Exercise 23A: Antidifferentiation | 练习23A：反导数与不定积分

📚 Exercise 23A: Antidifferentiation | 练习23A：反导数与不定积分

Exercise 23A introduces the fundamental concept of antidifferentiation, the reverse process of differentiation. It focuses on finding indefinite integrals using the power rule, handling constants of integration, and recovering original functions from their derivatives. This exercise typically appears early in the Integration chapter of IB Mathematics Analysis and Approaches (AA) and lays the groundwork for definite integration and area calculations.

练习23A介绍了反导数这一基本概念，即微分的逆过程。该练习重点在于运用幂函数积分法则求不定积分、处理积分常数，并从导数还原原函数。这部分习题通常出现在IB数学分析与方法课程中积分章节的开端，为定积分和面积计算打下基础。

1. Understanding Antidifferentiation | 理解反导数

ess of finding a function F(x) whose derivative is a given function f(x). In symbols, if F' constant. The symbol ∫ is the integral sign and dx indicates the variable of integration.

反导数，又称不定积分，是寻找一个函数 F(x)，使得其导数为给定的函数 f(x) 的过程。用符号表示，如果 F'(x) = f(x)，则我们写成 ∫ f(x) dx = F(x) + C，其中 C 为任意常数。符号 ∫ 是积分号，dx 表示积分变量。

The idea stems from the fact that differentiation is not a one‑to‑one operation: adding a constant to a function does not change its derivative. Therefore, the antiderivative of a function is a family of functions differing by a constant.

这一思想源于微分并非一一对应的运算：在一个函数上加一个常数并不会改变其导数。因此，一个函数的反导数是一族相差一个常数的函数。

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2. The Power Rule for Integration | 幂函数的积分法则

The most frequently used rule in Exercise 23A is the power rule for integration. For any real number n ≠ -1,

练习23A中最常用的法则是幂函数的积分法则。对于任何实数 n ≠ -1，有

∫ xⁿ dx = xⁿ⁺¹/(n+1) + C

u simply increase the exponent by 1 and divide by the new exponent, then add the constant of integration.

这是微分法则 d/dx (xⁿ⁺¹/(n+1)) = xⁿ 的逆运算。只需将指数加1，再除以新的指数，然后加上积分常数即可。

x⁴/4 + C. For ∫ x⁻² dx, the new exponent is -1, so the result is x⁻¹/(-1) + C = -1/x + C.

在练习23A中，常见积分如 ∫ x³ dx 或 ∫ x⁻² dx。对于 ∫ x³ dx，运用法则得到 x⁴/4 + C。对于 ∫ x⁻² dx，新指数为 -1，因此结果为 x⁻¹/(-1) + C = -1/x + C。

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3. The Constant of Integration | 积分常数 C

Every indefinite integral must include the constant of integration C. Omitting C is a common error that can cost marks in IB exams. The constant reflects the fact that infinitely many antiderivatives exist for a given function, all parallel vertical translations of each other.

每一个不定积分都必须加上积分常数 C。遗漏 C 是常见错误，在IB考试中可能丢分。这个常数反映了给定函数的反导数有无穷多个，它们彼此之间是垂直平移的关系。

When solving problems that involve finding a specific original function from a derivative and an initial condition, the constant C is determined uniquely. Without that condition, leaving the answer as a family of functions is required.

在需要通过导数和初始条件求特定原函数的问题中，常数 C 会被唯一确定。如果没有初始条件，答案就应该保留为函数族的形式。

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4. Integrating Sums and Differences | 和与差的积分

Integration is linear, meaning the integral of a sum is the sum of the integrals, and constant factors can be pulled outside the integral sign. For functions f(x) and g(x), and a constant k:

积分具有线性性质，即和的积分等于积分的和，常数因子可以提到积分号外。对于函数 f(x) 和 g(x)，以及常数 k，有：

∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx

∫ k·f(x) dx = k ∫ f(x) dx

(3x² + 2x - 5) dx becomes 3 ∫ x² dx + 2 ∫ x dx - 5 ∫ 1 dx, which yields x³ + x² - 5x + C.

这些性质可以将多项式和其它表达式分解为更简单的项。例如，∫ (3x² + 2x - 5) dx 变为 3 ∫ x² dx + 2 ∫ x dx - 5 ∫ 1 dx，得到 x³ + x² - 5x + C。

Note that we do not need a separate constant for each integral; one overall constant C at the end is sufficient because the sum of several constants is still a constant.

注意不必为每个积分单独写一个常数；最后写一个总的常数 C 已经足够，因为多个常数的和仍是一个常数。

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5. Integration of (ax + b)ⁿ | 线性函数的幂次积分

+ b)ⁿ dx, where a ≠ 0 and n ≠ -1. The chain rule in reverse gives an extra factor of 1/a:

练习23A常常将幂法则推广到形如 ∫ (ax + b)ⁿ dx 的积分，其中 a ≠ 0 且 n ≠ -1。逆向使用链式法则会带来一个额外的因子 1/a：

∫ (ax + b)ⁿ dx = (1/a) × (ax + b)ⁿ⁺¹/(n+1) + C

This formula can be verified by differentiating the right-hand side. The derivative of (ax + b)ⁿ⁺¹ is a·(n+1)·(ax + b)ⁿ, so dividing by a·(n+1) cancels the extra factors and leaves the original integrand.

这个公式可以通过对右边求导进行验证。(ax + b)ⁿ⁺¹ 的导数为 a·(n+1)·(ax + b)ⁿ，因此除以 a·(n+1) 就消去多余因子，还原到被积函数。

. Being comfortable with this extension is essential, as such integrals appear frequently.

例如，∫ (2x + 3)⁴ dx = (1/2) · (2x + 3)⁵/5 + C = (2x + 3)⁵/10 + C。熟练掌握这种扩展形式非常重要，因为这类积分经常出现。

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6. Finding the Original Function Given the Derivative | 已知导数求原函数

A typical question in Exercise 23A gives f'(x) and asks for f(x). You integrate f'(x) to obtain f(x) + C. If additional information is provided, such as the graph passing through a specific point, you can solve for C.

练习23A中典型的题型给出 f'(x)，要求 f(x)。通过对 f'(x) 积分得到 f(x) + C。如果还提供了额外信息，例如曲线经过某个特定点，就可以解出 C。

The process involves writing the general antiderivative and then substituting the known coordinates to find the particular solution. This technique bridges indefinite and definite contexts.

这一过程包括写出一般形式的反导数，然后代入已知坐标求出特解。这一技巧在不定积分与定积分之间建立起联系。

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7. Solving for the Constant Using Initial Conditions | 利用初始条件求常数

inding the general antiderivative with + C, substitute x₀ and set the expression equal to y₀ to solve for C.

当点 (x₀, y₀) 在曲线 y = f(x) 上时，我们需要 f(x₀) = y₀。在求出了带有 + C 的一般反导数后，代入 x₀ 并令表达式等于 y₀，即可解出 C。

For example, if f' tituting x = 1 gives 1⁴ + C = 3, so C = 2. Hence the particular solution is f(x) = x⁴ + 2.

例如，若 f'(x) = 4x³ 且 f(1) = 3，则 f(x) = x⁴ + C。代入 x = 1 得 1⁴ + C = 3，因此 C = 2。于是特解为 f(x) = x⁴ + 2。

This simple step is often tested, so always remember to find C when sufficient information is given.

这个简单步骤经常被考查，因此只要有充足信息，务必记住求出 C。

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8. Common Mistakes and How to Avoid Them | 常见错误及避免方法

+ C instead of x³/3 + C. Always increase the exponent and then divide by that new number.

一个常见错误是忘记除以新的指数。学生们常写出 ∫ x² dx = x³ + C，而不是 x³/3 + C。务必先增加指数，再除以这个新数字。

x should be written as ∫ x⁻² dx, giving x⁻¹/(-1) + C = -1/x + C. Writing it directly as ln|x| is incorrect unless the exponent is exactly -1.

另一个错误是处理负指数和分数指数时出现失误。例如，∫ 1/x² dx 应先写成 ∫ x⁻² dx，得到 x⁻¹/(-1) + C = -1/x + C。直接写成 ln|x| 是错误的，除非指数恰好为 -1。

Also, forgetting the absolute value in the integral of 1/x (which appears later) or forgetting the constant C are both penalised. In Exercise 23A the focus is mainly on the power rule, so ensure the form (ax + b)ⁿ follows the correct adjustment with the factor 1/a.

此外，∫ 1/x dx（稍后出现）中忘记绝对值，或忘记常数 C，都会被扣分。练习23A主要聚焦于幂法则，因此要确保 (ax + b)ⁿ 的形式配上正确的 1/a 因子。

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9. Worked Examples from Exercise 23A | 练习23A典型题解

Example 1: Find ∫ (6x² - 4x + 1) dx.

例1： 求 ∫ (6x² - 4x + 1) dx。

x³/3) = 2x³; ∫ -4x dx = -4·(x²/2) = -2x²; ∫ 1 dx = x. Add the constant: 2x³ - 2x² + x + C.

逐项积分：∫ 6x² dx = 6·(x³/3) = 2x³；∫ -4x dx = -4·(x²/2) = -2x²；∫ 1 dx = x。加上常数：2x³ - 2x² + x + C。

Example 2: Evaluate ∫ (5x + 2)³ dx.

例2： 计算 ∫ (5x + 2)³ dx。

Here a = 5, n = 3. Using the rule, ∫ (5x+2)³ dx = (1/5)·(5x+2)⁴/4 + C = (5x+2)⁴/20 + C.

这里 a = 5, n = 3。应用法则，∫ (5x+2)³ dx = (1/5)·(5x+2)⁴/4 + C = (5x+2)⁴/20 + C。

Example 3: Given f'(x) = 3x² - 2 and f(2) = 5, find f(x).

例3： 已知 f'(x) = 3x² - 2 且 f(2) = 5，求 f(x)。

2x + C. Substitute x=2: 2³ - 2·2 + C = 5 → 8 - 4 + C = 5 → C = 1. Thus f(x) = x³ - 2x + 1.

积分：f(x) = x³ - 2x + C。代入 x=2：2³ - 2·2 + C = 5 → 8 - 4 + C = 5 → C = 1。因此 f(x) = x³ - 2x + 1。

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10. Key Takeaways | 要点总结

The core of Exercise 23A is mastering the power rule for integration and the reverse chain rule for linear inner functions. Always add the constant C, simplify your answers, and use initial conditions to find a particular solution when needed.

练习23A的核心是掌握幂函数的积分法则和线性内层函数的逆向链式法则。一定要加常数 C，化简答案，并在需要时利用初始条件求出特解。

Build fluency with both positive and negative exponents, and remember that the constant of integration turns a derivative back into a family of curves. Consistent practice with these basic integrations will make the upcoming definite integration and area problems much easier.

熟练处理正指数和负指数，记住积分常数将导数还原为一族曲线。对这些基本积分进行持续练习，将使接下来的定积分和面积问题变得容易得多。

Function f(x)	Indefinite Integral ∫ f(x) dx
xⁿ (n ≠ -1)	xⁿ⁺¹/(n+1) + C
(ax + b)ⁿ (n ≠ -1)	(1/a)·(ax + b)ⁿ⁺¹/(n+1) + C
k (constant)	kx + C
Sum f(x) ± g(x)	∫ f(x) dx ± ∫ g(x) dx

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