题目
Problem
(a) Use the substitution u = 1 − x 2 u=1-x^2 u = 1 − x 2 ∫ x 1 − x 2 d x \displaystyle\int\frac{x}{\sqrt{1-x^2}}\,\mathrm{d}x ∫ 1 − x 2 x d x
(3)
The diagram shows the curve with equation for y = sin − 1 x y=\sin^{-1}x y = sin − 1 x 0 ≤ x ≤ 1 0\le x\le 1 0 ≤ x ≤ 1 n n n 1 n \dfrac{1}{n} n 1
(b) By considering the sum of the areas of these rectangles, show that
∑ r = 1 n 1 n sin − 1 ( r n ) ≥ 1 2 π − 1. \sum_{r=1}^{n}\frac{1}{n}\sin^{-1}\bigg(\frac{r}{n}\bigg)\ge\frac{1}{2}\pi-1. r = 1 ∑ n n 1 sin − 1 ( n r ) ≥ 2 1 π − 1. (5)
(c) Use a similar method to find, in terms of n n n ∑ r = 1 n 1 n sin − 1 ( r n ) \displaystyle\sum_{r=1}^{n}\frac{1}{n}\sin^{-1}\bigg(\frac{r}{n}\bigg) r = 1 ∑ n n 1 sin − 1 ( n r )
(3)
(d) Deduce the exact value of lim n → ∞ ∑ r = 1 n 1 n sin − 1 ( r n ) \displaystyle\lim_{n\to\infty}\sum_{r=1}^{n}\frac{1}{n}\sin^{-1}\bigg(\frac{r}{n}\bigg) n → ∞ lim r = 1 ∑ n n 1 sin − 1 ( n r )
(1)
题目中文翻译
(a) 用代换 u = 1 − x 2 u=1-x^2 u = 1 − x 2 ∫ x 1 − x 2 d x \displaystyle\int\frac{x}{\sqrt{1-x^2}}\,\mathrm{d}x ∫ 1 − x 2 x d x
(3)
图中显示了 0 ≤ x ≤ 1 0\le x\le 1 0 ≤ x ≤ 1 y = sin − 1 x y=\sin^{-1}x y = sin − 1 x n n n 1 n \dfrac{1}{n} n 1
(b) 通过考虑这些矩形的面积之和，证明
∑ r = 1 n 1 n sin − 1 ( r n ) ≥ 1 2 π − 1. \sum_{r=1}^{n}\frac{1}{n}\sin^{-1}\bigg(\frac{r}{n}\bigg)\ge\frac{1}{2}\pi-1. r = 1 ∑ n n 1 sin − 1 ( n r ) ≥ 2 1 π − 1. (5)
(c) 用类似的方法求 ∑ r = 1 n 1 n sin − 1 ( r n ) \displaystyle\sum_{r=1}^{n}\frac{1}{n}\sin^{-1}\bigg(\frac{r}{n}\bigg) r = 1 ∑ n n 1 sin − 1 ( n r ) n n n
(3)
(d) 由此推导出 lim n → ∞ ∑ r = 1 n 1 n sin − 1 ( r n ) \displaystyle\lim_{n\to\infty}\sum_{r=1}^{n}\frac{1}{n}\sin^{-1}\bigg(\frac{r}{n}\bigg) n → ∞ lim r = 1 ∑ n n 1 sin − 1 ( n r )
(1)
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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代换 u = 1 − x 2 u=1-x^2 u = 1 − x 2 x d x x\,\mathrm{d}x x d x − 1 2 d u -\frac12\mathrm{d}u − 2 1 d u
答题过程
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With u = 1 − x 2 u=1-x^2 u = 1 − x 2 d u = − 2 x d x \mathrm{d}u=-2x\,\mathrm{d}x d u = − 2 x d x
∫ x 1 − x 2 d x = − 1 2 ∫ u − 1 / 2 d u = − 1 − x 2 + C . \int\frac{x}{\sqrt{1-x^2}}\,\mathrm{d}x
=-\frac12\int u^{-1/2}\,\mathrm{d}u
=\boxed{-\sqrt{1-x^2}+C}. ∫ 1 − x 2 x d x = − 2 1 ∫ u − 1/2 d u = − 1 − x 2 + C .
(b)
解法一
思路
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反正弦在 [ 0 , 1 ] [0,1] [ 0 , 1 ] x arcsin x x\arcsin x x arcsin x
答题过程
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The right-endpoint rectangle sum for f ( x ) = arcsin x f(x)=\arcsin x f ( x ) = arcsin x
S n = 1 n [ arcsin 1 n + arcsin 2 n + arcsin 3 n + ⋯ + arcsin n − 1 n + arcsin 1 ] = 1 n ∑ r = 1 n arcsin ( r / n ) ≥ ∫ 0 1 arcsin x d x . \begin{align*}
S_n=&\,\frac1n\bigg[\arcsin\frac1n+\arcsin\frac2n+\arcsin\frac3n+\cdots\\
&\,\hspace{2pt}+\arcsin\frac{n-1}{n}+\arcsin1\bigg]\\
=&\,\frac1n\sum_{r=1}^n\arcsin(r/n)\\
\ge&\,\int_0^1\arcsin x\,\mathrm{d}x.
\end{align*} S n = = ≥ n 1 [ arcsin n 1 + arcsin n 2 + arcsin n 3 + ⋯ + arcsin n n − 1 + arcsin 1 ] n 1 r = 1 ∑ n arcsin ( r / n ) ∫ 0 1 arcsin x d x . Take u = arcsin x u=\arcsin x u = arcsin x d v = d x \mathrm{d}v=\mathrm{d}x d v = d x d u = d x / 1 − x 2 \mathrm{d}u=\mathrm{d}x/\sqrt{1-x^2} d u = d x / 1 − x 2 v = x v=x v = x
∫ arcsin x d x = x arcsin x − ∫ x 1 − x 2 d x . \int\arcsin x\,\mathrm{d}x=x\arcsin x-\int\frac{x}{\sqrt{1-x^2}}\,\mathrm{d}x. ∫ arcsin x d x = x arcsin x − ∫ 1 − x 2 x d x . Part (a) gives the antiderivative x arcsin x + 1 − x 2 x\arcsin x+\sqrt{1-x^2} x arcsin x + 1 − x 2 0 < x < 1 0<x<1 0 < x < 1
∫ 0 1 arcsin x d x = [ x arcsin x + 1 − x 2 ] 0 1 = π 2 − 1. \begin{align*}
\int_0^1\arcsin x\,\mathrm{d}x
=&\,[x\arcsin x+\sqrt{1-x^2}]_0^1\\
=&\,\frac\pi2-1.
\end{align*} ∫ 0 1 arcsin x d x = = [ x arcsin x + 1 − x 2 ] 0 1 2 π − 1. This gives the stated lower bound for S n S_n S n
(c)
解法一
思路
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左端点矩形给出积分的下界。它们与右端点矩形只在首尾取样点不同，右端点和多出 π / ( 2 n ) \pi/(2n) π / ( 2 n )
答题过程
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The left-endpoint sum is 1 n ∑ r = 0 n − 1 arcsin ( r / n ) \frac1n\sum_{r=0}^{n-1}\arcsin(r/n) n 1 ∑ r = 0 n − 1 arcsin ( r / n )
S n − arcsin 1 − arcsin 0 n = S n − π 2 n . S_n-\frac{\arcsin1-\arcsin0}{n}=S_n-\frac{\pi}{2n}. S n − n arcsin 1 − arcsin 0 = S n − 2 n π . Thus S n − π 2 n ≤ π 2 − 1 S_n-\frac{\pi}{2n}\le\frac\pi2-1 S n − 2 n π ≤ 2 π − 1 π 2 − 1 + π 2 n \boxed{\frac\pi2-1+\frac{\pi}{2n}} 2 π − 1 + 2 n π
(d)
解法一
思路
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上下界都趋于 π 2 − 1 \frac\pi2-1 2 π − 1
答题过程
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Since π 2 − 1 ≤ S n ≤ π 2 − 1 + π 2 n \frac\pi2-1\le S_n\le\frac\pi2-1+\frac{\pi}{2n} 2 π − 1 ≤ S n ≤ 2 π − 1 + 2 n π
lim n → ∞ ∑ r = 1 n 1 n arcsin ( r / n ) = π 2 − 1 . \lim_{n\to\infty}\sum_{r=1}^{n}\frac1n\arcsin(r/n)=\boxed{\frac\pi2-1}. n → ∞ lim r = 1 ∑ n n 1 arcsin ( r / n ) = 2 π − 1 . 
