Proof of the mean value theorem

Available under Creative Commons-ShareAlike 4.0 International License.

Suppose that the mean value theorem is violated. Let Lbe the set of all xin the interval from ato bsuch that y(x) < $\bar{y}$ , and likewise let Mbe the set with y(x) > $\bar{y}$ . If the theorem is violated, then the union of these two sets covers the entire interval from ato b. Neither one can be empty; if, for example, M was empty, then we would have y< $\bar{y}$ everywhere and also $\int_{a}^{b}y=\int_{a}^{b}\bar{y}$ , but it follows directly from the definition of the definite integral that when one function is less than another, its integral is also less than the other’s. Since y takes on values less than and greater than $\bar{y}$ , it follows from the intermediate value theorem that y takes on the value $\bar{y}$ somewhere (intuitively, at a boundary between L and M).

2223 reads

You are here

Proof of the mean value theorem