📚 AQA A Level Maths Question 12: Integration by Substitution | AQA A Level 数学第12题：换元积分法
Question 12 on AQA A Level Mathematics Paper 1 frequently tests core Pure topics such as integration by substitution, definite integrals and verification by differentiation. This article works through a mark-scheme style Question 12 in full, showing every line of working a candidate needs to secure all 10 marks, and explaining what examiners actually reward at each stage.
AQA A Level 数学卷一的第 12 题经常考查纯数核心主题，例如换元积分法、定积分以及用微分验证积分结果。本文完整讲解一道评分标准风格的第 12 题，展示考生拿到全部 10 分所需的每一行过程，并说明考官在每个环节实际给分的依据。
1. The Question | 题目呈现
Below is a typical AQA-style Question 12 from the Pure Mathematics component of Paper 1. It carries 10 marks and is split into three parts: an indefinite integral found by substitution, a definite integral based on the same result, and a verification by differentiation.
下面是一道典型 AQA 风格的卷一纯数第 12 题。该题共 10 分，分为三个小问：用换元法求不定积分、在同一结果基础上的定积分，以及用微分进行验证。
Question 12
(a) Use the substitution u = 2x + 1 to find ∫ x√2x+1 dx. (5 marks)
(b) Hence evaluate ∫₀⁴ x√2x+1 dx, giving your answer in exact form. (3 marks)
(c) Verify your answer to part (a) by differentiation. (2 marks)
= 0 and x = 4, so an exact evaluation is expected rather than a rounded decimal. The word "Hence" reuse the antiderivative you obtained in part (a) instead of starting again from scratch.
第 (b) 问中积分限为 x = 0 与 x = 4，因此应给出精确值而不是四舍五入的小数。"Hence"（由此）一词是提示：你应当重用第 (a) 问得到的原函数，而不是从头重做。
2. What the Examiner Wants | 考官意图
AQA examiners award method marks (M) for a clearly stated substitution, a correct expression for dx, a fully rewritten integrand in terms of u, and a correct attempt at integrating the resulting powers. Accuracy marks (A) are then given for the correct antiderivative and for the exact final value. In part (c) the verification must actually reach the original integrand, not merely state that it works.
AQA 考官对以下步骤给方法分（M）：清晰写出换元式、正确求出 dx 的表达式、把被积函数完全改写为 u 的形式，以及对所得幂函数进行正确的积分尝试。随后给准确分（A）：正确的原函数以及精确的最终数值。第 (c) 问的验证必须真正推到原来的被积函数，而不是只写一句"验证成立"。
You must show every line of working, especially when changing the limits in a definite integral. Missing the limit change, mixing x and u in the same line, or dropping the constant of integration are among the most common reasons for lost marks on this question.
你必须展示每一行过程，尤其是在定积分换限时。漏掉换限、在同一行里混用 x 和 u、或丢掉积分常数，都是这道题最常见的丢分原因。
3. Part (a): Choosing the Substitution | 第 (a) 问：选择换元
2x + 1, so there is no need to guess. Write it down clearly as your first line of working and immediately differentiate it to obtain du/dx. This single line earns the first method mark and makes everything that follows mechanical.
题目已经给出换元式：u = 2x + 1，因此不需要猜测。把它作为第一行过程清楚写出，并立刻对它求导得到 du/dx。这一行就能拿到第一个方法分，并使后续步骤变成机械操作。
u = 2x + 1
du/dx = 2 → dx = du/2
Notice that du/dx is a constant here, which is exactly why this substitution simplifies the integral. If the derivative had still contained x, you would need to be far more careful about eliminating every x from the integrand.
注意这里的 du/dx 是一个常数，这正是该换元能简化积分的原因。如果导数中仍然含有 x，你就必须更小心地把被积函数中的每一个 x 都消掉。
4. Part (a): Rewriting the Integral in Terms of u | 第 (a) 问：把积分完全改写为 u
x = (u − 1)/2. Every component of the integral must now be converted: x becomes (u − 1)/2, √2x+1 becomes u1/2, and dx becomes du/2.
被积函数中含有一个额外的 x 因子，因此还必须把 x 用 u 表示。由 u = 2x + 1 变形可得 x = (u − 1)/2。现在积分的每一个组成部分都要转换：x 变成 (u − 1)/2，√2x+1 变成 u1/2，dx 变成 du/2。
x = (u − 1)/2
∫ x√2x+1 dx = ∫ ((u − 1)/2) × u1/2 × (du/2)
= ¼ ∫ (u − 1)u1/2 du = ¼ ∫ (u3/2 − u1/2) du
x and a factor of ½ from dx, so the overall multiplier is ¼. Also check the powers: u × u1/2 = u3/2, and 1 × u1/2 = u1/2.
仔细核对常数：x 提供因子 ½，dx 提供另一个因子 ½，因此整体系数是 ¼。再核对指数：u × u1/2 = u3/2，而 1 × u1/2 = u1/2。
5. Part (a): Integrating and Back-Substituting | 第 (a) 问：积分与回代
Both terms are now simple powers of u, so integrate using the standard rule that increases each power by one and divides by the new power.
现在两项都只是 u 的幂函数，直接使用标准法则积分即可：指数加一，再除以新的指数。
¼ ∫ (u3/2 − u1/2) du = ¼ ((2/5)u5/2 − (2/3)u3/2) + C
= (1/10)u5/2 − (1/6)u3/2 + C
Finally replace u by 2x + 1 to return to the original variable. The constant of integration must be present because this is an indefinite integral, and it should stay as C rather than being given any particular value.
最后把 u 换回 2x + 1，回到原变量。由于这是不定积分，积分常数必须写出，并且应保留为 C，而不是取任何特定数值。
∫ x√2x+1 dx = (1/10)(2x + 1)5/2 − (1/6)(2x + 1)3/2 + C
6. Part (b): Changing the Limits | 第 (b) 问：更换积分限
For a definite integral found by substitution there are two valid routes: change the limits into u-values, or integrate in u and then convert back to x before substituting the x-limits. The first route is usually faster and less error-prone, and it is the one AQA mark schemes tend to expect.
对用换元法求的定积分，有两条都可行的路径：把积分限换成 u 的值，或者先对 u 积分再换回 x，然后代入 x 的上下限。第一条路径通常更快、更不容易出错，也是 AQA 评分标准更倾向的方式。
u = 2x + 1
When x = 0: u = 2(0) + 1 = 1
When x = 4: u = 2(4) + 1 = 9
∫₀⁴ x√2x+1 dx = ¼ ∫₁⁹ (u3/2 − u1/2) du
Write the new limits explicitly on the integral sign. If you change the limits but forget to remove the x-values from the integral, or vice versa, the answer will almost certainly be wrong and you will lose the accuracy mark.
要把新的积分限明确写在积分号上。如果你换了限却仍保留 x 的上下限，或者反过来，答案几乎一定错误，并且会丢掉准确分。
7. Part (b): Exact Evaluation | 第 (b) 问：精确求值
Use the antiderivative in u, substitute the u-limits, and evaluate in exact fractions. Powers such as 95/2 and 93/2 are best handled as (√9)⁵ = 3⁵ = 243 and (√9)³ = 3³ = 27.
使用以 u 表示的原函数，代入 u 的上下限，并用精确分数求值。像 95/2 与 93/2 这样的幂，最好按 (√9)⁵ = 3⁵ = 243 和 (√9)³ = 3³ = 27 处理。
[(1/10)u5/2 − (1/6)u3/2]₁⁹
= (243/10 − 27/6) − (1/10 − 1/6)
= (243/10 − 9/2) + 1/15 = 99/5 + 1/15 = 298/15
The exact value is therefore 298/15, which is approximately 19.87. Leaving the answer as a rounded decimal would not satisfy "giving your answer in exact form" and would lose the final accuracy mark.
因此精确值为 298/15，约等于 19.87。若把答案写成四舍五入的小数，就不满足"以精确形式给出答案"的要求，会丢掉最后一个准确分。
8. Part (c): Verification by Differentiation | 第 (c) 问：用微分验证
Verification is a structured two-mark task: differentiate your answer and show that the result simplifies to the original integrand. Use the chain rule on each term, remembering that the derivative of (2x + 1)n is 2n(2x + 1)n−1.
验证是一个结构化的两分任务：对你的答案求导，并证明结果可以化简为原来的被积函数。对每一项使用链式法则，记住 (2x + 1)n 的导数是 2n(2x + 1)n−1。
d/dx[(1/10)(2x + 1)5/2] = (1/10)(5/2)(2x + 1)3/2 × 2 = (1/2)(2x + 1)3/2
d/dx[(1/6)(2x + 1)3/2] = (1/6)(3/2)(2x + 1)1/2 × 2 = (1/2)(2x + 1)1/2
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