Question 8.T.9: (Mean Value Theorem) If f is continuous on [a, b], then ther...
(Mean Value Theorem)
If f is continuous on [a, b], then there exists a point c ∈ (a, b) such that
\int_{a}^{b}{f (x) dx = f (c) (b − a)} .
The case where f is constant on [a, b] can be dismissed, for we can then choose c to be any point in (a, b). Since f is continuous on the compact interval [a, b], it attains its maximum value M and its minimum value m on [a, b]; i.e., there are points x_{1}, x_{2} ∈ [a, b] such that
m = f (x_{1}) ≤ f (x) ≤ f (x_{2}) = M for all x ∈ [a, b].
Since f is not constant, there are points t_{1}, t_{2} ∈ [a, b] such that
f (t_{1}) > m, f (t_{2}) < M.
It follows from Remark 8.2 applied to the functions f(x) − m and M − f(x) that
\int_{a}^{b}{mdx} < \int_{a}^{b}{f (x) dx} < \int_{a}^{b}{M dx}.
Using the result of Example 8.1,
m (b − a) < \int_{a}^{b}{f (x) dx} < M (b − a)
m < \frac{1}{b − a} \int_{a}^{b}{f (x) dx} < M
f (x_{1}) < \frac{1}{b − a} \int_{a}^{b}{f (x) dx} < f (x_{2}).
By the intermediate value property for continuous functions (Theorem 6.6), there is a point c between x_{1} and x_{2} where
f (c) = \frac{1}{b − a} \int_{a}^{b}{f (x) dx}.