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Choosing Integration Order From Antiderivatives

Compare both orders of integration to choose bounds and inner antiderivatives that make a double integral easier to evaluate.

Intermediate 10 min read

What You'll Learn

  • Compare the inner antiderivatives produced by integrating with respect to different variables.
  • Use the region's slice bounds to assess the work required in each integration order.
  • Recognize when one order avoids an unavailable elementary antiderivative.
  • Evaluate the same double integral in two orders and check that the results agree.
  • Avoid treating a variable-dependent integrand as constant during inner integration.

Why the Integration Order Matters

A double integral over a region can often be written in either order: integrate with respect to \(y\) first, then \(x\), or integrate with respect to \(x\) first, then \(y\). If both iterated integrals describe the same region and the integrand satisfies the relevant conditions, they have the same value. But they need not be equally convenient to evaluate.

The choice affects two things at once: the limits imposed by the region's slices, and the antiderivative required at the inner step. A simple-looking pair of bounds may lead to a difficult inner integral. Conversely, an integrand that is easy to integrate in one variable may be paired with awkward bounds in that order.

Guiding principle: Compare the inner antiderivatives and the slice bounds in both orders. Prefer an order that makes the inner integration straightforward and leaves a manageable outer integral.

What to Compare

Suppose \(D\) is a bounded planar region and \(f\) is continuous on \(D\). If the region can be described by slices, its double integral can be expressed as an iterated integral. For example, a Type I description has the form \(a\leq x\leq b\), \(g_1(x)\leq y\leq g_2(x)\), while a Type II description has the form \(c\leq y\leq d\), \(h_1(y)\leq x\leq h_2(y)\).

Before selecting an order, check the following:

  • Inner antiderivative: Can you find an elementary antiderivative in the proposed inner variable?
  • Dependence on the other variable: Does integrating first produce an outer integrand that is manageable?
  • Bounds: Are the slice endpoints simple, or do they require complicated expressions or several pieces?
  • Verification: Can you differentiate the proposed antiderivative to check it?

An antiderivative with respect to \(y\) treats \(x\) as constant, and vice versa. That fact can make one order much simpler than the other. The region still matters: reversing the order can change the slice endpoints even though it does not change the region itself.

Order-of-integration principle: If \(f\) is continuous on a bounded region \(D\), and each iterated integral describes \(D\) with valid slice bounds, Fubini's theorem allows the two orders to be used to compute the same double integral. The bounds must describe the same region; changing the order does not permit changing the region.

Worked Example: Choose the Order That Avoids a Difficult Antiderivative

Worked Example: Choose the Order That Avoids a Difficult Antiderivative

Evaluate \(\iint_D e^{x^2}\,dA\) over the triangle \(D=\{(x,y):0\leq y\leq x\leq 1\}\). Compare the two possible orders.

First use vertical slices: for each \(x\) between \(0\) and \(1\), \(y\) ranges from \(0\) to \(x\). Thus

$$ \iint_D e^{x^2}\,dA = \int_0^1\int_0^x e^{x^2}\,dy\,dx. $$

For the inner integral, \(x\) is constant, so \(e^{x^2}\) is constant with respect to \(y\):

$$ \int_0^x e^{x^2}\,dy = \left[y e^{x^2}\right]_0^x = x e^{x^2}. $$

Therefore the remaining integral is

$$ \int_0^1 x e^{x^2}\,dx = \left[\frac{1}{2}e^{x^2}\right]_0^1 = \frac{e-1}{2}. $$

The antiderivative is checked by differentiation: \(\frac{d}{dx}\bigl(\frac12 e^{x^2}\bigr)=x e^{x^2}\), which is the outer integrand. The bounds give \(\frac12e-\frac12=\frac{e-1}{2}\).

Now reverse the order. The same triangle has \(0\leq y\leq 1\) and \(y\leq x\leq 1\), so the integral would be

$$ \int_0^1\int_y^1 e^{x^2}\,dx\,dy. $$

Its inner integral requires an antiderivative of \(e^{x^2}\) with respect to \(x\), which has no elementary expression. The first order is therefore preferable: its inner integral is immediate, and the remaining integral has an elementary antiderivative.

Worked Example: Compare Inverse-Trigonometric Antiderivatives

Worked Example: Compare Inverse-Trigonometric Antiderivatives

Evaluate \(\iint_D \frac{1}{1+x^2}\,dA\) over the same triangle \(D=\{(x,y):0\leq y\leq x\leq 1\}\). Here the integrand depends on \(x\) but not on \(y\).

With \(y\) as the inner variable, the bounds are \(0\leq y\leq x\):

$$ \begin{aligned} \iint_D \frac{1}{1+x^2}\,dA &= \int_0^1\int_0^x \frac{1}{1+x^2}\,dy\,dx\\ &= \int_0^1 \frac{x}{1+x^2}\,dx\\ &= \left[\frac12\ln(1+x^2)\right]_0^1\\ &= \frac12\ln 2. \end{aligned} $$

The first inner step uses the fact that \(1/(1+x^2)\) is constant with respect to \(y\). The outer antiderivative checks by differentiation: \(\frac{d}{dx}\bigl(\frac12\ln(1+x^2)\bigr)=x/(1+x^2)\).

In the reversed order, \(0\leq y\leq 1\) and \(y\leq x\leq 1\). This gives

$$ \begin{aligned} \iint_D \frac{1}{1+x^2}\,dA &= \int_0^1\int_y^1 \frac{1}{1+x^2}\,dx\,dy\\ &= \int_0^1\left(\frac{\pi}{4}-\arctan y\right)\,dy. \end{aligned} $$

To finish, use the antiderivative \(y\arctan y-\frac12\ln(1+y^2)\). Differentiating it gives \(\arctan y+y/(1+y^2)-y/(1+y^2)=\arctan y\). Hence

$$ \begin{aligned} \int_0^1\left(\frac{\pi}{4}-\arctan y\right)\,dy &= \frac{\pi}{4} - \left[y\arctan y-\frac12\ln(1+y^2)\right]_0^1\\ &= \frac{\pi}{4}-\left(\frac{\pi}{4}-\frac12\ln 2\right)\\ &= \frac12\ln 2. \end{aligned} $$

Both orders work, and they agree. In this example, integrating first with respect to \(y\) gives a shorter route: the inner integration simply multiplies by the slice length \(x\), leaving a single logarithmic antiderivative.

Worked Example: Let the Bounds and the Integrand Work Together

Worked Example: Let the Bounds and the Integrand Work Together

Evaluate \(\iint_D x^2\,dA\) over the region \(D=\{(x,y):0\leq x\leq 1,\ x^2\leq y\leq 1\}\). Compare vertical and horizontal slices.

For vertical slices, \(x\) ranges from \(0\) to \(1\), and \(y\) ranges from \(x^2\) to \(1\). Since \(x^2\) is constant with respect to \(y\),

$$ \begin{aligned} \iint_D x^2\,dA &= \int_0^1\int_{x^2}^1 x^2\,dy\,dx\\ &= \int_0^1 x^2(1-x^2)\,dx\\ &= \int_0^1(x^2-x^4)\,dx\\ &= \left[\frac{x^3}{3}-\frac{x^5}{5}\right]_0^1\\ &= \frac13-\frac15 = \frac{2}{15}. \end{aligned} $$

For horizontal slices, \(y\) ranges from \(0\) to \(1\). The condition \(x^2\leq y\), together with \(x\geq0\), gives \(0\leq x\leq\sqrt y\). Thus

$$ \begin{aligned} \iint_D x^2\,dA &= \int_0^1\int_0^{\sqrt y}x^2\,dx\,dy\\ &= \int_0^1\left[\frac{x^3}{3}\right]_0^{\sqrt y}\,dy\\ &= \int_0^1\frac{y^{3/2}}{3}\,dy\\ &= \left[\frac{2}{15}y^{5/2}\right]_0^1\\ &= \frac{2}{15}. \end{aligned} $$

The inner antiderivative checks because \(\frac{d}{dx}(x^3/3)=x^2\). The outer antiderivative in the horizontal calculation checks because \(\frac{d}{dy}(2y^{5/2}/15)=y^{3/2}/3\). The two descriptions yield the same value. For this integrand, the horizontal order leaves a single power in the outer integral; the vertical order also works but produces two terms.

A Reliable Selection Process

There is no rule that one order is always better. Use the geometry and the integrand together. A useful comparison can be made before doing the full calculation:

1
Write the region in both orders.
Use the slice descriptions to identify the inner-variable bounds in each order. Check that each description covers the same region.
2
Test the inner antiderivative.
Treat the other variable as constant. Decide whether the inner antiderivative is elementary and whether its endpoint values remain manageable.
3
Inspect the outer integral.
An easy inner antiderivative is not enough if it leaves a difficult outer integral. Compare the complete results of the inner step.
4
Choose and check.
Use the order with the simpler total calculation, then verify antiderivatives by differentiation and confirm the bounds describe the intended region.

Common Mistake: Guessing an Antiderivative

A common error is to assume that an inner integral is easy because a familiar pattern seems close. For instance, the calculation

$$ \int e^{x^2}\,dx \overset{\text{wrong}}{=} \frac12 e^{x^2} $$

is incorrect. Differentiating the proposed answer gives

$$ \frac{d}{dx}\left(\frac12e^{x^2}\right) = x e^{x^2}, $$

not \(e^{x^2}\). The missing factor \(x\) shows why this antiderivative cannot be used. In the first worked example, integrating with respect to \(y\) avoids this issue because \(e^{x^2}\) is constant in \(y\). The subsequent \(x\)-integral includes the factor \(x\), which makes the substitution \(u=x^2\) valid.

The correction is not to force an unsuitable antiderivative. Compare the two orders, use the one that gives a valid and manageable inner integral, and check each proposed antiderivative by differentiation.

Why the Choice Matters

Choosing an integration order is a practical part of evaluating a double integral, not merely a way to rewrite its bounds. A well-chosen order can replace a non-elementary inner antiderivative with a constant-factor integration, shorten the outer calculation, or avoid a more complicated description of the region. Sometimes both orders are reasonable; then the best choice may simply be the one with cleaner algebra.

Keep the two requirements distinct: the bounds must describe the region correctly, and the order should suit the integrand. A convenient antiderivative cannot compensate for incorrect bounds, and correct bounds do not guarantee an easy integral.

Check Your Understanding

For each question, compare the inner antiderivative and the remaining outer integral before deciding which order is preferable.

  1. For \(e^{x^2}\) on \(0\leq y\leq x\leq1\), which order gives an elementary inner antiderivative?
  2. Why is \(\frac12e^{x^2}\) not an antiderivative of \(e^{x^2}\) with respect to \(x\)?
  3. For the region \(0\leq y\leq x\leq1\), write the reversed bounds using \(y\) as the outer variable.
  4. What does Fubini's theorem require in the order-of-integration principle stated here?
  5. Give one reason why a simple inner integral might still leave an inconvenient overall calculation.

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