Consistency of definite integral and area in geometrical intuition

For the definite integral, the high school textbook has the following example:

(High school textbook itself is not rigorous, it doesn't matter, but I am too impressed with this example, so take it as an example to illustrate the problem)

Small rectangular area and limits, etc. not equal to the area below the curve? This is a very crazy question.

Some people would say that the area is not the definition of integral? But how do you make sure that you define the area as visually consistent with the geometry?

In other words, the sum of the rectangular area of the figure is less than the area below the curve, in terms of geometrical intuition. (Overall greater than part)

If the limit is present, then this limit (i.e. \ (\frac{1}{3}\)) may be equal to the area below the curve or less than the area below the curve. (i.e. less than the area below the curve)

How do you know it is equal, not less than?

In fact, it can be proved that it is equal. Here's how to borrow the ancient Greek Archimedes.

In the distant ancient Greece, there was no concept of calculus at all. But Archimedes still uses the idea of calculus to strictly derive the formula of the area of the circle and the volume of the ball.

How does he keep it strict? The idea is this. For example, now requires the area of the circle, it is first proved that the area of the circular inner regular polygon is less than \ (\pi r^2\) and convergence to the area of the circle, and then prove that the area of the round external regular polygon is greater than \ (\pi r^2\) and converge to the area of the circle. Thus, if the area of the circle is assumed to be greater than \ (\pi r^2\), then the first contradiction is assumed, and if the area of the circle is less than \ (\pi r^2\), it contradicts the second one. So the area of the circle can only be \ (\pi r^2\).

According to this idea, we have reached the limit is not greater than the area below the curve, so only need to re-proof that the limit value is not less than the area below the curve can be.

Take the function value of the right end point of each small interval as the height of the small rectangle and sum the area (the part of the dashed line in the graph), and then take the limit Again (\frac{1}{3}\) (note that this is equivalent to the limit obtained by using the previous method of high). And at this time the size of the small rectangle and from the geometric intuitive view is greater than the area below the curve, so the limit \ (\frac{1}{3}\) is not less than the area below the curve. In combination with the above point (the Limit \ (\frac{1}{3}\) is not larger than the area below the curve), the limit must be equal to the area below the curve. This corresponds to the integral and the area from the geometrical intuition.

Thus, it can be concluded that "integrable" means that the area can be obtained by means of integral method. Because the first necessary and sufficient condition of the integrable is that the upper and lower and the limit equal to the integrable (in turn), and this limit is equal to the value of the integral.

This is equivalent to saying: Upper and not less than the area, under and not greater than the area, the limit and the same, so the limit is equal to the area. If the limit is inconsistent, the area cannot be obtained by this method.

Consistency of definite integral and area in geometrical intuition