The Mean Value Theorem for Double Integrals

Why an Average Value Is More Than a Number

A double integral combines the values of a function across an entire region. Dividing that integral by the region’s area gives a single number: the function’s average value. But is that number merely an aggregate, or does the function actually equal it somewhere?

For a continuous function on a rectangle, the answer is yes. This is the mean value theorem for double integrals. It connects an integral over a whole region to the value of the function at at least one point in that region. The theorem guarantees that such a point exists; it does not generally identify a unique point or say that the point is the rectangle’s center.

Average Value on a Rectangle

Let \(R=[a,b]\times[c,d]\) be a rectangle with positive area, so \(a<b\) and \(c<d\). Its area is \(A=(b-a)(d-c)\). If \(f\) is integrable on \(R\), its average value on \(R\) is the double integral of \(f\) divided by the area.

The double integral adds the function’s contributions over the region; division by the area turns that total into a per-unit-area average. For instance, if \(f\) represents temperature over a flat plate, the average has temperature units, whereas the integral has temperature multiplied by area units.

The Mean Value Theorem for Double Integrals

The conclusion says that the average value is an actual value of the function somewhere on the rectangle. The point \((x_0,y_0)\) is sometimes called a mean-value point. The theorem guarantees at least one such point, but may not specify where it is or how many such points there are.

Why the Theorem Holds

The argument uses two facts. A continuous function on a closed, bounded rectangle has a minimum and a maximum, and the integral average lies between those extreme values. Then, because a rectangle contains the line segment joining any two of its points, continuity and the intermediate value theorem ensure the function takes every value between its minimum and maximum along that segment.

Here is the argument in full. Let \(m=\min_R f\) and \(M=\max_R f\), which exist because \(f\) is continuous and \(R\) is compact. For every \((x,y)\in R\), the inequalities \(m\leq f(x,y)\leq M\) hold. Integrating these inequalities over \(R\) and dividing by \(A>0\) yields

Write \(P=(p_1,p_2)\) for a point where \(f(P)=m\), and \(Q=(q_1,q_2)\) for a point where \(f(Q)=M\). The rectangle is convex, so \((1-t)P+tQ\in R\) for every \(t\in[0,1]\). Define \(g(t)=f((1-t)P+tQ)\). This is a continuous function on \([0,1]\), with \(g(0)=m\) and \(g(1)=M\). Since \(f_{\mathrm{avg}}\) lies between \(m\) and \(M\), the intermediate value theorem gives some \(t_0\in[0,1]\) for which \(g(t_0)=f_{\mathrm{avg}}\). The point \((1-t_0)P+t_0Q\) belongs to \(R\) and has function value \(f_{\mathrm{avg}}\), as required. If \(m=M\), then \(f\) is constant on \(R\), and every point gives the average value.

Worked Examples

What the Theorem Does—and Does Not—Guarantee

The result is useful whenever an integral-based average needs to be interpreted as a representative value actually present somewhere in the region. For a continuous temperature field, for example, the average temperature is not merely a number obtained by calculation: it is the temperature at some point on the plate. The theorem makes an existence claim, not a method for locating that point. In examples, a convenient point may be found by solving \(f(x,y)=f_{\mathrm{avg}}\), but that equation can have many solutions or be difficult to solve.

Continuity is an important hypothesis. Without it, an integrable function need not attain its average. For example, on \([0,1]\times[0,1]\), let \(f(x,y)=0\) when \(x\leq1/2\) and \(f(x,y)=1\) when \(x>1/2\). Its integral, and therefore its average on this unit-area rectangle, is \(1/2\). But the function takes only the values \(0\) and \(1\), so it never equals its average. This example is discontinuous along \(x=1/2\).

A Common Mistake: Assuming the Center Gives the Average

The theorem does not say that the average value occurs at the center of the rectangle. Consider \(f(x,y)=x^2\) on \(R=[0,2]\times[0,1]\). The rectangle’s center is \((1,1/2)\), where \(f(1,1/2)=1\). But the average is

A wrong conclusion would be: “The center has value \(1\), so the average is \(1\).” The error is treating the center value as if it determined the average. In fact, \(1\ne4/3\). The theorem guarantees some point with value \(4/3\); here, \((\sqrt{4/3},1/2)\) is such a point, since \(0\leq\sqrt{4/3}\leq2\) and \(f(\sqrt{4/3},1/2)=(\sqrt{4/3})^2=4/3\).