📚 Repeated Integrals with Constant Limits | 常数极限的累次积分
In the IB Mathematics Analysis and Approaches HL course, and particularly within the calculus option, you will encounter repeated integrals—also known as iterated integrals—as a natural extension of single-variable integration. When the limits of integration are constants, the region of integration is a rectangle. This simple geometry allows us to evaluate double integrals in a systematic way, often by integrating with respect to one variable at a time. Understanding repeated integrals with constant limits lays the foundation for computing areas, volumes, and other accumulated quantities over rectangular domains.
在IB数学分析与方法高水平课程中，特别是在微积分选修部分，你会遇到累次积分（也叫迭代积分），这是单变量积分的自然延伸。当积分限为常数时，积分区域是一个矩形。这种简单的几何图形使我们能够系统地计算二重积分，通常是一次对一个变量积分。理解常积分限下的累次积分为计算矩形区域上的面积、体积和其他累积量奠定了基础。
1. What is a Repeated Integral? | 什么是累次积分？
nction of two variables, f(x, y), a repeated integral with constant limits is written as ∫ab ∫cd f(x, y) dy dx or ∫cd ∫ab f(x, y) dx dy. The inner integral is computed first, treating the other variable as constant, and then the outer integral is evaluated.
累次积分是一个包含两个（或更多）积分符号依次作用的表达式。对于二元函数 f(x, y)，常数积分限的累次积分可以写成 ∫ab ∫cd f(x, y) dy dx 或 ∫cd ∫ab f(x, y) dx dy。先计算内层积分，将另一个变量视为常数，然后再计算外层积分。
2. Constant Limits of Integration | 常数积分限
When we say the limits are constant, we mean that a, b, c, and d are fixed real numbers. The inner integral's limits do not depend on the outer variable. This distinguishes constant-limit repeated integrals from more general double integrals where the limits can be functions of the other variable.
当我们说积分限为常数时，意思是 a、b、c、d 都是固定的实数。内层积分的上下限不依赖于外层变量。这使得常数限的累次积分区别于更一般的二重积分，后者的积分限可能是另一个变量的函数。
3. The Rectangular Region | 矩形区域
ight lines parallel to the axes, we can reverse the order of integration freely without changing the limits—only the order of the differentials changes.
常数限的累次积分在几何上对应于 xy 平面上矩形区域 R = [a, b] × [c, d] 上的二重积分。该区域由 a ≤ x ≤ b 和 c ≤ y ≤ d 定义。由于边界是平行于坐标轴的直线，我们可以自由交换积分次序而不改变积分限——只需改变微分的次序。
4. Iterated Integrals: Order of Integration | 迭代积分：积分次序
The order of integration tells you which variable is integrated first. In the symbol ∫ab ∫cd dy dx indicates the order: first y, then x. Changing the order to dx dy is equally valid for rectangles.
积分次序告诉你先对哪个变量积分。在符号 ∫ab ∫cd f(x, y) dy dx 中，内层积分是对 y 进行的，因此我们固定 x，将 f(x, y) 对 y 从 c 到 d 积分；然后将所得关于 x 的函数从 a 到 b 积分。记法 dA = dy dx 表示次序：先 y 后 x。对矩形区域而言，改变次序为 dx dy 同样有效。
5. Fubini's Theorem for Constant Limits | 常数极限下的富比尼定理
Fubini' , d], then the double integral over R can be evaluated by either repeated integral, and they are equal: ∬R f(x, y) dA = ∫ab ∫cd f(x, y) dy dx = ∫cd ∫ab f(x, y) dx dy. This powerful result justifies splitting a double integral into two single integrations.
富比尼定理断言，如果 f(x, y) 在矩形 R = [a, b] × [c, d] 上连续，那么该区域上的二重积分可以用任一次序的累次积分计算，且两者相等：∬R f(x, y) dA = ∫ab ∫cd f(x, y) dy dx = ∫cd ∫ab f(x, y) dx dy。这一强大结果使我们能把二重积分拆分成两次单积分。
6. Evaluating a Repeated Integral Step-by-Step | 逐步计算累次积分
To evaluate ∫ab ∫cd f(x, y) dy dx:
- Step 1: and find the inner antiderivative with respect to y. Evaluate it between y = c and y = d.
- Step 2: The result is a function of x only; simplify it algebraically.
- Step 3: Integrate this function with respect to x from a to b.
- Step 4: Substitute the limits to obtain a numerical value.
计算 ∫ab ∫cd f(x, y) dy dx 的步骤：
- 步骤1：将 x 视为常数，求出对 y 的内层原函数，并在 y = c 和 y = d 处求值。
- 步骤2：结果仅为 x 的函数；对其进行代数化简。
- 步骤3：将该函数对 x 从 a 到 b 积分。
- 步骤4：代入积分限求得数值。
7. Product of Single Integrals when f(x, y) = g(x)h(y) | 当函数可分离时单积分的乘积
its, the double integral separates into the product of two independent single integrals: ∫ab ∫cd g(x)h(y) dy dx = (∫ab g(x) dx) (∫cd h(y) dy). This simplification works because the constant limits ensure no cross-dependence.
当被积函数可以分解为只含 x 的函数与只含 y 的函数的乘积时会出现一种特殊情况：f(x, y) = g(x) h(y)。对于常数积分限的矩形区域，二重积分可以分离为两个独立的单积分的乘积：∫ab ∫cd g(x)h(y) dy dx = (∫ab g(x) dx) (∫cd h(y) dy)。这种简化之所以成立，是因为常数限确保了没有交叉依赖。
8. Physical Interpretation: Volume Under a Surface | 物理解释：曲面下的体积
If f(x, y) ≥ 0 over the rectangle R, the double integral ∬R repeated integral computes this volume by slicing the solid either perpendicular to the x‑axis or perpendicular to the y‑axis, summing cross-sectional areas.
如果在矩形 R 上 f(x, y) ≥ 0，那么二重积分 ∬R f(x, y) dA 表示以曲面 z = f(x, y)、xy 平面以及垂直平面 x = a、x = b、y = c、y = d 为界的立体体积。累次积分通过将立体沿垂直于 x 轴或垂直于 y 轴的方向切片，并累加截面面积来计算这一体积。
9. Worked Example 1: Polynomial Function | 例题1：多项式函数
Evaluate ∬R (x² + 3xy) dA over R = [0, 1] × [0, 2].
Solution using order dy dx:
∫01 ∫02 (x² + 3xy) dy dx
Inner integral: ∫02 (x² + 3xy) dy = [x²y + (3/2)xy²]02 = 2x² + (3/2)x(4) = 2x² + 6x.
Outer integral: ∫01 (2x² + 6x) dx = [(2/3)x³ + 3x²]01 = 2/3 + 3 = 11/3.
The volume under the surface is 11/3 cubic units. Reversing the order confirms the same result.
计算 ∬R (x² + 3xy) dA，其中 R = [0, 1] × [0, 2]。
采用次序 dy dx 求解：
∫01 ∫02 (x² + 3xy) dy dx
内层积分：∫02 (x² + 3xy) dy = [x²y + (3/2)xy²]02 = 2x² + 6x。
外层积分：∫01 (2x² + 6x) dx = [(2/3)x³ + 3x²]01 = 2/3 + 3 = 11/3。
曲面下的体积为 11/3 立方单位。交换积分次序可验证得到相同结果。
10. Worked Example 2: Trigonometric and Separable Function | 例题2：三角函数与可分离函数
Compute ∫0π/2 ∫13 (sin x)(ln y) dy dx.
Here the integrand is a product g(x)h(y), so we separate:
∫0π/2 sin x dx × ∫13 ln y dy
First integral: ∫0π/2 sin x dx = [−cos x]0π/2 = 0 − (−1) = 1.
Second integral: ∫13 y ln y − y. Evaluated from 1 to 3: (3 ln 3 − 3) − (1·0 − 1) = 3 ln 3 − 3 + 1 = 3 ln 3 − 2.
Product: 1 × (3 ln 3 − 2) = 3 ln 3 − 2. The integral evaluates to 3 ln 3 − 2.
计算 ∫0π/2 ∫13 (sin x)(ln y) dy dx。
这里被积函数是 g(x)h(y) 的乘积，因此可以分离：
∫0π/2 sin x dx × ∫13 ln y dy
第一个积分：∫0π/2 sin x dx = 1。
第二个积分：∫13 ln y dy，使用分部积分法得 y ln y − y，从 1 到 3 求值：(3 ln 3 − 3) − (−1) = 3 ln 3 − 2。
乘积：1 × (3 ln 3 − 2) = 3 ln 3 − 2。累次积分的值为 3 ln 3 − 2。
11. Common Mistakes and Tips | 常见错误与提示
- Mistake: Treating the outer variable as a constant when integrating the inner part. Always keep the inner integration variable clear and treat the outer variable as constant. Tip: Write a small note: 'treat x as constant' when integrating with respect to y.
- Mistake: Forgetting to distribute the limits correctly when the inner antiderivative has several terms. Tip: Use brackets and evaluate carefully term by term.
- Mistake: Confusing the order of differentials. The differentials dy dx tell you the order: integrate with respect to y first, then x. Tip: Match the inner integral sign with its differential.
- Mistake: Trying to separate a function that is not a product. Tip: If f(x, y) cannot be written as g(x)h(y), do not separate; integrate layer by layer.
- 错误：在积分内层时将外层变量当作常数。要始终明确内层积分变量，并将外层变量视为常数。提示：在对 y 积分时写上“视 x 为常数”的小注。
- 错误：当内层原函数含多项时忘记正确代入上下限。提示：使用括号并逐项仔细求值。
- 错误：混淆微分的次序。微分 dy dx 表示次序：先对 y 积分，再对 x。提示：使内层积分符号与它的微分相对应。
- 错误：尝试分离不是乘积的函数。提示：如果 f(x, y) 不能写成 g(x)h(y)，就不要分离；逐层积分即可。
12. Summary and Further Practice | 总结与进阶练习
Repeated integrals with constant limits provide a straightforward entry to multivariable calculus. Key points: the region is rectangular, Fubini's Theorem guarantees the order can be swapped, and when the integrand is separable the evaluation reduces to a product of two single integrals. Mastery of these ideas will prepare you for more complex regions where limits depend on variables, and for applications in probability, physics, and volume calculations. Practise with a variety of functions—polynomials, exponentials, trigonometric, and combinations—and always check your result by reversing the order when possible.
常积分限的累次积分为多变量微积分提供了一个直接的入口。要点是：区域为矩形，富比尼定理保证可以交换积分次序，而当被积函数可分离时，计算简化为两个单积分的乘积。掌握这些概念后，你将能够处理积分限依赖于变量的更复杂区域，以及在概率、物理和体积计算中的应用。请用各种函数进行练习——多项式、指数、三角函数及其组合——并在可能时通过交换次序来检验答案。
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