Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9231 2025 Nov Paper 21 Q5

A Level / CIE / FP2

CIE 9231 2025 Nov Paper 21 Paper · Question 5

题目

Problem

The diagram shows the curve with equation y=13x3+xy = \frac{1}{3}x^3 + x for 0x10 \leq x \leq 1, together with a set of nn rectangles of width 1n\frac{1}{n}.

(a) By considering the sum of the areas of these rectangles, show that

01(13x3+x)dx<Un,\int_0^1 \bigg(\frac{1}{3}x^3 + x\bigg)\,\mathrm{d}x < U_n,

where

Un=112(1+1n)(7+1n).U_n = \frac{1}{12}\bigg(1 + \frac{1}{n}\bigg) \bigg(7 + \frac{1}{n}\bigg).
[5]

(b) Use a similar method to find, in terms of nn, a lower bound LnL_n for

01(13x3+x)dx.\int_0^1 \bigg(\frac{1}{3}x^3 + x\bigg)\,\mathrm{d}x.
[4]

(c) Show that limn(UnLn)=0\lim_{n \to \infty}(U_n - L_n) = 0.

[2]
题目中文翻译

题图显示曲线 y=13x3+xy = \frac{1}{3}x^3 + x,其中 0x10 \leq x \leq 1,以及一组宽为 1n\frac{1}{n}nn 个矩形。

(a) 通过考虑这些矩形面积之和,证明

01(13x3+x)dx<Un,\int_0^1 \bigg(\frac{1}{3}x^3 + x\bigg)\,\mathrm{d}x < U_n,

其中

Un=112(1+1n)(7+1n).U_n = \frac{1}{12}\bigg(1 + \frac{1}{n}\bigg) \bigg(7 + \frac{1}{n}\bigg).

(b) 用类似方法,求出关于 nn

01(13x3+x)dx\int_0^1 \bigg(\frac{1}{3}x^3 + x\bigg)\,\mathrm{d}x

的下界 LnL_n

(c) 证明 limn(UnLn)=0\lim_{n \to \infty}(U_n - L_n) = 0

解答