Bounding Double Integrals by Extreme Values

Why Bound a Double Integral?

A double integral can be evaluated by an iterated integral, but a calculation may involve several steps and can be vulnerable to an arithmetic or setup error. A useful independent check is to ask how large or small the answer could possibly be, based on the values of the integrand.

If a continuous function never falls below $m$ and never rises above $M$ on a rectangle, its double integral cannot be smaller than $m$ times the rectangle’s area or larger than $M$ times that area. The same principle applies when some or all of the function’s values are negative.

Minimum, Maximum, and Rectangle Area

Let $R=[a,b]\times[c,d]$, where $a<b$ and $c<d$. Its area is $(b-a)(d-c)$. If $f$ is continuous on $R$, the extreme value theorem guarantees that $f$ attains both a minimum and a maximum there. Write these values as $m$ and $M$, respectively.

Theorem: If $f$ is continuous on $R=[a,b]\times[c,d]$, and $m$ and $M$ are its minimum and maximum on $R$, then $$ m(b-a)(d-c) \leq \iint_R f(x,y)\,dA \leq M(b-a)(d-c). $$

The bounds use the area of the entire rectangle, not the area of a region where the function happens to be positive. The integral here is a signed double integral: contributions from negative function values count negatively.

Why the Bounds Hold

At every point of $R$, the definitions of the minimum and maximum give $m\leq f(x,y)\leq M$. The integral preserves this order: if one continuous function is everywhere no greater than another, its integral over the same rectangle is no greater. This follows, for example, from the nonnegativity of the integral of a continuous, nonnegative function.

Subtract the minimum.
Since $f(x,y)-m\geq 0$ throughout $R$, its double integral is nonnegative.

Translate that inequality.
By linearity and the fact that the integral of $1$ over $R$ is its area, $\iint_R f\,dA-m(b-a)(d-c)\geq 0$. This gives the lower bound.

Subtract the function from the maximum.
Since $M-f(x,y)\geq 0$, its integral is nonnegative. Thus $M(b-a)(d-c)-\iint_R f\,dA\geq 0$, giving the upper bound.

These steps prove both inequalities. They do not require $f$ itself to be nonnegative: the functions used in the argument are $f-m$ and $M-f$, which are nonnegative by construction.

Worked Examples

Worked Example: Bound and Check an Integral of a Nonnegative Function

Find the extreme-value bounds for $f(x,y)=x^2+y$ on $R=[0,2]\times[1,3]$, then evaluate the integral to check the bounds.

On this rectangle, $x^2$ ranges from $0$ to $4$, and $y$ ranges from $1$ to $3$. Both terms increase as their respective coordinates increase. Therefore the minimum is $m=0+1=1$, attained at $(0,1)$, and the maximum is $M=4+3=7$, attained at $(2,3)$.

The rectangle has area $(2-0)(3-1)=2\cdot2=4$. The theorem gives

1\cdot4 \leq \iint_R (x^2+y)\,dA \leq 7\cdot4, \qquad 4\leq \iint_R (x^2+y)\,dA\leq28.

Now evaluate using the iterated integral over the rectangle:

\begin{aligned} \iint_R (x^2+y)\,dA &=\int_1^3\int_0^2(x^2+y)\,dx\,dy\\ &=\int_1^3\left(\left[\frac{x^3}{3}\right]_0^2+[yx]_0^2\right)\,dy\\ &=\int_1^3\left(\frac{8}{3}+2y\right)\,dy\\ &=\left[\frac{8y}{3}+y^2\right]_1^3\\ &=\left(8+9\right)-\left(\frac{8}{3}+1\right) =\frac{51}{3}-\frac{11}{3} =\frac{40}{3}. \end{aligned}

Since $4=\frac{12}{3}\leq\frac{40}{3}\leq\frac{84}{3}=28$, the computed value passes the bound check.

Worked Example: Bound an Integral When the Function Changes Sign

Find the bounds for $f(x,y)=x-y$ on $R=[0,2]\times[1,3]$, and check them against the integral.

The function increases with $x$ and decreases with $y$. Its minimum occurs at the smallest $x$ and largest $y$: $m=0-3=-3$, attained at $(0,3)$. Its maximum occurs at the largest $x$ and smallest $y$: $M=2-1=1$, attained at $(2,1)$. The area is $2\cdot2=4$, so

(-3)\cdot4 \leq \iint_R(x-y)\,dA \leq 1\cdot4, \qquad -12\leq\iint_R(x-y)\,dA\leq4.

Evaluate the integral to verify the check:

\begin{aligned} \iint_R(x-y)\,dA &=\int_1^3\int_0^2(x-y)\,dx\,dy\\ &=\int_1^3\left(\left[\frac{x^2}{2}\right]_0^2-[yx]_0^2\right)\,dy\\ &=\int_1^3(2-2y)\,dy\\ &=[2y-y^2]_1^3\\ &=(6-9)-(2-1)=-3-1=-4. \end{aligned}

The result satisfies $-12\leq-4\leq4$. In particular, the integral is negative even though the upper bound is positive; that is entirely consistent with the theorem.

Worked Example: Use Bounds to Reject an Impossible Result

For $f(x,y)=3-2x+y$ on $R=[0,1]\times[0,2]$, determine the integral’s allowed range. Then decide whether a reported value of $12$ can be correct.

The function decreases as $x$ increases and increases as $y$ increases. Thus its minimum is $3-2(1)+0=1$, at $(1,0)$, and its maximum is $3-2(0)+2=5$, at $(0,2)$. The rectangle’s area is $(1-0)(2-0)=2$, so the integral must satisfy

1\cdot2 \leq \iint_R(3-2x+y)\,dA \leq 5\cdot2, \qquad 2\leq\iint_R(3-2x+y)\,dA\leq10.

Because $12>10$, the reported value is impossible. To check by direct evaluation, integrate first with respect to $x$:

\begin{aligned} \iint_R(3-2x+y)\,dA &=\int_0^2\int_0^1(3-2x+y)\,dx\,dy\\ &=\int_0^2[3x-x^2+yx]_0^1\,dy\\ &=\int_0^2(3-1+y)\,dy\\ &=\int_0^2(2+y)\,dy\\ &=[2y+y^2/2]_0^2\\ &=(4+2)-0=6. \end{aligned}

The value $6$ lies in the required interval $[2,10]$; the reported value $12$ does not. The bounds diagnose that something is wrong without needing to identify which step produced the error.

A Common Pitfall: Forgetting the Area Factor

The minimum and maximum are bounds on the function values, not usually direct bounds on the integral. For the first example, $1\leq f(x,y)\leq7$ throughout the rectangle. It would be incorrect to conclude that $1\leq\iint_R f\,dA\leq7$: those inequalities omit the area of $R$.

The correct calculation multiplies both extreme values by the area $4$, giving $4\leq\iint_R f\,dA\leq28$. This factor matters because a double integral accumulates function values across the whole rectangle. For a constant function $f(x,y)=k$, for instance, the integral is $k$ times the area, not merely $k$.

How to Use a Bound as a Check

To check a computed integral, first determine the actual minimum and maximum of the continuous integrand on the entire rectangle. Then multiply each by the rectangle’s area, preserving the order of the inequalities, and compare the computed value with the resulting interval.

If the computed value is outside the interval, it is wrong: revisit the integral setup or arithmetic.
If the value lies inside the interval, it is consistent with the bound, but that alone does not prove the calculation is correct.
A wide interval may provide only a weak check. More precise information about the integrand can sometimes produce tighter bounds, but the extreme-value bounds are a straightforward first test.

Finding the extrema is an essential part of the check. The minimum and maximum must be taken over the full rectangle, including its boundary. A value from just one point supplies neither extreme unless it has been shown to be the smallest or largest value on the whole domain.

Key takeaway: For a continuous function on a rectangle, the double integral lies between the function’s minimum times the rectangle’s area and its maximum times the rectangle’s area. These bounds handle signed functions as well as nonnegative ones, and they can rule out impossible computed answers.

Check Your Understanding

Use the extreme-value bounds and the rectangle’s area to answer the following questions.

If $f$ has minimum $-2$ and maximum $4$ on a rectangle of area $3$, what interval must contain its double integral?
Why does the extreme-value bound still apply when the function takes negative values?
On $R=[0,1]\times[0,2]$, the function $g(x,y)=x+y$ has minimum $0$ and maximum $3$. What bounds does the theorem give for its integral?
If a computed value falls outside the extreme-value bounds, what can you conclude? If it falls inside them, what can you conclude?
What common error results from treating the minimum and maximum function values as the integral bounds without multiplying by the area?