题目
Problem
Let f ( x ) = 5 − x + 6 x 2 ( 3 − x ) ( 1 + 3 x 2 ) \mathrm{f}(x)=\dfrac{5-x+6x^2}{(3-x)(1+3x^2)} f ( x ) = ( 3 − x ) ( 1 + 3 x 2 ) 5 − x + 6 x 2 .
(a) Express f ( x ) \mathrm{f}(x) f ( x ) in partial fractions.
(5)
(b) Find the exact value of ∫ 0 1 f ( x ) d x \displaystyle\int_0^1\mathrm{f}(x)\,\mathrm{d}x ∫ 0 1 f ( x ) d x , simplifying your answer.
(5)
题目中文翻译
设 f ( x ) = 5 − x + 6 x 2 ( 3 − x ) ( 1 + 3 x 2 ) \mathrm{f}(x)=\dfrac{5-x+6x^2}{(3-x)(1+3x^2)} f ( x ) = ( 3 − x ) ( 1 + 3 x 2 ) 5 − x + 6 x 2 。
(a) 将 f ( x ) \mathrm{f}(x) f ( x ) 表示为部分分式。
(5)
(b) 求 ∫ 0 1 f ( x ) d x \displaystyle\int_0^1\mathrm{f}(x)\,\mathrm{d}x ∫ 0 1 f ( x ) d x 的精确值,并化简答案。
(5)
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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二次因子的分子设一次式,即使最终一次项为零也应从完整形式开始。清分母后用特殊值和系数比较求解。
答题过程
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Write
f ( x ) = A 3 − x + B x + C 1 + 3 x 2 . \mathrm f(x)=\frac{A}{3-x}+\frac{Bx+C}{1+3x^2}. f ( x ) = 3 − x A + 1 + 3 x 2 B x + C .
Then
5 − x + 6 x 2 = A ( 1 + 3 x 2 ) + ( B x + C ) ( 3 − x ) . \begin{align*}
5-x+6x^2=&\,A(1+3x^2)\\[4mm]
&\,\hspace{2pt}+(Bx+C)(3-x).
\end{align*} 5 − x + 6 x 2 = A ( 1 + 3 x 2 ) + ( B x + C ) ( 3 − x ) .
At x = 3 x=3 x = 3 , 56 = 28 A 56=28A 56 = 28 A , so A = 2 A=2 A = 2 .
Comparing constant terms gives 5 = A + 3 C 5=A+3C 5 = A + 3 C , so C = 1 C=1 C = 1 . Comparing coefficients of x 2 x^2 x 2 gives 6 = 3 A − B 6=3A-B 6 = 3 A − B , so B = 0 B=0 B = 0 .
Thus
f ( x ) = 2 3 − x + 1 1 + 3 x 2 . \boxed{\mathrm f(x)=\frac2{3-x}+\frac1{1+3x^2}}. f ( x ) = 3 − x 2 + 1 + 3 x 2 1 .
(b)
解法一
思路
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利用(a)分别得到对数与反正切积分,代入上下限后合并对数。
答题过程
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Using part (a), the derivative of tan − 1 ( 3 x ) \tan^{-1}(\sqrt3x) tan − 1 ( 3 x ) is
3 / ( 1 + 3 x 2 ) \sqrt3/(1+3x^2) 3 / ( 1 + 3 x 2 ) , so its integral coefficient is 1 / 3 1/\sqrt3 1/ 3 .
Also 3 − x > 0 3-x>0 3 − x > 0 on 0 ≤ x ≤ 1 0\le x\le1 0 ≤ x ≤ 1 . Hence
I = [ − 2 ln ( 3 − x ) + 1 3 tan − 1 ( 3 x ) ] 0 1 = − 2 ln 2 + 1 3 π 3 + 2 ln 3 = 2 ln 3 2 + π 3 3 . \begin{align*}
I
=&\,\bigg[-2\ln(3-x)
+\frac1{\sqrt3}\tan^{-1}(\sqrt3x)\bigg]_0^1\\[4mm]
=&\,-2\ln2+\frac1{\sqrt3}\frac\pi3+2\ln3\\[4mm]
=&\,\boxed{2\ln\frac32+\frac{\pi}{3\sqrt3}}.
\end{align*} I = = = [ − 2 ln ( 3 − x ) + 3 1 tan − 1 ( 3 x ) ] 0 1 − 2 ln 2 + 3 1 3 π + 2 ln 3 2 ln 2 3 + 3 3 π .