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Double Integrals on Rectangles · Tutorial 507 of 999

Basic Algebraic Properties of Double Integrals

Use rectangular Riemann sums to justify and apply the algebraic properties of double integrals over the same rectangle.

Beginner 9 min read

What You'll Learn

  • State the linearity, positivity, and comparison properties for integrable functions on one rectangle.
  • Explain how each property follows from the corresponding rectangular Riemann sums.
  • Compute exact Riemann sums for linear functions on the unit square.
  • Use pointwise inequalities to compare double integrals.
  • Recognize why multiplying an inequality by a negative number reverses its direction.

Why Algebraic Properties Matter

Double integrals are defined through limits of rectangular Riemann sums. The algebraic properties of those integrals are not separate rules imposed on the definition: they follow from the way the sums behave. In particular, sums preserve addition and scalar multiplication, and a sum of nonnegative terms cannot be negative.

These observations lead to useful conclusions. One can split an integrand into simpler terms, determine the sign of an integral from the sign of its integrand, and compare two integrals by comparing their functions point by point. The functions being combined or compared must be integrable on the same rectangle.

Rectangular Riemann Sums

Let \(R=[a,b]\times[c,d]\) be a rectangle. A partition divides each side into intervals, producing smaller rectangles \(R_{ij}\). Choose a sample point \((x_{ij}^*,y_{ij}^*)\) in each small rectangle, and let \(\Delta A_{ij}\) be the area of that small rectangle. The rectangular Riemann sum for a function \(f\) is

$$ S(f;P)=\sum_{i,j}f(x_{ij}^*,y_{ij}^*)\,\Delta A_{ij}. $$

Here \(P\) denotes the partition together with its sample points. The mesh of the partition is the largest diameter of any small rectangle. If \(f\) is integrable on \(R\), its double integral is the limit of these sums as the mesh tends to zero, independent of the choices of sample points:

$$ \iint_R f(x,y)\,dA = \lim_{\lVert P\rVert\to 0} S(f;P). $$

In this tutorial, assume the functions under discussion are Riemann integrable on \(R\). In particular, continuous functions on a closed rectangle are integrable. The limit definition lets us justify each algebraic property by first examining the finite sums.

Linearity

Theorem: If \(f\) and \(g\) are integrable on the same rectangle \(R\), and \(\alpha,\beta\) are real numbers, then \(\alpha f+\beta g\) is integrable on \(R\) and $$ \iint_R(\alpha f+\beta g)\,dA = \alpha\iint_R f\,dA+\beta\iint_R g\,dA. $$

For a fixed partition with fixed sample points, the sum for \(\alpha f+\beta g\) separates term by term:

$$ \begin{aligned} S(\alpha f+\beta g;P) &=\sum_{i,j}\bigl(\alpha f(x_{ij}^*,y_{ij}^*)+\beta g(x_{ij}^*,y_{ij}^*)\bigr)\Delta A_{ij}\\ &=\alpha S(f;P)+\beta S(g;P). \end{aligned} $$

As the mesh tends to zero, the sums for \(f\) and \(g\) tend to their respective integrals. Taking limits in the last equality gives the stated formula. This also explains why both functions must be considered on the same rectangle: the sums must use the same subrectangles and sample points for the term-by-term equality.

Worked Example: A Linear Combination on the Unit Square

Find the double integral of \(h(x,y)=2x-3y+4\) over \(R=[0,1]\times[0,1]\), using rectangular Riemann sums.

Divide each side into \(n\) equal intervals. The sample coordinates are \(x_i^*=(i-\tfrac12)/n\) and \(y_j^*=(j-\tfrac12)/n\), for \(1\leq i,j\leq n\). Every small rectangle has area \(1/n^2\).

For the function \(x\), the sum has \(n\) identical contributions in the \(y\)-direction:

$$ \begin{aligned} S_n(x) &=\frac{1}{n^2}\,n\sum_{i=1}^{n}\frac{i-\tfrac12}{n}\\ &=\frac{1}{n^2}\left(\frac{n(n+1)}{2}-\frac{n}{2}\right)\\ &=\frac{1}{n^2}\left(\frac{n^2}{2}\right) =\frac12. \end{aligned} $$

For \(y\), the sum is the same, with the repeated contributions now in the \(x\)-direction:

$$ \begin{aligned} S_n(y) &=\frac{1}{n^2}\,n\sum_{j=1}^{n}\frac{j-\tfrac12}{n}\\ &=\frac{1}{n^2}\left(\frac{n(n+1)}{2}-\frac{n}{2}\right) =\frac12. \end{aligned} $$

The constant function \(4\) contributes \(4\) in every sum, since there are \(n^2\) small rectangles, each of area \(1/n^2\). By linearity of finite sums,

$$ S_n(h)=2S_n(x)-3S_n(y)+4 =2\left(\frac12\right)-3\left(\frac12\right)+4 =\frac72. $$

This value is independent of \(n\), so its limit is also \(\frac72\). The result is checked by substituting the component values into the linearity formula: \(2(\frac12)-3(\frac12)+4=1-\frac32+4=\frac72\).

Positivity

Theorem: If \(f\) is integrable on \(R\) and \(f(x,y)\geq 0\) at every point of \(R\), then $$ \iint_R f(x,y)\,dA\geq 0. $$

For every partition and every choice of sample points, each term \(f(x_{ij}^*,y_{ij}^*)\Delta A_{ij}\) is nonnegative: the function value is nonnegative, and an area is positive. Thus every Riemann sum is nonnegative. Its limit cannot be negative, which proves the result.

Positivity means the integral is nonnegative, not necessarily strictly positive. For example, the zero function has integral zero. Even when a function is nonnegative and not identically zero, the positivity property alone only concludes that the integral is at least zero.

Worked Example: Checking Positivity with Riemann Sums

Evaluate the integral of \(p(x,y)=1+x+y\) on \(R=[0,1]\times[0,1]\), and check that its sign agrees with positivity.

Every point in the square has \(x\geq0\) and \(y\geq0\), so \(p(x,y)=1+x+y\geq1>0\). Use the same \(n\)-by-\(n\) partition as before. The constant term contributes \(1\), and the sums for \(x\) and \(y\) each contribute \(\frac12\). Therefore, for every \(n\),

$$ S_n(p)=1+S_n(x)+S_n(y) =1+\frac12+\frac12=2. $$

Taking the limit gives \(\iint_R p(x,y)\,dA=2\), which is positive as predicted. As a check, the Riemann-sum calculation gives the same value: the two coordinate contributions total \(1\), and the constant contributes another \(1\).

Comparison

Theorem: If \(f\) and \(g\) are integrable on the same rectangle \(R\), and \(f(x,y)\leq g(x,y)\) at every point of \(R\), then $$ \iint_R f(x,y)\,dA\leq\iint_R g(x,y)\,dA. $$

For any fixed partition and sample points, the pointwise inequality gives \(f(x_{ij}^*,y_{ij}^*)\leq g(x_{ij}^*,y_{ij}^*)\). Multiplication by the positive area \(\Delta A_{ij}\) preserves the inequality, and adding the terms gives \(S(f;P)\leq S(g;P)\). Taking the limits of these sums preserves the inequality and proves the comparison property.

Equivalently, \(g-f\geq0\) on \(R\), so positivity gives \(\iint_R(g-f)\,dA\geq0\). Linearity rewrites this as \(\iint_R g\,dA-\iint_R f\,dA\geq0\). This derivation shows how comparison follows from positivity and linearity.

Worked Example: Comparing \(x^2\) and \(x\)

On \(R=[0,1]\times[0,1]\), the inequality \(0\leq x^2\leq x\) holds because \(0\leq x\leq1\). Calculate both integrals from rectangular sums and verify their order.

For \(x\), the sum is \(\frac12\), as computed above. For \(x^2\), again there are \(n\) repeated terms in the \(y\)-direction:

$$ \begin{aligned} S_n(x^2) &=\frac{1}{n^2}\,n\sum_{i=1}^{n}\left(\frac{i-\tfrac12}{n}\right)^2\\ &=\frac{1}{n^3}\sum_{i=1}^{n}(i-\tfrac12)^2. \end{aligned} $$

Expanding the square and using \(\sum_{i=1}^n i=n(n+1)/2\) and \(\sum_{i=1}^n i^2=n(n+1)(2n+1)/6\), the sum inside is

$$ \begin{aligned} \sum_{i=1}^{n}(i-\tfrac12)^2 &=\frac{n(n+1)(2n+1)}{6} -\frac{n(n+1)}{2}+\frac{n}{4}\\ &=\frac{4n^3-n}{12}. \end{aligned} $$

Consequently, \(S_n(x^2)=\frac13-\frac{1}{12n^2}\), which tends to \(\frac13\), while \(S_n(x)=\frac12\). Thus \(\iint_R x^2\,dA=\frac13\) and \(\iint_R x\,dA=\frac12\). The computed values satisfy \(\frac13\leq\frac12\), in agreement with the pointwise comparison. Their difference is \(\frac12-\frac13=\frac16\geq0\), which also checks the comparison through positivity of \(x-x^2\).

A Sign Pitfall

Multiplication by a negative number reverses an inequality. This matters when using positivity or comparison: multiplying a nonnegative function by a negative scalar does not preserve nonnegativity.

For a concrete check on the unit square, \(x\geq0\), but \(-x\leq0\). The Riemann-sum calculation above gives \(\iint_R x\,dA=\frac12\). By linearity, \(\iint_R(-x)\,dA=-\frac12\), not a nonnegative value. The negative answer is consistent with the fact that \(-x\) is nonpositive.

Common mistake: From \(f\leq g\), it is incorrect to conclude that \(-f\leq-g\). Multiplying by \(-1\) reverses the order: \(f\leq g\) implies \(-f\geq-g\). For instance, \(0\leq x\) on the unit square, but \(0\geq -x\), and the integral of \(-x\) is \(-\frac12\).

Using the Properties Together

Linearity, positivity, and comparison are often used in sequence. If \(f\leq g\), first observe that \(g-f\geq0\). Positivity gives a nonnegative integral of the difference, and linearity turns that statement into the comparison of the two original integrals. If an integrand is a sum or scalar multiple, linearity can simplify it before a sign or comparison argument is applied.

Each property depends on its hypotheses. The functions must be integrable on the same rectangle, and the inequalities in the positivity and comparison statements must hold throughout that rectangle. A numerical check at a few sample points does not establish a pointwise inequality. The justification comes from the inequality at every possible sample point, which makes every relevant Riemann-sum comparison valid.

Check Your Understanding

Use the definitions and properties in this tutorial to answer the following questions.

  1. For integrable \(f\) and \(g\) on the same rectangle, what does linearity say about the integral of \(3f-2g\)?
  2. Why is every rectangular Riemann sum for a nonnegative function nonnegative?
  3. If \(f(x,y)\leq g(x,y)\) throughout a rectangle, what inequality holds between their double integrals, and why?
  4. On the unit square, verify that the midpoint Riemann sum for \(x\) is \(\frac12\) for every positive integer \(n\).
  5. If \(f\leq g\), which way does the inequality point after multiplying both sides by \(-1\)?

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