题目
Problem
The diagram shows the curve with equation y=2x for 0≤x≤1, together with a set of N rectangles each of width N1.
(a) By considering the sum of the areas of these rectangles, show that ∫012xdx<UN, where
UN=N(2N1−1)2N1.
(4)
(b) Use a similar method to find, in terms of N, a lower bound LN for ∫012xdx.
(4)
(c) Find the least value of N such that UN−LN<10−4.
(2)
题目中文翻译
图中显示了方程为 y=2x 的曲线(0≤x≤1),以及一组 N 个矩形,每个矩形的宽均为 N1。
(a) 通过考虑这些矩形的面积之和,证明 ∫012xdx<UN,其中
UN=N(2N1−1)2N1.
(4)
(b) 用类似的方法求 ∫012xdx 的一个下界 LN,用含 N 的式子表示。
(4)
(c) 求满足 UN−LN<10−4 的 N 的最小值。
(2)
原卷通用作答要求
Answer all questions. You should use a calculator where appropriate.
You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。
解答
(a)
解法一
思路
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指数函数递增,题图取右端点,所以面积和给出严格上界。相邻矩形高度的比是 21/N。
答题过程
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Let r=21/N, so rN=2. The right-endpoint rectangle sum is
UN===N1(r+r2+r3+⋯+rN−1+rN)N1⋅r−1r(rN−1)N(21/N−1)21/N.
Since 2x is strictly increasing, the curve lies below the top of each rectangle on the interval interior, so ∫012xdx<UN.
(b)
解法一
思路
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改用左端点。矩形高度从 1 开始,共 N 项,仍可用等比求和。
答题过程
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The left-endpoint rectangles give a lower bound since 2x is increasing. Their area sum is
LN===N1(1+r+r2+⋯+rN−2+rN−1)N1⋅r−1rN−1N(21/N−1)1.
The rectangles lie strictly below the curve on each interval interior, so LN<∫012xdx.
(c)
解法一
思路
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上下界相减后约掉 21/N−1。题设是不含等号的严格不等式,取整数时需再进一位。
答题过程
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We have
UN−LN=N(21/N−1)21/N−1=N1.
The inequality 1/N<10−4 gives N>10000, so the least integer is 10001.