Definite Integrals: Concepts and Computation | 定积分的概念与计算

📚 Definite Integrals: Concepts and Computation | 定积分的概念与计算

The definite integral is one of the most powerful tools in mathematics, allowing us to compute areas, accumulate quantities, and solve real-world problems that would otherwise be impossible to tackle with simple algebra. For IB Mathematics students, mastering the concept and computation of definite integrals is essential for success in both Analysis and Approaches (AA) and Applications and Interpretation (AI) courses.

定积分是数学中最强大的工具之一，它使我们能够计算面积、累积数量，并解决那些仅靠简单代数无法处理的实际问题。对于IB数学学生而言，掌握定积分的概念与计算，是在分析与方法（AA）以及应用与解释（AI）课程中取得好成绩的关键。

1. From Area to Integral: The Riemann Sum | 从面积到积分：黎曼和

Before we define the definite integral, we must first understand how to approximate the area under a curve. Consider a continuous function f(x) on a closed interval [a, b]. To estimate the area between the curve and the x-axis, we can divide the interval into n subintervals of equal width Δx = (b − a)/n. On each subinterval, we choose a sample point xᵢ* and construct a rectangle with height f(xᵢ*) and width Δx. The sum of the areas of all such rectangles is called a Riemann sum.

在定义定积分之前，我们首先需要理解如何逼近曲线下方的面积。考虑闭区间 [a, b] 上的连续函数 f(x)。为了估算曲线与 x 轴之间的面积，我们可以将区间分成 n 个等宽的子区间，宽度为 Δx = (b − a)/n。在每个子区间内，我们选择一个采样点 xᵢ*，构造一个高度为 f(xᵢ*)、宽度为 Δx 的矩形。所有这些矩形面积之和称为黎曼和。

The Riemann sum is denoted as:

黎曼和的记号为：

Sₙ = Σᵢ₌₁ⁿ f(xᵢ*) · Δx

When we use the left endpoint of each subinterval as the sample point, we obtain the left Riemann sum. When we use the right endpoint, we obtain the right Riemann sum. For an increasing function, the left Riemann sum underestimates the true area while the right Riemann sum overestimates it. The trapezoidal rule, which uses the average of the left and right heights, often provides a better approximation.

当我们以每个子区间的左端点为采样点时，得到左黎曼和；以右端点为采样点时，得到右黎曼和。对于增函数而言，左黎曼和会低估真实面积，而右黎曼和会高估真实面积。梯形法则使用左右高度的平均值，通常能提供更好的逼近效果。

2. The Definition of the Definite Integral | 定积分的定义

The definite integral is defined as the limit of the Riemann sum as the number of subintervals approaches infinity. This is the precise way to express the “area under a curve” as a rigorous mathematical concept.

定积分被定义为黎曼和在子区间数量趋于无穷大时的极限。这是将“曲线下方面积”表达为严谨数学概念的精确方式。

∫ₐᵇ f(x) dx = limₙ→∞ Σᵢ₌₁ⁿ f(xᵢ*) · Δx

In this notation, the integral sign ∫ is an elongated S, representing a sum. The lower and upper limits a and b indicate the interval of integration. The expression f(x) dx suggests the height f(x) multiplied by an infinitesimal width dx. The definite integral produces a number, not a function — this distinguishes it from the indefinite integral, which produces a family of antiderivatives.

在此记号中，积分号 ∫ 是拉长了的 S，代表“和”。上下限 a 和 b 表示积分的区间。表达式 f(x) dx 表示高度 f(x) 乘以无穷小的宽度 dx。定积分得出的是一个数，而非一个函数——这使它与得出反导数函数族的不定积分相区别。

For the definite integral to exist, the function must be bounded on [a, b] and the interval must be finite. If f(x) is continuous on [a, b], then the definite integral is guaranteed to exist — this is a fundamental theorem in analysis.

要使定积分存在，函数在 [a, b] 上必须是有界的，且积分区间必须是有限的。如果 f(x) 在 [a, b] 上连续，则定积分必然存在——这是分析学中的基本定理。

3. Geometric Interpretation: Area with Sign | 几何意义：带符号的面积

The definite integral ∫ₐᵇ f(x) dx represents the net signed area between the graph of f and the x-axis, from x = a to x = b. Regions above the x-axis contribute positive area, while regions below the x-axis contribute negative area.

定积分 ∫ₐᵇ f(x) dx 表示函数 f 的图像与 x 轴之间从 x = a 到 x = b 的净有符号面积。x 轴上方的区域贡献正面积，而 x 轴下方的区域贡献负面积。

For example, consider the function f(x) = sin(x) on the interval [0, 2π]. The integral over [0, π] is positive, the integral over [π, 2π] is negative, and the net integral over [0, 2π] equals 0 because the positive and negative areas cancel out. However, the total actual area (without signs) would be 4. This distinction is crucial when solving problems involving area — if the question asks for “area enclosed by the curve and the x-axis,” you must split the integral at the x-intercepts and take the absolute value of each part.

例如，考虑函数 f(x) = sin(x) 在区间 [0, 2π] 上的情形。[0, π] 上的积分为正，[π, 2π] 上的积分为负，而整个 [0, 2π] 上的净积分为 0，因为正负面积相互抵消。然而，实际总面积（不带符号）应为 4。在解决涉及面积的问题时，这个区别至关重要——如果题目问“曲线与 x 轴所围成的面积”，你必须在 x 轴交点处拆开积分，并对每一部分取绝对值。

To compute the total enclosed area between a curve y = f(x) and the x-axis on [a, b], we use:

要计算曲线 y = f(x) 与 x 轴在 [a, b] 上围成的总面积，我们使用：

Total Area = ∫ₐᵇ |f(x)| dx

4. Properties of the Definite Integral | 定积分的基本性质

The definite integral satisfies several important properties that simplify computation considerably. These properties appear frequently in IB exam questions and are essential for efficient problem-solving.

定积分满足若干重要的性质，这些性质能显著简化计算。它们在IB考试题目中频繁出现，是高效解题的关键。

Linearity: ∫ₐᵇ [f(x) ± g(x)] dx = ∫ₐᵇ f(x) dx ± ∫ₐᵇ g(x) dx

线性性质：∫ₐᵇ [f(x) ± g(x)] dx = ∫ₐᵇ f(x) dx ± ∫ₐᵇ g(x) dx
Constant multiple: ∫ₐᵇ c · f(x) dx = c · ∫ₐᵇ f(x) dx, where c is a constant

常数倍性质：∫ₐᵇ c · f(x) dx = c · ∫ₐᵇ f(x) dx，其中 c 为常数
Interval reversal: ∫ₐᵇ f(x) dx = −∫ᵇₐ f(x) dx

区间反向性质：∫ₐᵇ f(x) dx = −∫ᵇₐ f(x) dx
Additivity of intervals: ∫ₐᵇ f(x) dx = ∫ₐᶜ f(x) dx + ∫ᶜᵇ f(x) dx, where a ≤ c ≤ b

区间可加性：∫ₐᵇ f(x) dx = ∫ₐᶜ f(x) dx + ∫ᶜᵇ f(x) dx，其中 a ≤ c ≤ b
Zero width: ∫ₐᵃ f(x) dx = 0

零宽度性质：∫ₐᵃ f(x) dx = 0
Comparison: If f(x) ≥ g(x) on [a, b], then ∫ₐᵇ f(x) dx ≥ ∫ₐᵇ g(x) dx

比较性质：若在 [a, b] 上 f(x) ≥ g(x)，则 ∫ₐᵇ f(x) dx ≥ ∫ₐᵇ g(x) dx

5. The Fundamental Theorem of Calculus | 微积分基本定理

The Fundamental Theorem of Calculus (FTC) establishes the deep connection between differentiation and integration. It has two parts, both of which are essential for IB students to know thoroughly.

微积分基本定理（FTC）建立了微分与积分之间的深刻联系。它包含两个部分，IB学生必须熟练掌握。

Part 1 (FTC1): If f is continuous on [a, b] and F(x) = ∫ₐˣ f(t) dt, then F'(x) = f(x). This states that differentiation undoes integration — that the derivative of an integral with respect to its upper limit is the original integrand evaluated at that limit.

第一部分（FTC1）：若 f 在 [a, b] 上连续，且 F(x) = ∫ₐˣ f(t) dt，则 F'(x) = f(x)。这意味着微分是积分的逆运算——关于上限求导的积分，等于原被积函数在该上限处的取值。

Part 2 (FTC2): If f is continuous on [a, b] and F is any antiderivative of f, then:

第二部分（FTC2）：若 f 在 [a, b] 上连续，F 是 f 的任一原函数，则：

∫ₐᵇ f(x) dx = F(b) − F(a)

This is the fundamental tool for computing definite integrals. Instead of evaluating limits of Riemann sums, we simply find an antiderivative and evaluate it at the bounds. The notation F(b) − F(a) is often written as [F(x)]ₐᵇ. For example, ∫₀¹ x² dx = [x³/3]₀¹ = 1/3 − 0 = 1/3.

这是计算定积分的基本工具。我们无需计算黎曼和的极限，只需找到一个原函数并在上下界处取值即可。记号 F(b) − F(a) 通常写作 [F(x)]ₐᵇ。例如，∫₀¹ x² dx = [x³/3]₀¹ = 1/3 − 0 = 1/3。

6. Basic Integration Rules and Standard Results | 基本积分法则与标准结果

A solid command of standard integrals is essential. The following table lists the most important results that IB students should memorize. These appear on the formula booklet in some cases, but instant recall greatly speeds up problem-solving.

熟练掌握标准积分至关重要。下表列出了IB学生应当记忆的最重要的积分公式。部分公式会出现在公式册中，但快速回忆能极大地提高解题速度。

Function \| 函数	Integral \| 积分
xⁿ (n ≠ −1)	xⁿ⁺¹/(n + 1) + C
1/x = x⁻¹	ln\|x\| + C
eˣ	eˣ + C
sin(x)	−cos(x) + C
cos(x)	sin(x) + C
sec²(x)	tan(x) + C
1/(1 + x²)	arctan(x) + C
1/√(1 − x²)	arcsin(x) + C

Note that for a general exponential function aˣ, the integral is aˣ/ln(a) + C. This result follows from the fact that d/dx (aˣ) = aˣ ln(a).

注意，对于一般指数函数 aˣ，其积分为 aˣ/ln(a) + C。这一结果来自于 d/dx (aˣ) = aˣ ln(a)。

7. Integration by Substitution | 换元积分法

When the integrand contains a composite function, the method of substitution, also known as the chain rule in reverse, is often the most efficient approach. For a definite integral, we have two options: either change the limits of integration along with the variable, or use the substitution only to find the antiderivative and then return to the original variable.

当被积函数包含复合函数时，换元法——即链式法则的逆用——通常是最有效的方法。对于定积分，我们有两种选择：一是同时更换积分上下限和变量；二是仅用换元求原函数，然后回到原变量。

Example: Compute ∫₀¹ 2x · (x² + 1)⁴ dx.

示例：计算 ∫₀¹ 2x · (x² + 1)⁴ dx。

Solution: Let u = x² + 1. Then du/dx = 2x, so du = 2x dx. The integrand becomes u⁴ du. When x = 0, u = 1; when x = 1, u = 2. Therefore:

解答：令 u = x² + 1。则 du/dx = 2x，所以 du = 2x dx。被积函数变为 u⁴ du。当 x = 0 时，u = 1；当 x = 1 时，u = 2。因此：

∫₀¹ 2x · (x² + 1)⁴ dx = ∫₁² u⁴ du = [u⁵/5]₁² = (32/5) − (1/5) = 31/5

One common pitfall is forgetting to change the limits when using substitution on a definite integral. If you decide not to change the limits, you must substitute the expression for u back into the antiderivative and evaluate using the original x-limits. Both approaches yield the same result, but consistency is key.

一个常见错误是在对定积分使用换元法时忘记更换上下限。如果你决定不更换上下限，则必须将 u 的表达式代回原函数，并用原始的 x 上下限求值。两种方法结果相同，但保持一致性至关重要。

8. Integration by Parts | 分部积分法

The method of integration by parts is derived from the product rule in differentiation and is used when the integrand is a product of two functions, one of which is easy to differentiate and the other easy to integrate. The formula for definite integrals takes the form:

分部积分法由微分的乘积法则推导而来，适用于被积函数为两个函数乘积、其中一个易于求导而另一个易于积分的情形。定积分的分部积分公式为：

∫ₐᵇ u dv = [u · v]ₐᵇ − ∫ₐᵇ v du

Here, the term [u · v]ₐᵇ = u(b)v(b) − u(a)v(a). The key to applying this method successfully is choosing u and dv appropriately. As a general guide, choose u to be the function that becomes simpler when differentiated (such as x, x², ln(x)), and choose dv to be the function that is readily integrable (such as eˣ dx, sin(x) dx, cos(x) dx).

在此，[u · v]ₐᵇ = u(b)v(b) − u(a)v(a)。成功应用此方法的关键在于恰当地选择 u 和 dv。一般经验法则是：选择 u 为求导后变得更简单的函数（如 x、x²、ln(x)），选择 dv 为容易积分的部分（如 eˣ dx、sin(x) dx、cos(x) dx）。

Example: Compute ∫₀¹ x · eˣ dx.

示例：计算 ∫₀¹ x · eˣ dx。

Solution: Let u = x, dv = eˣ dx. Then du = dx, v = eˣ. Applying the formula:

解答：令 u = x，dv = eˣ dx。则 du = dx，v = eˣ。代入公式：

[x · eˣ]₀¹ − ∫₀¹ eˣ dx = (e − 0) − [eˣ]₀¹ = e − (e − 1) = 1

Occasionally, integration by parts may need to be applied more than once, such as for integrals of the form ∫ x² eˣ dx. The technique of tabular integration can streamline such repeated applications.

有时，分部积分需要多次应用，例如对于形如 ∫ x² eˣ dx 的积分。列表积分法可以简化这种重复应用的过程。

9. Applications: Area Between Curves | 应用：曲线之间的面积

One of the most common exam applications of the definite integral is computing the area enclosed by two curves. To find the area between the curves y = f(x) and y = g(x) on an interval where f(x) ≥ g(x), we compute:

定积分最常见的考试应用之一是计算两条曲线围成的面积。要求 f(x) ≥ g(x) 区间上曲线 y = f(x) 与 y = g(x) 之间的面积，我们计算：

Area = ∫ₐᵇ [f(x) − g(x)] dx

where a and b are the x-coordinates of the intersection points of the two curves. These points are found by solving the equation f(x) = g(x).

其中 a 和 b 是两条曲线交点处的 x 坐标。这些交点通过解方程 f(x) = g(x) 得到。

Example: Find the area enclosed by the parabola y = x² and the line y = x + 2.

示例：求抛物线 y = x² 与直线 y = x + 2 所围成的面积。

Solution: First, find the intersection points: x² = x + 2 ⟹ x² − x − 2 = 0 ⟹ (x − 2)(x + 1) = 0, so x = −1 and x = 2. On the interval [−1, 2], the line lies above the parabola. Therefore:

解答：首先求交点：x² = x + 2 ⟹ x² − x − 2 = 0 ⟹ (x − 2)(x + 1) = 0，所以 x = −1 和 x = 2。在区间 [−1, 2] 上，直线位于抛物线上方。因此：

Area = ∫₋₁² [(x + 2) − x²] dx = [x²/2 + 2x − x³/3]₋₁² = (2 + 4 − 8/3) − (1/2 − 2 + 1/3) = 9/2

10. Applications: Volume of Revolution | 应用：旋转体体积

Beyond areas, definite integrals are also used to compute volumes of solids of revolution. When a region under a curve y = f(x) from x = a to x = b is rotated about the x-axis, the resulting volume is given by the disk method:

除了面积，定积分还可用于计算旋转体的体积。当曲线 y = f(x) 下方从 x = a 到 x = b 的区域绕 x 轴旋转时，所得体积由圆盘法给出：

V = π ∫ₐᵇ [f(x)]² dx

If the region is instead rotated about the y-axis, and f can be expressed as x = f⁻¹(y), the analogous formula applies with respect to y. If the region between two curves y = f(x) and y = g(x) is rotated about the x-axis, we use the washer method:

如果区域绕 y 轴旋转，且 f 可以表示为 x = f⁻¹(y)，则类似公式关于 y 变量适用。如果两条曲线 y = f(x) 和 y = g(x) 之间的区域绕 x 轴旋转，我们使用垫圈法：

V = π ∫ₐᵇ [f(x)² − g(x)²] dx

Example: The region bounded by y = √x, y = 0, and x = 4 is rotated about the x-axis. Find the volume of the resulting solid.

示例：由 y = √x、y = 0 和 x = 4 围成的区域绕 x 轴旋转，求所得旋转体的体积。

V = π ∫₀⁴ (√x)² dx = π ∫₀⁴ x dx = π [x²/2]₀⁴ = π · 8 = 8π

Volume problems are a staple of IB Mathematics HL and AA HL papers. Be attentive to which axis the rotation occurs about — this determines whether you integrate with respect to x or y.

体积问题是IB数学HL和AA HL试卷中的常考内容。注意旋转绕哪条轴进行——这决定了对 x 还是对 y 积分。

11. Definite Integrals in Kinematics | 定积分在运动学中的应用

In kinematics, definite integrals connect displacement, velocity, and acceleration. If v(t) represents the velocity of a particle at time t, then the displacement between t = a and t = b is given by ∫ₐᵇ v(t) dt. The total distance travelled, which accounts for changes in direction, is given by ∫ₐᵇ |v(t)| dt.

在运动学中，定积分将位移、速度和加速度联系起来。如果 v(t) 表示粒子在时刻 t 的速度，则从 t = a 到 t = b 之间的位移由 ∫ₐᵇ v(t) dt 给出。而考虑方向变化的总路程由 ∫ₐᵇ |v(t)| dt 给出。

Similarly, if a(t) represents acceleration, then the change in velocity over the interval [a, b] is ∫ₐᵇ a(t) dt = v(b) − v(a). These relationships are frequently tested in AI and AA applications papers, often combined with other topics such as kinematics graphs.

类似地，如果 a(t) 表示加速度，则区间 [a, b] 上的速度变化量为 ∫ₐᵇ a(t) dt = v(b) − v(a)。这些关系在AI和AA应用卷中经常出现，通常与运动学图形等主题结合考查。

Example: A particle moves with velocity v(t) = t² − 4t + 3 (m/s) for t ∈ [0, 5]. Find the total distance travelled.

示例：某质点以速度 v(t) = t² − 4t + 3（m/s）在 t ∈ [0, 5] 内运动，求总路程。

Solution: First, find when the velocity changes sign: t² − 4t + 3 = 0 ⟹ (t − 1)(t − 3) = 0, so t = 1 and t = 3. The total distance is:

解答：首先求速度变号的时刻：t² − 4t + 3 = 0 ⟹ (t − 1)(t − 3) = 0，所以 t = 1 和 t = 3。总路程为：

∫₀¹ (t² − 4t + 3) dt + ∫₁³ −(t² − 4t + 3) dt + ∫₃⁵ (t² − 4t + 3) dt

= 4/3 + 4/3 + 20/3 = 28/3 ≈ 9.33 m

12. Common Mistakes and Exam Tips | 常见错误与应试技巧

Students frequently lose marks on definite integral questions due to avoidable errors. Here are the most common pitfalls and how to avoid them.

学生在定积分题目中常因可避免的错误而丢分。以下是最常见的陷阱及规避方法。

Forgetting the constant of integration — although the constant C is not needed when evaluating definite integrals via FTC2, it is still required for indefinite integrals.

忘记积分常数——虽然在使用FTC2计算定积分时不需要常数 C，但在不定积分中仍然需要。
Incorrectly handling negative areas — when computing actual area, always split the integral at x-intercepts and take absolute values.

错误处理负面积——在计算实际面积时，务必在 x 轴交点处拆分积分并取绝对值。
Differential errors in substitution — do not forget to solve for dx in terms of du when performing a substitution.

换元中的微分错误——在进行换元时，不要忘记将 dx 用 du 表示。
Swapping limits without adjusting the negative sign — remember that reversing the order of a and b introduces a factor of −1.

交换上下限时没有调整符号——记住交换 a 和 b 的顺序会引入一个 −1 因子。
Applying ln|x| to negative arguments — for integrals involving 1/x over intervals that include negative values, use the natural logarithm of the absolute value.

对负数使用 ln|x|——对包含负值的区间上 1/x 的积分，应使用绝对值的自然对数。

In the IB exam, clearly show your steps when evaluating a definite integral. This includes stating the antiderivative before substituting the limits, and simplifying intermediate expressions carefully. In particular, with technology-based questions, use your GDC to check numeric answers but do not rely solely on it — examiners reward correct analytic reasoning.

在IB考试中，计算定积分时应清晰展示步骤，包括在代入上下限之前写出原函数，并仔细简化中间表达式。特别是对于使用技术工具的问题，可以用图形计算器（GDC）检查数字答案，但不要完全依赖——阅卷老师会奖励正确的分析推理。

Finally, remember that regular practice is the key to mastery. Work through a variety of problems — from straightforward FTC2 computations to multi-step applications involving substitution, parts, and geometry. As your familiarity with the definite integral grows, so will your confidence and your IB score.

最后，请记住，规律练习是精通的关键。广泛练习不同类型的题目——从直接的FTC2计算到涉及换元、分部积分和几何的多步应用题。随着你对定积分熟悉度的提高，你的信心和IB成绩也会随之提高。

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Definite Integrals: Concepts and Computation | 定积分的概念与计算