题目
Problem
The diagram shows the curve with equation y=2x−x2 for 0≤x≤1, together with a set of n rectangles of width n1.
(a) By considering the sum of the areas of these rectangles, show that
∫01(2x−x2)dx<Un,
where
Un=(1+n1)(32−6n1).
[5]
(b) Use a similar method to find, in terms of n, a lower bound Ln for
∫01(2x−x2)dx.
[4]
(c) Show that
n→∞lim(Un−Ln)=0.
[2]
题目中文翻译
图中显示曲线 y=2x−x2,其中 0≤x≤1,以及一组宽度均为 n1 的 n 个矩形。
(a) 通过考虑这些矩形的面积之和,证明
∫01(2x−x2)dx<Un
其中
Un=(1+n1)(32−6n1)
(b) 使用类似方法,求关于 n 的积分
∫01(2x−x2)dx
的下界 Ln。
(c) 证明
n→∞lim(Un−Ln)=0
解答