Exercise 8D.1: Definite Integrals | 习题8D.1：定积分

📚 Exercise 8D.1: Definite Integrals | 习题8D.1：定积分

This article provides a comprehensive guide to Exercise 8D.1, which consolidates the core techniques of definite integration in the IB Mathematics curriculum. Whether you are studying Analysis and Approaches or Applications and Interpretation, mastering these problems will strengthen your understanding of area, accumulation, and the Fundamental Theorem of Calculus. We break down the essential concepts, step-by-step solution strategies, and the most common pitfalls so that you can approach every integral with confidence.

本文为 IB 数学课程中定积分习题 8D.1 提供全面指导。无论你学习的是分析与方法还是应用与解释，掌握这些题目都将加深你对面积、累积量以及微积分基本定理的理解。我们将分解核心概念、分步解题策略和最常见失误，让你自信应对每一道积分题。

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1. Understanding Definite Integrals | 理解定积分

The definite integral of a function f(x) from a to b, written ∫_a^b xis over the interval [a,b]. Areas above the x-axis count positively, while areas below count negatively. This geometric interpretation is the foundation for many real-world applications in physics, economics and probability.

函数 f(x) 从 a 到 b 的定积分，记作 ∫_a^b f(x) dx，表示曲线 y = f(x) 与 x 轴在区间 [a,b] 之间的有正负的面积。x轴上方的面积取正，下方的面积取负。这一几何解释是物理学、经济学和概率论中许多实际应用的基础。

Unlike an indefinite integral, which produces a family of functions plus an arbitrary constant, a definite integral is a single real number. The precise definition involves Riemann sums and the limit of increasingly fine partitions, but in practice we rely on the Fundamental Theorem of Calculus to evaluate it efficiently.

与不定积分产生一族函数加任意常数不同，定积分是一个确定的实数。其精确定义涉及黎曼和及不断细分的极限，但在实际计算中我们借助微积分基本定理来高效求值。

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

ion. It states that if F is any antiderivative of a continuous function f on [a,b], then ∫_a^b mula transforms the problem of finding a limit of sums into a search for an antiderivative.

微积分基本定理架起了微分与积分的桥梁。该定理指出：如果 F 是连续函数 f 在 [a,b] 上的任一原函数，那么 ∫_a^b f(x) dx = F(b) - F(a)。这一简洁公式将求极限和的问题转化为寻找原函数的问题。

∫_a^b f(x) dx = [F(x)]_a^b = F(b) - F(a)

A crucial condition is that f must be continuous on the closed interval. If the integrand has a vertical asymptote or a jump inside [a,b], the integral is improper and must be treated with limits. In Exercise 8D.1, all functions will be carefully chosen to satisfy continuity, so you can safely apply the theorem.

一个关键条件是 f 必须在闭区间上连续。如果被积函数在 [a,b] 内有垂直渐近线或跳跃，该积分就是反常积分，需用极限处理。在习题 8D.1 中，所有函数都经过精心挑选以确保连续，因此你可以放心使用该定理。

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3. Basic Integration Formulas | 基本积分公式

To use the Fundamental Theorem effectively, you must be fluent with a set of standard antiderivatives. Below are the most common ones you will encounter in Exercise 8D.1. For definite integrals, the constant of integration cancels out, so you can use the simplest antiderivative.

要有效运用基本定理，你必须熟练一组标准原函数。以下是你将在习题 8D.1 中遇到的最常见公式。对于定积分，积分常数会抵消，因此可使用最简单的原函数。

Key formulas include: ∫ xⁿ dx = xⁿ⁺¹/(n+1) for n ≠ -1, ∫ 1/x dx = ln|x|, ∫ e^x dx = e^x, ∫ sin x dx = -cos x, ∫ cos x dx = sin x, and ∫ sec² x dx = tan x. For linear combinations, use ∫ [cf(x) + dg(x)] dx = c∫ f(x) dx + d∫ g(x) dx.

关键公式包括：当 n≠-1 时，∫ xⁿ dx = xⁿ⁺¹/(n+1)；∫ 1/x dx = ln|x|；∫ eˣ dx = eˣ；∫ sin x dx = -cos x；∫ cos x dx = sin x；∫ sec² x dx = tan x。对于线性组合，使用 ∫ [cf(x) + dg(x)] dx = c∫ f(x) dx + d∫ g(x) dx。

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4. Definite Integral Evaluation Steps | 定积分计算步骤

A structured approach will help you avoid careless errors. First, find an antiderivative F(x) of the integrand. Second, evaluate F at the upper limit and the lower limit. Third, compute F(upper) - F(lower). Finally, simplify the result, combining fractions or rationalising if necessary.

有条理的方法能帮你避免粗心错误。第一步，找出被积函数的一个原函数 F(x)。第二步，计算 F 在上限和下限的值。第三步，计算 F(上限) - F(下限)。最后化简结果，必要时通分或有理化。

For example, to evaluate ∫₁⁴ (2x + 3) dx, write the antiderivative F(x) = x² ys check your answer by differentiating F(x) to see if you recover the original integrand.

例如，计算 ∫₁⁴ (2x + 3) dx，写出原函数 F(x) = x² + 3x。然后 F(4) - F(1) = (16+12) - (1+3) = 28 - 4 = 24。始终通过求导 F(x) 来检验是否得到原来的被积函数。

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5. Integration by Substitution | 换元积分法

When the integrand contains a composite function multiplied by a derivative of the inner function, substitution (also called u-substitution) becomes powerful. For definite integrals, remember to change the limits of integration according to the substitution, which saves you from converting back to the original variable.

当被积函数包含复合函数乘以内函数的导数时，换元法（也称 u 代换）非常有效。对于定积分，记住要根据换元改变积分限，这样就无需换回原变量，直接计算即可。

∫_a^b f(g(x)) g'(x) dx = ∫_g(a)^g(b) f(u) du

Consider ∫₀¹ 2x e^x² u = x², so du = 2x dx. When x = 0, u = 0; when x = 1, u = 1. The integral transforms to ∫₀¹ e^u du = e¹ - e⁰ = e - 1. This technique is especially handy for trigonometric integrals such as ∫₀^π/2 sin x cos x dx.

考虑 ∫₀¹ 2x eˣ² dx。令 u = x²，则 du = 2x dx。当 x = 0 时，u = 0；当 x = 1 时，u = 1。积分变换为 ∫₀¹ eᵘ du = e¹ - e⁰ = e - 1。这一技巧对三角积分如 ∫₀^π/2 sin x cos x dx 尤其方便。

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6. Integration by Parts | 分部积分法

types, such as a polynomial times an exponential. The formula for definite integrals is ∫_a^b u dv = [uv]_a^b - ∫_a^b v du, where u and dv are appropriately chosen parts of the original integrand.

分部积分法是乘积求导法则的逆运算，用于两种不同类型函数的乘积，例如多项式乘以指数函数。定积分公式为 ∫_a^b u dv = [uv]_a^b - ∫_a^b v du，其中 u 和 dv 从原被积函数中恰当选取。

To compute ∫₀¹ x e^x dx, let u = x, dv = e^x dx. Then du = dx and v = e^x. The integral becomes [x e^x]₀¹ - ∫₀¹ e^x dx = (1·e¹ - 0) - (e¹ - e⁰ nction that becomes simpler when differentiated, usually a polynomial or logarithmic term.

计算 ∫₀¹ x eˣ dx，令 u = x，dv = eˣ dx，则 du = dx，v = eˣ。积分变为 [x eˣ]₀¹ - ∫₀¹ eˣ dx = (1·e¹ - 0) - (e¹ - e⁰) = e - (e - 1) = 1。通常选择求导后会变得更简单的函数作为 u，一般是多项式或对数项。

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7. Properties of Definite Integrals | 定积分的性质

Several algebraic properties make evaluating definite integrals easier and are frequently tested alongside the mechanics of integration. Knowing them allows you to break down complex problems into manageable pieces.

几条代数性质能让定积分的计算更加容易，并且经常与积分技巧一起考查。掌握这些性质可以将复杂问题分解为可处理的部分。

• Reversal of limits: ∫_a^b f(x) dx = -∫_b^a f(x) dx
• Zero interval: ∫_a^a f(x) dx = 0
• Sum/Difference: ∫_a^b [f(x) ± g(x)] dx = ∫_a^b f(x) dx ± ∫_a^b g(x) dx
• Constant multiple: ∫_a^b k·f(x) dx = k·∫_a^b f(x) dx
• Additivity of intervals: ∫_a^b f(x) dx + ∫_b^c f(x) dx = ∫_a^c f(x) dx

For example, if you need ∫₁⁵ f(x) dx and you know ∫₁³ f(x) dx = 4 and ∫₃⁵ t any explicit integration. Reversing limits is especially useful when a given integral has the larger number as the lower limit.

例如，若需求 ∫₁⁵ f(x) dx，且已知 ∫₁³ f(x) dx = 4，∫₃⁵ f(x) dx = 7，则总和为 11，无需具体积分。当已知积分的下限大于上限时，交换限号的性质尤为有用。

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8. Definite Integrals with Absolute Values | 含绝对值的定积分

When the integrand involves an absolute value, such as |x - 2| or |sin x|, the function changes expression at the points where its argument is zero. To evaluate ∫_a^b |g(x)| dx, find all roots of g(x) in [a,b] and split the integral into subintervals where g(x) keeps a constant sign, dropping the absolute value with appropriate sign changes.

当被积函数含有绝对值，如 |x - 2| 或 |sin x|，该函数在其参数为零的点处转变表达式。要计算 ∫_a^b |g(x)| dx，需找出 g(x) 在 [a,b] 内的所有根，并将积分拆分为 g(x) 保持符号不变的子区间，去掉绝对值并配以适当的正负号。

Evaluate ∫₀³ |x - 2| dx. The root is x = 2. On [0,2], x - 2 ≤ 0, so |x - 2| = -(x - 2) = 2 - x. On [2,3], x - 2 ≥ 0, so |x - 2| = x - 2. Hence the integral becomes ∫₀² (2 - x) dx + ∫₂³ (x - 2) dx = [2x - x²/2]₀² + [x²/2 - 2x]₂³ = (4 - 2) + ((9/2 - 6) - (2 - 4)) = 2 + (-1.5 + 2) = 2.5.

计算 ∫₀³ |x - 2| dx。根为 x=2。在 [0,2] 上，x - 2 ≤ 0，故 |x - 2| = -(x - 2) = 2 - x。在 [2,3] 上，x - 2 ≥ 0，故 |x - 2| = x - 2。因此积分变为 ∫₀² (2 - x) dx + ∫₂³ (x - 2) dx，计算结果为 2.5。这种分段技巧在处理含绝对值的面积问题时必不可少。

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9. Area Between Curves | 曲线间的面积

grals also measure the area enclosed between two curves. If f(x) ≥ g(x) on [a,b], the area

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