IB Mathematics: Worked Examples of Iterated Integrals with Constant Limits | IB数学：常数限累次积分的典型例题解析

📚 IB Mathematics: Worked Examples of Iterated Integrals with Constant Limits | IB数学：常数限累次积分的典型例题解析

In IB Higher Level Mathematics, an iterated integral with constant limits is a way of evaluating double integrals over a rectangle. This article explains the idea through carefully chosen worked examples, and highlights common mistakes to avoid.

在 IB 高级水平数学中，常数限累次积分是计算矩形区域上二重积分的方法。本文通过精心挑选的典型例题解释这一概念，并指出常见错误。

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1. What Is an Iterated Integral with Constant Limits? | 什么是常数限累次积分？

d. The double integral of a function f(x,y) over R can be written as an iterated integral:

假设 R 是矩形区域 a ≤ x ≤ b，c ≤ y ≤ d。函数 f(x,y) 在该矩形上的二重积分可以写成累次积分：

I = ∫_a^b ∫_c^d f(x,y) dy dx

The word “iterated” means one integration is performed inside another. The notation tells us to integrate with respect to y from c to d first, treating x as a constant, and then to integrate the result with respect to x from a to b.

“累次”意味着进行嵌套积分。上面的记号表示：先把 y 看作积分变量，从 c 积到 d，同时把 x 当作常数；然后再把结果对 x 从 a 积到 b。

The phrase “constant limits” is important. Because a, b, c and d are fixed numbers, the limits of the inner integral do not depend on the outer variable. The integration region is therefore exactly a rectangle.

“常数限”这一说法很关键。因为 a、b、c、d 都是固定常数，内层积分的上下限不依赖于外层变量，所以积分区域是一个矩形。

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2. Fubini's Theorem and the Order of Integration | 富比尼定理与积分顺序

For a continuous function f(x,y) on a rectangle, Fubini's theorem states that the order of integration can be swapped without changing the value of the integral:

对于矩形上的连续函数 f(x,y)，富比尼定理说明：交换积分顺序不会改变积分值：

∫_a^b ∫_c^d f(x,y) dy dx = ∫_c^d ∫_a^b f(x,y) dx dy

This result is a special case of Fubini's theorem because the region is a rectangle. When the limits are variables, we would need to change the region when swapping the order, but here the simple constant limits allow us to choose either order freely.

由于积分区域是矩形，这是富比尼定理的一种特殊情况。当上下限是变量时，交换积分顺序需要重新描述积分区域；但常数限非常简单，我们可以自由选择先对 x 或先对 y 积分。

In practice, one order may be much easier than the other. Always check both orders before starting the calculation.

实际计算中，某一种积分顺序可能比另一种简单得多。开始计算前，最好先比较两种顺序。

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3. Step-by-Step Method | 分步计算方法

The method for evaluating a constant-limit iterated integral can be summarised in three steps.

常数限累次积分的计算方法可以概括为三步。

• Step 1: Identify the inner and outer variables. If the integral is written as ∫_a^b ∫_c^d f(x,y) dy dx, then y is the inner variable and x is the outer variable.

  步骤一：确定内层和外层变量。如果积分写成 ∫_a^b ∫_c^d f(x,y) dy dx，那么 y 是内层变量，x 是外层变量。

• Step 2: Evaluate the inner integral. During the inner integration, treat the outer variable as a constant. Then substitute the inner limits.

  步骤二：计算内层积分。在内层积分时，把外层变量当作常数，然后代入内层积分的上下限。

• Step 3: Evaluate the outer integral. Simplify the resulting expression and integrate it with respect to the outer variable over the outer limits.

  步骤三：计算外层积分。化简上一步得到的表达式，再对外层变量进行积分，并代入外层上下限。

The final answer must be a single number if both integrals have constant limits.

如果两层积分的上下限都是常数，最终答案一定是一个数值。

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4. Worked Example 1: Polynomial Integrand | 典型例题一：多项式被积函数

Evaluate the iterated integral

计算累次积分

I = ∫₀¹ ∫₀² (x² y + y³) dx dy

Solution. The inner integration is with respect to x, so y is treated as a constant during that step.

解：内层积分变量是 x，因此在计算内层时把 y 当作常数。

∫₀² (x² y + y³) dx = [ y x³ / 3 + y³ x ]₀²

Substituting x = 2 and x = 0 gives

代入 x = 2 和 x = 0，得到

y · 8/3 + y³ · 2 = (8/3)y + 2y³

Now integrate this expression with respect to y from 0 to 1:

现在对这个表达式关于 y 从 0 到 1 积分：

I = ∫₀¹ ((8/3)y + 2y³) dy = [ (8/3) y²/2 + 2 y⁴/4 ]₀¹ = (4/3) + (1/2) = 11/6

The value of the iterated integral is 11/6.

因此该累次积分的值为 11/6。

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5. Worked Example 2: Exponential Integrand | 典型例题二：指数函数被积函数

Evaluate

计算

I = ∫₀¹ ∫₀² x e^xy dy dx

Solution. Here the inner integration is with respect to y, so x is treated as a constant. Notice that the derivative of e^xy with respect to y is x e^xy.

解：这里内层积分变量是 y，因此把 x 当作常数。注意 e^xy 对 y 求导的结果正好是 x e^xy。

∫₀² x e^xy dy = [ e^xy ]₀² = e^2x − 1

Then integrate with respect to x from 0 to 1:

然后对 x 从 0 到 1 积分：

I = ∫₀¹ (e^2x − 1) dx = [ (1/2)e^2x − x ]₀¹ = (e² / 2 − 1/2) − 1 = (e² − 3)/2

The final answer is (e² − 3)/2.

最终答案为 (e² − 3)/2。Published by TutorHao | IB Mathematics Revision Series | aleveler.com

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