题目
Problem
The diagram shows the curve y = x e − 1 4 x 2 y=x\mathrm{e}^{-\frac14x^2} y = x e − 4 1 x 2 , for x ≥ 0 x\ge0 x ≥ 0 , and its maximum point M M M .
(a) Find the exact coordinates of M M M .
(4)
(b) Using the substitution x = u x=\sqrt u x = u , or otherwise, find by integration the exact area of the shaded region bounded by the curve, the x x x -axis and the line x = 3 x=3 x = 3 .
(5)
题目中文翻译
图中显示了曲线 y = x e − 1 4 x 2 y=x\mathrm{e}^{-\frac14x^2} y = x e − 4 1 x 2 (x ≥ 0 x\ge0 x ≥ 0 )及其极大值点 M M M 。
(a) 求 M M M 的精确坐标。
(4)
(b) 使用代换 x = u x=\sqrt u x = u 或其他方法,通过积分求由该曲线、x x x 轴和直线 x = 3 x=3 x = 3 围成的阴影区域的精确面积。
(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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用乘积法则与链式法则求导。指数因子恒正,因此导数符号由 1 − x 2 / 2 1-x^2/2 1 − x 2 /2 决定。
答题过程
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Differentiating,
d y d x = e − x 2 / 4 + x e − x 2 / 4 ( − x 2 ) = e − x 2 / 4 ( 1 − x 2 2 ) . \begin{align*}
\frac{\mathrm{d}y}{\mathrm{d}x}
=&\,e^{-x^2/4}
+x e^{-x^2/4}\bigg(-\frac x2\bigg)\\[4mm]
=&\,e^{-x^2/4}\bigg(1-\frac{x^2}{2}\bigg).
\end{align*} d x d y = = e − x 2 /4 + x e − x 2 /4 ( − 2 x ) e − x 2 /4 ( 1 − 2 x 2 ) .
At a stationary point, x 2 = 2 x^2=2 x 2 = 2 . Since x ≥ 0 x\ge0 x ≥ 0 , x = 2 x=\sqrt2 x = 2 .
The derivative is positive for 0 ≤ x < 2 0\le x<\sqrt2 0 ≤ x < 2 and negative for x > 2 x>\sqrt2 x > 2 , so this point is a maximum. Its ordinate is
y = 2 e − 2 / 4 = 2 e − 1 / 2 . y=\sqrt2\,e^{-2/4}=\sqrt2\,e^{-1/2}. y = 2 e − 2/4 = 2 e − 1/2 .
Thus M = ( 2 , 2 e − 1 / 2 ) \boxed{M=(\sqrt2,\sqrt2\,e^{-1/2})} M = ( 2 , 2 e − 1/2 ) .
(b)
解法一
思路
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按题设代换 x = u x=\sqrt u x = u ,用 x d x = 1 2 d u x\,\mathrm{d}x=\tfrac12\,\mathrm{d}u x d x = 2 1 d u 消去根式。上下限也要换成 u u u 。
答题过程
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Since y ≥ 0 y\ge0 y ≥ 0 for x ≥ 0 x\ge0 x ≥ 0 , the area is
A = ∫ 0 3 x e − x 2 / 4 d x . A=\int_0^3 xe^{-x^2/4}\,\mathrm{d}x. A = ∫ 0 3 x e − x 2 /4 d x .
Put u = x 2 u=x^2 u = x 2 , so x = u x=\sqrt u x = u and d x = 1 2 u − 1 / 2 d u \mathrm{d}x=\tfrac12u^{-1/2}\,\mathrm{d}u d x = 2 1 u − 1/2 d u . The limits x = 0 , 3 x=0,3 x = 0 , 3 become u = 0 , 9 u=0,9 u = 0 , 9 . Hence
A = 1 2 ∫ 0 9 e − u / 4 d u = [ − 2 e − u / 4 ] 0 9 = 2 − 2 e − 9 / 4 . \begin{align*}
A=&\,\frac12\int_0^9 e^{-u/4}\,\mathrm{d}u\\[4mm]
=&\,\big[-2e^{-u/4}\big]_0^9\\[4mm]
=&\,\boxed{2-2e^{-9/4}}.
\end{align*} A = = = 2 1 ∫ 0 9 e − u /4 d u [ − 2 e − u /4 ] 0 9 2 − 2 e − 9/4 .
解法二
思路
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观察到指数的导数为 − x / 2 -x/2 − x /2 ,因此被积函数与 e − x 2 / 4 e^{-x^2/4} e − x 2 /4 的导数只差常数,可直接逆用链式法则。
答题过程
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The area is
A = ∫ 0 3 x e − x 2 / 4 d x . A=\int_0^3 xe^{-x^2/4}\,\mathrm{d}x. A = ∫ 0 3 x e − x 2 /4 d x .
Since
d d x e − x 2 / 4 = − x 2 e − x 2 / 4 , \frac{\mathrm{d}}{\mathrm{d}x}e^{-x^2/4}
=-\frac x2e^{-x^2/4}, d x d e − x 2 /4 = − 2 x e − x 2 /4 ,
an antiderivative of x e − x 2 / 4 xe^{-x^2/4} x e − x 2 /4 is − 2 e − x 2 / 4 -2e^{-x^2/4} − 2 e − x 2 /4 . Therefore
A = [ − 2 e − x 2 / 4 ] 0 3 = − 2 e − 9 / 4 + 2. A=\big[-2e^{-x^2/4}\big]_0^3
=-2e^{-9/4}+2. A = [ − 2 e − x 2 /4 ] 0 3 = − 2 e − 9/4 + 2.
Thus A = 2 − 2 e − 9 / 4 \boxed{A=2-2e^{-9/4}} A = 2 − 2 e − 9/4 .