📂Analysis

Mean Value Theorem for Integrals

Theorem

If a function $f$ is continuous on a closed interval $[a,b]$, there exists at least one $c$ in $(a,b)$ that satisfies $\displaystyle f(c) = {{1}\over {b-a} } \int_{a}^{b} f(x) dx$.

Description

Similar to the Mean Value Theorem but as it is used for integration, it is named as such. The usage is very similar, and its utility is by no means inferior to the Mean Value Theorem. On the other hand, considering defining the average value of a function as on the right side makes this theorem more likely to be the mean value theorem, and the widely known Mean Value Theorem might be more appropriately called the ‘mean value theorem for derivatives’.

Proof

Strategy: The continuity of $f$ is assumed, thus we use both the Extreme Value Theorem and the Intermediate Value Theorem.

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Since $f$ is continuous on $[a,b]$, and by the Extreme Value Theorem, the minimum value $m$ and maximum value $M$ exist, then

$$\int_{a}^{b} m dx \le \int_{a}^{b} f(x) dx \le \int_{a}^{b} M dx$$

$$\implies m \le {{1}\over {b-a} } \int_{a}^{b} f(x) dx \le M$$

Once again, as $f$ is continuous on $[a,b]$, by the Intermediate Value Theorem, for $m$ and $M$, there exists at least one $c$ between $a$ and $b$ that satisfies $f(c) = \displaystyle {{1}\over {b-a} } \int_{a}^{b} f(x) dx$ for $\displaystyle {{1}\over {b-a} } \int_{a}^{b} f(x) dx$.

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Likewise, the same method can generalize to weighted $w$. The form introduced above is for $w(x) = 1$, and it is well covered by the theorem below as $\displaystyle \int_{a}^{b} dx = b - a$.

Corollary

If a function $f$ is continuous on a closed interval $[a,b]$ and $w(x) \ge 0$ is integrable, then there exists at least one $\xi$ in $(a,b)$ that satisfies $\displaystyle \int_{a}^{b} f(x) w(x) dx = f( \xi ) \int_{a}^{b} w(x) dx$.

See Also

• Mean Value Theorem
• Cauchy’s Mean Value Theorem
• Gauss’s Mean Value Theorem