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Functions defined by definite integrals (accumulation functions)

Understanding that a function can be defined using a definite integral. Thinking about how to evaluate functions defined this way.

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Video transcript

- [Instructor] 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 function's 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 x odd. And you could say it's equal to x to the third otherwise, otherwise. So if it's an odd integer, it's an odd integer, you just 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 a 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 to, 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, 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 one, what is g of x going to be equal to? All right, so g of one is going to be equal to the definite integral, going from negative two. Now x is going to be equal to one in this situation. That's what we're inputting into the function. So one 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 two and t is equal to one. So it's going to be this area here. And we, 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 two 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 two? What is g of two going to be equal to? Pause this video, and try to figure that out. Well, g of two is going to be equal to the definite integral from negative two, and now our upper bound's going to be our input into the function to two, 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 was 16 square units, plus another one, two, three, four, five square units. So 16 plus five, 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.