Example 6.9.2: Integration by Substitution | 例6.9.2：换元积分法

📚 Example 6.9.2: Integration by Substitution | 例6.9.2：换元积分法

This worked example, taken from the AQA A-Level Mathematics syllabus, demonstrates how to evaluate a definite integral using the substitution method. It reinforces the key skill of changing the variable and the limits of integration simultaneously, a technique frequently tested in exam papers.

这个来自 AQA A-Level 数学大纲的例题，展示了如何使用换元法计算定积分。它强化了同时改变变量和积分上下限这一关键技巧，该技巧在考试中经常出现。

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1. The Problem | 1. 题目

Use the substitution u = 1+x² to find the exact value of the definite integral:

使用代换 u = 1+x² 求定积分的精确值：

∫₀¹ x√1+x² dx

You are expected to express your final answer as an exact value, without using a calculator.

要求最终答案为精确值，不得使用计算器。

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2. The Idea of Substitution | 2. 换元的基本思想

Substitution is a powerful technique that simplifies an integrand by replacing a composite expression with a single variable. It is essentially the reverse of the chain rule for differentiation. When we choose u = 1+x² , we aim to transform the integral into a simpler one that we can integrate directly.

换元是一种有效的技巧，它用一个变量代替复合表达式来化简被积函数。它本质上就是微分链式法则的逆运算。当我们选择 u = 1+x² 时，目的是将积分转化为一个可以直接积分的更简单的形式。

u = 1 + x²

Notice that the derivative of u with respect to x is 2x . This is closely related to the factor x in the original integrand.

注意 u 关于 x 的导数是 2x ，这与原被积函数中的因子 x 密切相关。

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3. Choosing the Substitution | 3. 选择代换

In this example, the substitution is given. However, it is important to understand why it is effective. The integrand contains the composite function √1+x² times x . The derivative of the inner function 1+x² is 2x , so the factor x is, up to a constant, the derivative of the inner function.

在此例中，代换是给定的。然而，理解它为何有效很重要。被积函数包含复合函数 √1+x² 乘以 x 。内函数 1+x² 的导数是 2x ，所以因子 x 在常数意义下就是内函数的导数。

Therefore, we set:

因此，我们令：

u = 1 + x²

This choice will make the square root become simply √u .

这个选择会使平方根简化为 √u 。

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4. Changing the Variable: dx to du | 4. 改变变量：从 dx 到 du

To express the entire integral in terms of u , we must also replace dx . Differentiate both sides of the substitution:

为了用 u 表示整个积分，我们还需要替换 dx 。对代换等式两边求导：

du/dx = 2x

Rearranging to make dx the subject gives:

整理以 dx 为主项，得到：

dx = du / (2x)

Now substitute xdx in the original integral. Since dx = du/(2x) , we have:

现在替换原积分中的 xdx 。因为 dx = du/(2x) ，我们有：

x√1+x² dx = x · √u · du/(2x) = ½√u du

The factor x cancels, leaving an integrand in u only.

因子 x 消去，只留下关于 u 的被积函数。

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5. Changing the Limits | 5. 改变积分上下限

Because we now integrate with respect to u , the original limits in x must be converted to limits in u . Use the substitution u = 1+x² :

由于我们现在对 u 积分，原来的 x 上下限必须转化为 u 的上下限。使用代换 u = 1+x² ：

• Lower limit: when x = 0 , u = 1 + 0² = 1 .
• 上（下）限：当 x = 0 时， u = 1 + 0² = 1 。
• Upper limit: when x = 1 , u = 1 + 1² = 2 .
• 上限：当 x = 1 时， u = 1 + 1² = 2 。

Thus the integral becomes:

因此积分变为：

∫₀¹ x√1+x² dx = ∫₁² ½√u du

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6. Evaluating the Integral | 6. 计算积分

We can now integrate ½√u with respect to u . Write √u = u^1/2 . Then:

我们现在对 ½√u 关于 u 积分。将 √u = u^1/2 ，则：

∫ ½ u^1/2 du = ½ × (2/3) u^3/2 = (1/3) u^3/2

In the form without a fractional exponent:

换成不含分数指数的形式：

∫ ½√u du = (1/3) u√u

Now apply the limits u = 1 and u = 2 :

现在代入上下限 u = 1 和 u = 2 ：

[ (1/3) u√u ]₁² = (1/3) · 2√2 − (1/3) · 1√1

Simplify:

简化：

= (2√2 − 1) / 3

This is the exact value of the integral.

这就是该积分的精确值。

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7. Verifying the Answer by Differentiation | 7. 通过求导验证答案

To confirm the result, we can check that the original integrand is the derivative of 23(1+x²)^3/2 (the antiderivative before substitution). But since we used the u -substitution and changed the limits, it is simpler to differentiate the final expression with respect to x using the chain rule.

为了确认结果，我们可以检查原被积函数是 23(1+x²)^3/2 的导数（未代换前的不定积分）。但因为我们使用了代换并改变了上下限，更简单的方法是用链式法则对最终表达式求导。

Consider F(x) = 23(1+x²)^3/2 . Its derivative is:

考虑 F(x) = 23(1+x²)^3/2 ，其导数为：

F'(x) = 23 × (3/2) (1+x²)^1/2 × 2x = 2x√1+x²

Wait, this is 2x√1+x² , but our original integrand is x√1+x² . That suggests the antiderivative should be 13(1+x²)^3/2 because the derivative of 13(1+x²)^3/2 is:

等等，这是 2x√1+x² ，而原被积函数是 x√1+x² 。这表明原函数应为 13(1+x²)^3/2 ，因为 13(1+x²)^3/2 的导数为：

d/dx [ (1/3)(1+x²)^3/2 ] = (1/3) × (3/2)(1+x²)^1/2 × 2x = x√1+x²

Thus the definite integral equals F(1) - F(0) = 13(1+1²)^3/2 - 13(1+0²)^3/2 = 13(2√2 - 1) . This matches our result.

因此定积分等于 F(1) - F(0) = 13(1+1²)^3/2 - 13(1+0²)^3/2 = 13(2√2 - 1) ，与我们的结果一致。

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8. Common Mistakes | 8. 常见错误

• Forgetting to change the limits: When evaluating a definite integral with substitution, the limits must be converted to the new variable. Using the original x -limits in the u -integral will give a wrong answer.
• 忘记改变上下限：当用代换计算定积分时，必须将上下限转换为新变量。在原 u -积分中使用原来的 x -上下限会产生错误结果。
• Incorrectly differentiating the substitution: Make sure du/dx is correct. For u = 1+x² , du/dx = 2x , not x .
• 错误地求代换的导数：确保 du/dx 正确。对于 u = 1+x² ， du/dx = 2x ，而不是 x 。
• Omitting the dx : Always remember to replace dx completely in terms of du . Leaving dx in the transformed integral is incorrect.
• 遗漏 dx ：始终记住用 du 完全替换 dx 。在变换后的积分中保留 dx 是错误的。

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9. Practice Exercise | 9. 练习

Try the following similar problem on your own:

试自行完成以下类似问题：

Use the substitution u = 9 - x² to evaluate

使用代换 u = 9 - x² 计算

∫₀³ x√9 - x² dx

Answer: 9 . (Hint: after substitution the integral becomes -12∫₉⁰ √udu , then swap the limits to change the sign.)

答案： 9 。（提示：代换后积分变为 -12∫₉⁰ √udu ，然后交换上下限以改变符号。）

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10. Conclusion | 10. 总结

Substitution is a cornerstone technique in A-Level mathematics. This example shows the step-by-step process: select u , differentiate to find dx , change limits, integrate, and simplify. Mastering this method will help you solve a wide range of integrals confidently in the exam.

换元法是 A-Level 数学中的核心技术之一。本例题展示了完整步骤：选择 u ，求导以得到 dx ，改变上下限，积分，再化简。掌握这种方法将帮助你在考试中自信地解决各种积分问题。

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