Proof of the integral mean value theorem

The proof combines two theorems on continuous functions: Weierstrass (existence of a maximum and minimum) and the intermediate value theorem. The guiding idea is to “wedge” the integral mean between the minimum and the maximum of $f$.

Proof

Step 1 (Weierstrass). $f$ is continuous on $[a,b]$, so by Weierstrass’s theorem it admits an absolute minimum value $f_{\min}$ and an absolute maximum value $f_{\max}$: $f_{\min} \le f(x) \le f_{\max} \qquad \text{per ogni } x\in[a,b].$

Step 2 (integrating the inequalities). If we integrate functions ordered by an inequality, the integrals preserve the same inequality. Integrating from $a$ to $b$: $\int_a^b f_{\min}\,dx \le \int_a^b f(x)\,dx \le \int_a^b f_{\max}\,dx.$ But $f_{\min}$ and $f_{\max}$ are constants, so: $f_{\min}\cdot(b-a) \le \int_a^b f(x)\,dx \le f_{\max}\cdot(b-a).$

Step 3 (division). Dividing everything by $(b-a) > 0$: $f_{\min} \le \frac{1}{b-a}\int_a^b f(x)\,dx \le f_{\max},$ that is $f_{\min} \le \bar f_{ab} \le f_{\max}$.

Step 4 (intermediate values). Let $x_{\min}$ and $x_{\max}$ be the points where $f$ attains the minimum and the maximum: $f_{\min} = f(x_{\min})$, $f_{\max} = f(x_{\max})$. On the interval $[x_{\min}, x_{\max}]$ (or $[x_{\max}, x_{\min}]$), by the intermediate value theorem, $f$ takes all the values between $f(x_{\min})$ and $f(x_{\max})$. But $\bar f_{ab}$ lies between these two values: hence there exists $c$ such that $f(c) = \bar f_{ab}$. $\blacksquare$

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Topics: Calculus theorems
Concepts: Intermediate value theorem · Integral mean value theorem · Weierstrass’s theorem
Skills: Proving · Integrating
People: Karl Weierstrass