Exercise 17B.2: Integrating (ax + b)^n | 练习17B.2：积分 (ax + b)^n

📚 Exercise 17B.2: Integrating (ax + b)^n | 练习17B.2：积分 (ax + b)^n

In IB Mathematics: Analysis and Approaches, integrating functions of the form (ax + b)^n is a fundamental skill. Exercise 17B.2 provides practice in finding both indefinite and definite integrals of such expressions, preparing you for area calculations and real-world applications. The topic rests on reversing the chain rule and understanding the special logarithmic case.

在IB数学：分析与方法的课程中，对形如 (ax + b)^n 的函数进行积分是一项基础技能。练习17B.2提供了此类表达式的不定积分和定积分练习，为面积计算和实际应用做好准备。这一主题的核心在于逆向运用链式法则，以及理解特殊的对数情形。

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1. Review: The Power Rule for Differentiation | 回顾：幂法则求导

Before integrating, you must be comfortable with differentiating (ax + b)^n. Using the chain rule, the derivative is n·a·(ax + b)^{n−1}. For example, d/dx (2x+3)^4 = 4·2·(2x+3)^3 = 8(2x+3)^3. Similarly, if y = (5−x)^7, then dy/dx = 7·(−1)·(5−x)^6 = −7(5−x)^6. Pay close attention to the factor a, which comes from the inner derivative.

在学习积分之前，必须熟练掌握对 (ax+b)^n 的求导。利用链式法则，导数为 n·a·(ax+b)^{n−1}。例如，d/dx (2x+3)^4 = 4·2·(2x+3)^3 = 8(2x+3)^3。类似地，若 y = (5−x)^7，则 dy/dx = 7·(−1)·(5−x)^6 = −7(5−x)^6。要特别注意由内层导数所带来的因子 a。

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2. Reversing Differentiation to Obtain an Integral | 逆向求导得到积分公式

Integration reverses differentiation. If the derivative of (ax+b)^{n+1} is a(n+1)(ax+b)^n, then the antiderivative of (ax+b)^n involves dividing by a(n+1). This yields the general indefinite integral formula, valid for all real n except n = −1.

积分是求导的逆运算。若 (ax+b)^{n+1} 的导数是 a(n+1)(ax+b)^n，那么 (ax+b)^n 的原函数就需要除以 a(n+1)。由此得到一般的不定积分公式，它对除 n = −1 外的所有实数 n 均成立。

∫ (ax + b)ⁿ dx = (1/a) · (ax + b)ⁿ⁺¹/(n+1) + C,   n ≠ −1

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3. Why Divide by a? The Role of the Inner Derivative | 为何除以 a？内层导数的作用

When differentiating (ax+b)^{n+1}, the chain rule multiplies by the derivative of the inside function, a. To compensate during integration, we must divide by a. This is essentially the reverse chain rule. You can verify by differentiating the antiderivative: the a in the denominator cancels the a that appears from the chain rule, leaving the original integrand.

对 (ax+b)^{n+1} 求导时，链式法则会乘以内层函数的导数 a。为在积分时补偿这一因子，我们必须除以 a。这本质上是逆向链式法则。你可以通过对原函数求导来验证：分母中的 a 会与链式法则产生的 a 相互抵消，从而还原出被积函数。

d/dx [(ax+b)^{n+1}] = a(n+1)(ax+b)^n  ⇒  ∫ (ax+b)^n dx = (1/a)·(ax+b)^{n+1}/(n+1) + C

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4. The Special Case n = −1 | 特殊情况 n = −1

If n = −1, the formula would involve division by zero, so it is not applicable. Instead, ∫ (ax+b)^{−1} dx = (1/a) ln |ax+b| + C. This result comes from recognising that the derivative of ln|ax+b| is a/(ax+b). For definite integrals, you must still use absolute values inside the logarithm to ensure the expression is defined over the interval.

若 n = −1，公式将出现除以零的情况，因此不适用。此时 ∫ (ax+b)^{−1} dx = (1/a) ln |ax+b| + C。该结果源于 ln|ax+b| 的导数为 a/(ax+b)。对于定积分，仍须在对数内使用绝对值，以确保表达式在整个区间上有定义。

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5. Worked Example: Simple Indefinite Integral | 例题：简单不定积分

Find ∫ (5x − 2)^3 dx. Here a=5, b=−2, n=3. Apply the formula: (1/5)·(5x−2)^4 /4 + C = (1/20)(5x−2)^4 + C. Always differentiate your result to check: d/dx[(1/20)(5x−2)^4] = (1/20)·4·5·(5x−2)^3 = (5x−2)^3, which matches the original integrand.

求 ∫ (5x − 2)^3 dx。这里 a=5, b=−2, n=3。套用公式：(1/5)·(5x−2)^4 /4 + C = (1/20)(5x−2)^4 + C。务必通过求导检验：d/dx[(1/20)(5x−2)^4] = (1/20)·4·5·(5x−2)^3 = (5x−2)^3，与被积函数一致。

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6. Definite Integrals and the Fundamental Theorem | 定积分与微积分基本定理

Exercise 17B.2 also includes definite integrals. Using the Fundamental Theorem of Calculus, ∫_a^b f(x) dx = F(b) − F(a), where F is any antiderivative of f. The constant C cancels out, so you can omit it when evaluating definite integrals. For (ax+b)^n, you simply plug the limits into the antiderivative obtained from the formula.

练习17B.2也包含定积分。利用微积分基本定理，∫_a^b f(x) dx = F(b) − F(a)，其中 F 是 f 的任意一个原函数。常数 C 会相互抵消，因此在计算定积分时可将其略去。对于 (ax+b)^n，只需把积分上下限代入由公式得到的原函数即可。

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7. Worked Example: Definite Integral | 例题：定积分

Evaluate ∫₀¹ (2x+1)^4 dx. The antiderivative is (1/2)·(2x+1)^5/5 = (1/10)(2x+1)^5. Then F(1)−F(0) = (1/10)(3^5) − (1/10)(1^5) = (243/10) − (1/10) = 242/10 = 24.2. This value can also be verified on your calculator.

计算 ∫₀¹ (2x+1)^4 dx。原函数为 (1/2)·(2x+1)^5/5 = (1/10)(2x+1)^5。则 F(1)−F(0) = (1/10)(3^5) − (1/10)(1^5) = (243/10) − (1/10) = 242/10 = 24.2。该结果也可用计算器验证。

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8. Using Your GDC to Verify Integrals | 使用图形计算器验证积分

On the TI-84 or TI-Nspire, you can compute definite integrals directly. On the TI-84, press MATH then 9: fnInt(, and enter the expression, the variable, and the limits. On the TI-Nspire, use the integral template. It is good practice to check your manual results with technology, especially in Paper 2 where GDC use is permitted.

在 TI-84 或 TI-Nspire 上，你可以直接计算定积分。在 TI-84 上，按 MATH 键然后选择 9: fnInt(，输入表达式、变量和积分限。在 TI-Nspire 上，使用积分模板即可。使用技术手段检查手算结果是个好习惯，尤其是在允许使用图形计算器的试卷二中。

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

Many students forget to divide by the coefficient a, or they mis-apply the power increment (using n instead of n+1). Another frequent mistake is treating n = −1 with the power rule, leading to division by zero. Always check n before applying the formula. When evaluating definite integrals, ensure you substitute the limits carefully, especially when the lower limit is negative; a sign error in (ax+b)^m is easy to make.

许多学生忘记除以系数 a，或者在增加幂次时算错（误用 n 而不是 n+1）。另一个常见错误是对 n = −1 误用幂法则，从而导致除以零。应用公式前务必检查 n。计算定积分时，要确保仔细代入上下限，尤其是当下限为负数时；在计算 (ax+b)^m 的值时很容易出现符号错误。

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10. Extension: Integrating Expressions with Constants and Negative Exponents | 拓展：带常数因子与负指数的积分

Sometimes the integrand is a constant multiple, such as 6(4x−1)^2. You can factor out the constant: ∫ 6(4x−1)^2 dx = 6 ∫ (4x−1)^2 dx = 6·(1/4)·(4x−1)^3/3 + C = (1/2)(4x−1)^3 + C. For negative exponents other than −1, the same power rule applies. Example: ∫ (3x+2)^{−3} dx = (1/3)·(3x+2)^{−2}/(−

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