Riemann Sums — Mark Scheme Patterns

Mark Allocation Overview

Question Type	Total Marks	M marks	A marks	B marks
Upper bound (rectangles)	4	2	2	0
Lower bound (rectangles)	4	2	2	0
Both bounds (combined)	8	4	4	0
Stirling approximation	8	4	4	0

Pattern: Upper/Lower Bound Using Rectangles (4 marks each)

• M1: $\Delta x = \frac{b-a}{n}$ and correct endpoint identification
• M1: Form sum $\sum f(x_i)\Delta x$ with correct $x_i$
• A1: Correct sigma expression (unsimplified)
• A1: Correct simplified expression in terms of $n$

评分细节

步骤	标记	示例
$\Delta x = \frac{1}{n}$ + 判断递增/递减	M1	$\Delta x = \frac{1}{n}$, 递增用右端点
$\sum_{i=1}^n f(x_i)\Delta x$	M1	$\sum_{i=1}^n \left(\frac{i}{n}\right)^2 \cdot \frac{1}{n}$
$\frac{1}{n^3}\sum i^2 = \frac{1}{n^3}\cdot\frac{n(n+1)(2n+1)}{6}$	A1	代入求和公式
$\frac{(n+1)(2n+1)}{6n^2}$	A1	化简完成

Pattern: Stirling Approximation (8 marks)

• M1: Express $\ln N! = \sum_{r=1}^N \ln r$
• M1: Set $f(x) = \ln x$, identify increasing function
• M1: Left endpoint inequality (lower bound for increasing function)
• M1: Right endpoint inequality (upper bound for increasing function)
• A1: Evaluate $\int_1^N \ln x\,dx = N\ln N - N + 1$
• A1: Correct lower bound inequality (may be for $\ln(N-1)!$)
• A1: Correct upper bound inequality
• A1: Final inequality for $\ln N!$

不等式的方向

对递增函数 $f(x) = \ln x$：

$\text{左端点：} \sum_{r=1}^{N-1} \ln r \le \int_1^N \ln x\,dx \le \sum_{r=2}^N \ln r$

$\text{即 } \ln(N-1)! \le N\ln N - N + 1 \le \ln N!$

关键扣分点

• 将左右端点弄反 → 扣 M1
• 求和下标范围错误（如从 $r=1$ 到 $N$ 而非 $N-1$）→ 扣 A1
• 最终不等式方向错误 → 扣 A1

Follow-Through Rules

• 如果求和公式用错（如 $\sum r^2$ 公式写错），后续使用可 ft
• 上下界判断错误但计算正确，最多扣 M1
• Stirling 近似中，积分计算错可 ft 后续不等式