 So how can we actually find the value of a definite integral? For that, we'll try and figure out what properties this definite integral has. So since we've learned how to differentiate, let's find the derivative of the definite integral. So for that, we'll need to define it as a function. So let capital F of x be the definite integral from 0 to x of f of t dt. So it'll be helpful to consider this geometrically. Capital F of x represents the area under the graph of y equals f of t above the t axis from t equals 0 to t equals x. Now if we don't make any assumptions about what f of t looks like, the problem becomes very complicated. So to begin our analysis, we'll assume that f of t is increasing. And so this region might look something like this. Now if we want to find the derivative of capital F of x, we need to use the definition of the derivative. And so we're looking for the limit as h goes to 0 of this difference quotient. Well let's consider the geometry. F of capital X plus h is the area under the curve out to t equals x plus h. So this difference f of x plus h minus f of x represents this sliver of area between t equals x and t equals x plus h. Well let's try to bound this area. Because we assume that f of t is increasing, this sliver of area is going to be larger than the left rectangle, but smaller than the right rectangle. But we know these two areas. The left rectangle has height f of x and width h. So its area is going to be f of x times h. Meanwhile, the dimensions of the right rectangle are height f of x plus h and width h, which gives us an area of f of x plus h times h. Now we can do a little bit of algebra and our difference quotient is bounded between f of x and f of x plus h. Well since we want to take the limit as h goes to 0 and we have trapped our difference quotient in between two functions, we can apply the sandwich theorem. So as h goes to 0, the limit as h goes to 0 of f of x is going to be f of x. And as h goes to 0, the limit as h goes to 0 of f of x plus h is going to be f of x. And so this difference quotient also goes to f of x. What's important to recognize is nothing important changes as f of t is decreasing. And if f of t is sometimes increasing and sometimes decreasing, we can split it up into parts that are increasing and parts that are decreasing. Or can we? Again, that's a topic for a later course. This gives us the following important result. Provided certain conditions are met, let capital f of x be the integral from 0 to x of some function. Then the derivative of capital f of x is going to be the same as the integrand f of x. And so now we're ready to find some definite integrals. Let's find the definite integral from 0 to x of 2t dt. Our fundamental theorem says that if capital f of x is this integral, then the derivative is going to be the integrand where we use x instead of t. And so our derivative is going to be 2x, which means that capital f of x is going to be x squared plus c. What's c? In order to find c, we need to calibrate our function. Well, we're at luck because f of 0 is going to be 0. And that's because f of 0 is going to be the integral from 0 to 0 of something. And our theorem says that if our beginning and ending points are the same, the definite integral is 0. And so we can evaluate f of 0 and find c. And we find that c is equal to 0. So that gives us our definite integral is equal to x squared. How about the integral from 0 to x of sine t? So again, by the fundamental theorem, I know that the derivative is going to be the integrand. So f prime of x is going to be sine of x. And so f of x is minus cosine x plus c. And once again, we know that f of 0 is 0, so we can calibrate and find c, which turns out to be equal to 1, and that gives us our definite integral.