The Mean Value Theorem for Double Integrals

# The Mean Value Theorem for Double Integrals

Recall The Mean Value Theorem in single variable calculus says that if $f$ is a continuous single variable function on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, then there exists a point $c \in (a, b)$ such that:

(1)
\begin{align} \quad f'(c)(b - a) = f(b) - f(a) \end{align}

We will now extend The Mean Value Theorem to double integrals.

 Theorem 1 (The Mean Value Theorem for Double Integrals): Let $z = f(x, y)$ be a continuous function on the closed, bounded, and connected subset $D \subseteq D(f)$, and let $A$ be the positive area of $D$. Then there exists a point $(x_0, y_0) \in D$ such that $\iint_D f(x, y) \: dA = A f(x_0, y_0)$.
• Proof: Let $z = f(x, y)$ be a continuous two variable real-valued function, and let $D \subseteq D(f)$ be a closed, bounded, and connected set that has a positive area $A$. Since $f$ is closed and bounded over $D$, it follows by The Extreme Value Theorem for Functions of Several Variables that there exists points $(x_1, y_1), (x_2, y_2) \in D$ such that $f(x_1, y_1)$ is an absolute minimum over $D$ and $f(x_2, y_2)$ is an absolute maximum over $D$, and so for all $(x, y) \in D$:
(2)
\begin{align} \quad f(x_1, y_1) ≤ f(x, y) ≤ f(x_2, y_2) \end{align}
• Now we will integrate both sides of this inequality over $D$, and divide by $A$ ($A \neq 0$ since we described $A$ to be a positive area) and we have that:
(3)
\begin{align} \quad \iint_D f(x_1, y_1) \: dA ≤ \iint_D f(x, y) \: dA ≤ \iint_D f(x_2, y_2) \: dA \\ \quad A f(x_1, y_1) ≤ \iint_D f(x, y) \: dA ≤ A f(x_2, y_2) \\ \quad f(x_1, y_1) ≤ \frac{1}{A} \iint_D f(x, y) \: dA ≤ f(x_2, y_2) \end{align}
• Now let $x = x(t)$ and $y = y(t)$ for $0 ≤ t ≤ 1$ be continuous parametric equations that are contained in $D$, and suppose that for $t = 0$ we have $(x(0), y(0)) = (x_1, y_1)$ and for $t = 1$ we have $(x(1), y(1)) = (x_2, y_2)$. Then the parametric curves $x(t)$, and $y(t)$ joins the point $(x_1, y_1)$ to $(x_2, y_2)$, which is possible since $D$ is connected. Now suppose that $g(t) = f(x(t), y(t))$ for $0 ≤ t ≤ 1$. Then $g$ traces this parametric curve and is continuous, and $g(0) = f(x(0), y(0)) = f(x_1, y_1)$ and $g(1) = f(x(1), y(1)) = f(x_2, y_2)$. Therefore, by the Intermediate Value Theorem there exists a $t_0$ such that $0 ≤ t_0 ≤ 1$ for which $x(t_0) = x_0$ and $y(t_0) = y_0$ where:
(4)
\begin{align} \quad \frac{1}{A} \iint_D f(x, y) \: dA = g(t_0) = f(x(t_0), y(t_0)) = f(x_0, y_0) \\ \quad \iint_D f(x, y) \: dA = A f(x_0, y_0) \quad \blacksquare \end{align}