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Double Integrals on Rectangles · Tutorial 514 of 999

Integrating Separable Products Over Rectangles

Factor a separable double integral into two single integrals, and check that the factors and bounds are matched correctly.

Beginner 9 min read

What You'll Learn

  • Recognize when a rectangular integrand is a product of an x-only factor and a y-only factor
  • State the continuity and rectangular-bound conditions for separating the integral
  • Derive the factorization from iterated integration
  • Evaluate separable products with polynomial, exponential, and trigonometric factors
  • Identify why a sum of separate-variable terms cannot automatically be factored as one product

Why Separable Products Are Useful

A double integral over a rectangle can sometimes be evaluated with less work than a general iterated integral. The key is to recognize when the integrand is a product of two factors, one depending only on \(x\) and the other depending only on \(y\). In that case, the two variables can be integrated separately, and the resulting single integrals are multiplied.

This is a specific rule for a specific form of integrand. It does not say that every double integral is a product of two single integrals. The rectangular domain and the separation of the factors both matter.

Central rule: If \(f\) and \(g\) are continuous on the intervals \([a,b]\) and \([c,d]\), respectively, then \[ \iint_{[a,b]\times[c,d]} f(x)g(y)\,dA = \left(\int_a^b f(x)\,dx\right) \left(\int_c^d g(y)\,dy\right). \]

What Counts as a Separable Product?

A function of two variables has the separable-product form considered here if it can be written as \(F(x,y)=f(x)g(y)\), where \(f\) depends only on \(x\) and \(g\) depends only on \(y\). For example, \(x^2e^y\) has this form, with \(f(x)=x^2\) and \(g(y)=e^y\). So does \((x+1)\cos y\).

The rectangle must also use independent bounds: \(a\le x\le b\) and \(c\le y\le d\), with the \(x\)-bounds not changing as \(y\) changes and the \(y\)-bounds not changing as \(x\) changes. This is what lets each factor be integrated over its own fixed interval.

Definition: A separable product on a rectangle is an integrand expressible as \(f(x)g(y)\), with \(f\) a function of \(x\) alone and \(g\) a function of \(y\) alone. In this tutorial, the factors are assumed continuous on their respective intervals, and the rectangle is \([a,b]\times[c,d]\).

The factor \(f(x)\) is integrated using the \(x\)-bounds, and \(g(y)\) is integrated using the \(y\)-bounds. Keeping those pairings straight is part of applying the rule correctly.

Why the Factorization Works

The justification follows from iterated integration. Since \(f\) and \(g\) are continuous on their intervals, their product \(f(x)g(y)\) is continuous on the rectangle. Fubini’s theorem therefore allows the double integral to be evaluated as an iterated integral. Integrate first with respect to \(x\), treating \(y\) as fixed.

1
Hold \(y\) fixed.
The value \(g(y)\) is constant while integrating with respect to \(x\), so it can be taken outside that single integral.
2
Evaluate the inner integral.
The result is \(g(y)\int_a^b f(x)\,dx\). The integral involving \(f\) is a constant with respect to \(y\).
3
Integrate with respect to \(y\).
Take the constant \(\int_a^b f(x)\,dx\) outside the outer integral. What remains is \(\int_c^d g(y)\,dy\).

In equations, this argument gives the factorization directly:

$$ \begin{aligned} \iint_{[a,b]\times[c,d]} f(x)g(y)\,dA &=\int_c^d\left(\int_a^b f(x)g(y)\,dx\right)dy\\ &=\int_c^d g(y)\left(\int_a^b f(x)\,dx\right)dy\\ &=\left(\int_a^b f(x)\,dx\right)\left(\int_c^d g(y)\,dy\right). \end{aligned} $$

Continuity supplies the conditions needed to apply Fubini’s theorem here. The algebraic separation then follows because each factor is constant with respect to the other variable’s integration.

Worked Example: A Polynomial Product

Worked Example: A Polynomial Product

Evaluate \(\iint_R (x^2+1)(y+2)\,dA\), where \(R=[0,2]\times[0,1]\).

The integrand is \(f(x)g(y)\), with \(f(x)=x^2+1\) and \(g(y)=y+2\). The separate integrals use their corresponding intervals:

$$ \begin{aligned} \int_0^2 (x^2+1)\,dx &=\left[\frac{x^3}{3}+x\right]_0^2\\ &=\left(\frac{8}{3}+2\right)-0\\ &=\frac{14}{3}, \end{aligned} \qquad \begin{aligned} \int_0^1 (y+2)\,dy &=\left[\frac{y^2}{2}+2y\right]_0^1\\ &=\left(\frac12+2\right)-0\\ &=\frac52. \end{aligned} $$

Multiplying the two values gives the double integral:

$$ \iint_R (x^2+1)(y+2)\,dA = \frac{14}{3}\cdot\frac52 = \frac{70}{6} = \frac{35}{3}. $$

Check: The antiderivatives differentiate correctly: \(\frac{d}{dx}(\frac{x^3}{3}+x)=x^2+1\), and \(\frac{d}{dy}(\frac{y^2}{2}+2y)=y+2\). Evaluating in the \(x\)-then-\(y\) order gives the same result:

$$ \begin{aligned} \int_0^1\int_0^2 (x^2+1)(y+2)\,dx\,dy &=\int_0^1 (y+2)\left(\frac{14}{3}\right)dy\\ &=\frac{14}{3}\cdot\frac52\\ &=\frac{35}{3}. \end{aligned} $$

Worked Example: An Exponential Product

Worked Example: An Exponential Product

Evaluate \(\iint_R e^{2x}e^{-y}\,dA\), where \(R=[0,\ln 2]\times[0,\ln 2]\).

Here \(f(x)=e^{2x}\) and \(g(y)=e^{-y}\). Evaluate each factor over its own interval:

$$ \begin{aligned} \int_0^{\ln 2} e^{2x}\,dx &=\left[\frac12e^{2x}\right]_0^{\ln 2}\\ &=\frac12\left(e^{2\ln 2}-e^0\right)\\ &=\frac12(4-1)\\ &=\frac32, \end{aligned} \qquad \begin{aligned} \int_0^{\ln 2} e^{-y}\,dy &=\left[-e^{-y}\right]_0^{\ln 2}\\ &=-e^{-\ln 2}-(-e^0)\\ &=-\frac12+1\\ &=\frac12. \end{aligned} $$

Thus the double integral is

$$ \iint_R e^{2x}e^{-y}\,dA = \frac32\cdot\frac12 = \frac34. $$

Check: Differentiation verifies the antiderivatives: \(\frac{d}{dx}(\frac12e^{2x})=e^{2x}\), and \(\frac{d}{dy}(-e^{-y})=e^{-y}\). Integrating first with respect to \(x\) gives \(\frac32e^{-y}\); integrating that from \(0\) to \(\ln 2\) gives \(\frac32\cdot\frac12=\frac34\), as required.

Worked Example: A Trigonometric Product

Worked Example: A Trigonometric Product

Evaluate \(\iint_R \cos x\sin y\,dA\), where \(R=[0,\frac{\pi}{2}]\times[0,\pi]\).

Use \(f(x)=\cos x\) and \(g(y)=\sin y\). The \(x\)-integral is

$$ \int_0^{\pi/2}\cos x\,dx = [\sin x]_0^{\pi/2} = 1-0 = 1. $$

The \(y\)-integral is

$$ \int_0^\pi\sin y\,dy = [-\cos y]_0^\pi = -\cos\pi-(-\cos 0) = 1-(-1) = 2. $$

Therefore,

$$ \iint_R \cos x\sin y\,dA = 1\cdot 2 = 2. $$

Check: The antiderivatives are correct because \(\frac{d}{dx}\sin x=\cos x\) and \(\frac{d}{dy}(-\cos y)=\sin y\). In the iterated integral, the inner integral of \(\cos x\) is \(1\), leaving \(\int_0^\pi\sin y\,dy=2\). This agrees with the factored answer.

Conditions and Common Pitfalls

The rule applies when the integrand is a product \(f(x)g(y)\) on a rectangle and the factors meet the stated continuity conditions. It is not a rule for splitting every expression involving both variables. In particular, the sum \(x+y\) is not itself one product of an \(x\)-only factor and a \(y\)-only factor.

For a concrete example, on \([0,1]\times[0,1]\), incorrectly treating \(x+y\) as a single separable product might lead to the calculation \((\int_0^1 x\,dx)(\int_0^1 y\,dy)=\frac12\cdot\frac12=\frac14\). That calculation does not represent the integral of \(x+y\). Direct iterated integration gives

$$ \int_0^1\int_0^1(x+y)\,dy\,dx = \int_0^1\left(x+\frac12\right)dx = \frac12+\frac12 = 1. $$

The correction is to check the algebraic form before applying the product rule: \(x+y\) is a sum, not a single product \(f(x)g(y)\). Also check that each single integral uses the interval for its own variable. The \(x\)-factor must use the \(x\)-bounds, and the \(y\)-factor must use the \(y\)-bounds.

When to Use the Method

For an integral over a rectangle, first inspect the integrand for a product of an \(x\)-only expression and a \(y\)-only expression. If it has that form and the factors are continuous on their intervals, compute the two one-variable integrals and multiply. This can avoid carrying both variables through an antiderivative.

The result also provides a useful check on an iterated-integral calculation: evaluating in either order must agree with the product of the separate integrals. If the integrand does not separate as a product, this particular shortcut is not justified; evaluate the iterated integral using an appropriate method instead.

Takeaway: On a rectangle, a continuous integrand \(f(x)g(y)\) has a double integral equal to the product of the \(x\)-integral of \(f\) and the \(y\)-integral of \(g\). Match each factor to its own bounds, and verify that the integrand really has the required product form.

Check Your Understanding

Use the factorization rule where its conditions apply, and identify when they do not.

  1. For \(R=[0,1]\times[0,2]\), identify the two factors and the two intervals for \((x^2)(y+1)\).
  2. What continuity and domain conditions allow the product factorization rule to be used?
  3. Evaluate \(\iint_{[0,1]\times[0,1]} e^x\cos y\,dA\) by separating the factors.
  4. Why does the rule not directly turn the integral of \(x+y\) into a product of two single integrals?
  5. In what order should the bounds be matched to the factors in \(f(x)g(y)\)?

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