题目
Problem
The equation of a curve is y=xtan−13x for −π≤x≤π.
(a) Find the exact value of the gradient of the curve when x=31.
(4)
(b) Find the exact value of ∫031ydx.
(6)
题目中文翻译
一条曲线的方程为 y=xtan−13x,其中 −π≤x≤π。
(a) 求当 x=31 时曲线斜率的精确值。
(4)
(b) 求 ∫031ydx 的精确值。
(6)
原卷通用作答要求
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解答
(a)
解法一
思路
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使用乘积法则与反正切的链式求导。最后代入 x=1/3。
答题过程
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Differentiating gives
dxdy=arctan(3x)+1+9x23x.
At x=1/3, the gradient is π/4+1/2.
(b)
解法一
思路
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以反正切为需要求导的部分,使用分部积分。剩余有理式先作除法,再积分。
答题过程
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Set u=arctan(3x) and dv=xdx, so du=3/(1+9x2)dx and v=x2/2. Integration by parts gives
I==∫01/3xarctan(3x)dx[2x2arctan(3x)]01/3−23∫01/31+9x2x2dx.
The remaining fraction is
1+9x2x2=91−9(1+9x2)1.
Using ∫(1+9x2)−1dx=31arctan(3x)+C, we obtain
I=[2x2arctan(3x)−6x+181arctan(3x)]01/3.
Evaluating at 1/3 and 0 gives
72π−181+72π=36π−181.