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。

1. The Problem | 1. 题目

Use the substitution \( u = 1+x^2 \) to find the exact value of the definite integral:

使用代换 \( u = 1+x^2 \) 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。

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 \( u = 1+x^2 \), 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 \( u = 1+x^2 \) 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。

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 \( \sqrt{1+x^2} \) times \( x \). The derivative of the inner function \( 1+x^2 \) 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 \( \sqrt{1+x^2} \) 乘以 \( x \)。内函数 \( 1+x^2 \) 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 \( \sqrt{u} \).

这个选择会使平方根简化为 \( \sqrt{u} \)。

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

现在替换原积分中的 \( x\,dx \)。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。

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 \( u = 1+x^2 \):

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 \( u = 1+x^2 \)：

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

Thus the integral becomes:

因此积分变为：

∫₀¹ x√(1+x²) dx = ∫₁² ½√u du

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。

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 \( \frac{2}{3}(1+x^2)^{3/2} \) (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 \( \frac{2}{3}(1+x^2)^{3/2} \) 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 \( F(x) = \frac{2}{3}(1+x^2)^{3/2} \). Its derivative is:

考虑 \( F(x) = \frac{2}{3}(1+x^2)^{3/2} \)，其导数为：

F'(x) = \frac{2}{3} × (3/2) (1+x^2)^{1/2} × 2x = 2x√(1+x^2)

Wait, this is \( 2x\sqrt{1+x^2} \), but our original integrand is \( x\sqrt{1+x^2} \). That suggests the antiderivative should be \( \frac{1}{3}(1+x^2)^{3/2} \) because the derivative of \( \frac{1}{3}(1+x^2)^{3/2} \) is:

等等，这是 \( 2x\sqrt{1+x^2} \)，而原被积函数是 \( x\sqrt{1+x^2} \)。这表明原函数应为 \( \frac{1}{3}(1+x^2)^{3/2} \)，因为 \( \frac{1}{3}(1+x^2)^{3/2} \) 的导数为：

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

Thus the definite integral equals \( F(1) – F(0) = \frac{1}{3}(1+1^2)^{3/2} – \frac{1}{3}(1+0^2)^{3/2} = \frac{1}{3}(2√2 – 1) \). This matches our result.

因此定积分等于 \( F(1) – F(0) = \frac{1}{3}(1+1^2)^{3/2} – \frac{1}{3}(1+0^2)^{3/2} = \frac{1}{3}(2√2 – 1) \)，与我们的结果一致。

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 \( u = 1+x^2 \), \( du/dx = 2x \), not \( x \).
错误地求代换的导数：确保 \( du/dx \) 正确。对于 \( u = 1+x^2 \)，\( du/dx = 2x \)，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。

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 \( u = 9 – x^2 \) to evaluate

使用代换 \( u = 9 – x^2 \) 计算

∫₀³ x√(9 – x²) dx

Answer: \( 9 \). (Hint: after substitution the integral becomes \( -\frac{1}{2}\int_{9}^{0} \sqrt{u}\,du \), then swap the limits to change the sign.)

答案：\( 9 \)。（提示：代换后积分变为 \( -\frac{1}{2}\int_{9}^{0} \sqrt{u}\,du \)，然后交换上下限以改变符号。）

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。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

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