of F of T DT and what is that equal to well that's going to be the area under the curve and above the T axis between T equals negative 2 and T is equal to 1 so it's going to be this area here and week since it's on a grid we can actually figure this out we can actually break this up into two sections this rectangular section is three wide and five high so it has an area of 15 square units and this little triangular section up here is too wide and one high two times one times one half area of a triangle this is going to be another one so that area is going to be equal to 16 what if X is equal to 2 what is G of 2 going to be equal to pause this video and try to figure that out well G of 2 is going to be equal to the definite integral from negative 2 and now our upper bound is going to be our input into the function to 2 of F of T DT so that's going to be going from here all the way now to here and so it's the area we just calculated it's all of this stuff which we figured out with 16 square units plus another 1 2 3 4 5 square units so 16 plus 5 this is going to be equal to 21 so hopefully that helps and the key thing to appreciate here is that we can define valid functions by using definite integrals
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