Why can the median point be taken to the endpoint in the integral median value theorem?

There is also a closed-interval version of the mediator theorem, such as the following:

Let f(x) be continuous at [a,b], and if t satisfies f(a) t f(b), then there is c a,b], so that f(c) = t

The usual proof of the integrative first median theorem is essentially using this version.

Because the conditions for use are: the minimum value of f, the mean of integrals, the maximum value of f, not a strict inequality sign.

In fact, the open-interval version of the integral first median theorem is also true (it is recommended to try to prove it yourself).

But this does not contradict the closed-interval version of the integral-first median theorem, since c a,b) can deduce c a,b], so this is just a reinforcement.

The median point is generally not unique, so sometimes the endpoint can also be taken to the median.

The simplest example is that f(x) is a constant function, and all points are taken as the median.

For example, f(x) = sin(x) takes the median value at the integral of [-, at - ,0.

Regardless of whether it is the differential median theorem or the integral median theorem, the median point is guaranteed by the theorem to be taken in the open interval (the median point is not the only one). However, it is not excluded that the endpoint of the interval can also be taken as the median point, such as a constant function, and any point in the interval can be used as the median point.

This is higher mathematics.

The basic concept of .

Original function. It is known that the function f(x) is a function defined in an interval, and if there is a function f(x) such that any point in the interval has df(x)=f(x)dx, then the function f(x) is said to be the original function of the function f(x) in that interval. Integrating f(x) gives us both the original function f(x) and the differentiation of f(x) gives f(x).

Antiderivative. Relative to the definite integral, there is an indefinite constant in the final solution of the expression. Differential of sinx+c yields cosx, where c is an arbitrary constant, and indefinite integral of cosx yields sinx+c.

If the definite integral is performed then there is no indefinite constant, then the finite condition will be given in the problem, for example, when the original function is 1 at x=0, then the integral of cosx is sinx+c, and when x=0, sinx+c=1, so c=1, so the definite integral of cosx is sinx+1