Stealing chickens is enough. Take the limit, in short, a≈b, and then transfer the equation to get:

Let it go first, and then use it later.

Now let's look at another function g(x). Now the area of curve g(x) is filled with many, many thin rectangles. First of all, let's assume that the width of the rectangle attached to the X axis is xn-xn- 1, where n is the serial number (0, 1, 2, 3, 4, ..., n ..., n), and input n=2, where n is the height of g(xn). Note that I have raised the right side of the rectangle here, and part of it is divided into rectangles, so among many rectangles, the rectangular area of one rectangle is (xn-xn- 1)g(xn). If you add up all the rectangles, it is:

(x1-x0) (GX1)+(x2-x1) g (x2)+... (xk-xk-1) g (xk)

The above is the area of any interval under the g(x) curve.

I'm telling you now, actually,

G(x) is the derivative of f(x)

So, considering the relationship derived from the basic principle at the beginning:

Let a=x0, b=x 1, and get

In fact, this is an integral with x 1 as the upper limit and x0 as the lower limit. The reason why the area of the definite integral you asked here is the interval difference of the original derivative function has been solved.

Further promotion details:

Let's sum a larger area, such as the sum from x0 to x3.

Below you will find that there will be such a situation:

The first bracket and the last bracket have cancelable items.

I think the conclusion is very obvious here.

Maybe that's the idea. I have never studied advanced mathematics. I heard it from a teacher. I think the answer is quite intuitive, so I'll share it with you.