Polynomial Integrands on Rectangles

From One-Variable Power Rules to Rectangular Integrals

A polynomial in two variables combines familiar powers of $x$ and $y$. To integrate one over a rectangle, integrate with respect to one variable while treating the other as constant, then integrate the resulting expression in the remaining variable. The one-variable power rule handles each step.

For a polynomial, this method is especially direct: each term is a constant times a power of $x$ times a power of $y$. Integrate each term, keep the limits attached to the variable being integrated, and add the results. The method applies to rectangles with any finite endpoints, including negative endpoints, and the final value is a signed double integral.

Definition: A polynomial in $x$ and $y$ is a finite sum of terms of the form $c_{ij}x^iy^j$, where each coefficient $c_{ij}$ is a real number and each exponent $i,j$ is a nonnegative integer. A rectangle is $R=[a,b]\times[c,d]$, with $a\le b$ and $c\le d$.

A polynomial is continuous on every rectangle, so Fubini’s theorem for continuous functions allows its double integral to be evaluated as an iterated integral in either order. For example, integrating with respect to $x$ first gives

\iint_R f(x,y)\,dA = \int_c^d\left(\int_a^b f(x,y)\,dx\right)dy.

The inner integral uses $x$-limits $a$ and $b$; $y$ is held constant there. The outer integral then uses $y$-limits $c$ and $d$. The other order is also valid, with the variables and their corresponding limits exchanged.

The Formula for a Polynomial Term

Consider one term, $c_{ij}x^iy^j$. In the inner integral with respect to $x$, both the coefficient and the factor $y^j$ are constant. Apply the power rule to $x^i$, evaluate at $a$ and $b$, and then integrate the remaining power of $y$:

\begin{aligned} \int_c^d\left(\int_a^b c_{ij}x^iy^j\,dx\right)dy &=\int_c^d c_{ij}y^j \left[\frac{x^{i+1}}{i+1}\right]_a^b\,dy\\ &=\int_c^d c_{ij}\frac{b^{i+1}-a^{i+1}}{i+1}y^j\,dy\\ &=c_{ij} \frac{b^{i+1}-a^{i+1}}{i+1} \frac{d^{j+1}-c^{j+1}}{j+1}. \end{aligned}

The denominators $i+1$ and $j+1$ are nonzero because the exponents are nonnegative integers. This calculation also explains why the answer is unchanged if the integration order is reversed: integrating $y^j$ first contributes the factor $(d^{j+1}-c^{j+1})/(j+1)$, and integrating $x^i$ then contributes $(b^{i+1}-a^{i+1})/(i+1)$. The same two factors are multiplied together.

Theorem: If $f(x,y)=\sum_{(i,j)\in S}c_{ij}x^iy^j$ is a polynomial, where $S$ is a finite set of pairs of nonnegative integers, then $$ \iint_{[a,b]\times[c,d]} f(x,y)\,dA = \sum_{(i,j)\in S} c_{ij} \frac{b^{i+1}-a^{i+1}}{i+1} \frac{d^{j+1}-c^{j+1}}{j+1}. $$

To justify the sum, use linearity of the integral: the integral of a finite sum is the sum of the integrals of its terms. The calculation above evaluates each term, giving the stated formula. This term-by-term rule is not limited to products with just one $x$-power and one $y$-power; every polynomial is a finite sum of such products.

Worked Examples

Worked Example: A Single Product of Powers

Evaluate $\iint_R x^2y^3\,dA$ on $R=[0,2]\times[1,2]$, integrating with respect to $x$ first.

The inner integral treats $y^3$ as constant. Apply the power rule in $x$, substitute both $x$-limits, and then integrate in $y$:

\begin{aligned} \int_1^2\left(\int_0^2 x^2y^3\,dx\right)dy &=\int_1^2 y^3\left[\frac{x^3}{3}\right]_0^2\,dy\\ &=\int_1^2 y^3\left(\frac{8}{3}-0\right)dy\\ &=\frac{8}{3}\left[\frac{y^4}{4}\right]_1^2\\ &=\frac{8}{3}\left(\frac{16}{4}-\frac{1}{4}\right)\\ &=\frac{8}{3}\cdot\frac{15}{4}=10. \end{aligned}

As a check, reverse the order. The $y$-integral is $[y^4/4]_1^2=15/4$, and the $x$-integral is $[x^3/3]_0^2=8/3$. Their product is $(15/4)(8/3)=10$, as before.

Worked Example: A Sum of Polynomial Terms

Evaluate $f(x,y)=2x^2+3xy+y$ on $R=[0,1]\times[0,2]$, integrating with respect to $x$ first. Since the integral is linear, the terms can be integrated together or separately.

For the inner integral, $y$ is constant. Integrating each term in $x$ and substituting $0$ and $1$ gives:

\begin{aligned} \int_0^1(2x^2+3xy+y)\,dx &=\frac{2}{3}+ \frac{3y}{2}+y\\ &=\frac{2}{3}+\frac{5y}{2}. \end{aligned}

Now integrate this result from $y=0$ to $y=2$:

\begin{aligned} \int_0^2\left(\frac{2}{3}+\frac{5y}{2}\right)dy &=\left[\frac{2y}{3}+\frac{5y^2}{4}\right]_0^2\\ &=\left(\frac{4}{3}+5\right)-0\\ &=\frac{19}{3}. \end{aligned}

Check by integrating the original terms separately. The term $2x^2$ contributes $2(1/3)(2)=4/3$. The term $3xy$ contributes $3(1/2)(2)=3$. The term $y$ contributes $1\cdot 2=2$. Adding gives $4/3+3+2=19/3$, confirming the result.

Worked Example: Nonzero Lower Limits and a Negative Coefficient

Evaluate $f(x,y)=x^2y-2xy^2+4y$ on $R=[1,2]\times[0,1]$, integrating with respect to $y$ first. Here $x$ is held constant in the inner integral.

\begin{aligned} \int_1^2\left(\int_0^1(x^2y-2xy^2+4y)\,dy\right)dx &=\int_1^2\left(\left[\frac{x^2y^2}{2} -\frac{2xy^3}{3}+2y^2\right]_0^1\right)dx\\ &=\int_1^2\left(\frac{x^2}{2}-\frac{2x}{3}+2\right)dx\\ &=\left[\frac{x^3}{6}-\frac{x^2}{3}+2x\right]_1^2\\ &=\left(\frac{8}{6}-\frac{4}{3}+4\right) -\left(\frac{1}{6}-\frac{1}{3}+2\right)\\ &=\frac{4}{1}-\frac{11}{6}\\ &=\frac{13}{6}. \end{aligned}

Check term by term using the power-rule factors. The first term contributes $(7/3)(1/2)=7/6$, since $\int_1^2x^2\,dx=7/3$ and $\int_0^1y\,dy=1/2$. The second contributes $-2(3/2)(1/3)=-1$. The third contributes $4(1)(1/2)=2$. Their sum is $7/6-1+2=13/6$, matching the iterated calculation.

A Reliable Calculation Process

For a polynomial on a rectangle, a consistent sequence of steps reduces the chance of mixing up variables, powers, or limits.

Identify the rectangle and the order.
Match each pair of limits to its variable. Decide which variable is integrated in the inner integral.

Keep the other coordinate constant.
In an inner $x$-integral, factors involving only $y$ act as constants, and conversely.

Apply the power rule and evaluate the limits.
For a power $u^n$, use the antiderivative $u^{n+1}/(n+1)$, then substitute the upper and lower limits in that order.

Integrate the remaining expression.
After evaluating the inner integral, integrate the resulting polynomial in the outer variable.

Check the value.
Recompute term by term with the power-rule formula, or reverse the integration order and compare.

A Common Variable Error

A frequent mistake is to apply the power rule to the coordinate that is being held constant. For example, in the inner integral below, $x^2$ is constant because the integration is with respect to $y$:

\int x^2y^3\,dy.

The incorrect calculation $x^2y^3\mapsto x^3y^4/12$ changes the $x$-power even though $x$ is not the integration variable. It also gives the wrong coefficient. The correct calculation is:

\int x^2y^3\,dy = x^2\frac{y^4}{4} = \frac{x^2y^4}{4}+C,

where $C$ is constant with respect to $y$ and may depend on $x$. Differentiating with respect to $y$ confirms the antiderivative: $\frac{\partial}{\partial y}(x^2y^4/4)=x^2y^3$. In a definite inner integral, substitute only the $y$-limits at this stage; the outer $x$-integration comes afterward.

Why the Method Is Useful

The power-rule formula turns a two-variable polynomial integral into a finite collection of ordinary one-variable calculations. It also makes the role of the rectangle’s side limits explicit: each monomial contributes one factor from the $x$-interval and one from the $y$-interval. That structure is useful both for efficient evaluation and for checking a result.

The exact same rules apply when a polynomial has many terms. A negative coefficient remains negative throughout the calculation, and nonzero lower limits must be subtracted rather than discarded. No special geometric interpretation is needed to evaluate the integral: the repeated-integral calculation and the specified limits determine the value.

Check Your Understanding

For each question, keep track of the integration variable and evaluate the limits carefully.

What is the contribution of $5x^3y^2$ to an integral over $[a,b]\times[c,d]$, where $a\le b$ and $c\le d$?
In an inner integral with respect to $x$, which factors in $7x^2y^4$ are treated as constants?
Evaluate $\iint_{[0,1]\times[0,1]}(x+y)\,dA$.
Why can the two integration orders be used for a polynomial on a rectangle?
What is the correct antiderivative of $3xy^2$ with respect to $y$, treating $x$ as constant?