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CIE 9709 2022 June Paper 12 Q6

A Level / CIE / P1

CIE 9709 2022 June Paper 12 Paper · Question 6

题目

Problem

The diagram shows the curve with equation y=5x12y=5x^{\frac12} and the line with equation y=2x+2y=2x+2.

Find the exact area of the shaded region which is bounded by the line and the curve.

(5)
题目中文翻译

图中所示为方程为 y=5x12y=5x^{\frac12} 的曲线和方程为 y=2x+2y=2x+2 的直线。

求由直线与曲线围成的阴影区域面积的精确值。

(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

解法一

思路

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联立后用 u=xu=\sqrt x 求交点。图中根式曲线在上,面积是纵坐标之差的积分。

答题过程

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The intersection equation 5x=2x+25\sqrt x=2x+2 becomes (2u−1)(u−2)=0(2u-1)(u-2)=0 for u=xu=\sqrt x. Thus the endpoints are x=1/4x=1/4 and x=4x=4. The area is

 ∫1/44(5x−2x−2) dx= [103x3/2−x2−2x]1/44= 83+748=4516.\begin{align*} &\,\int_{1/4}^4(5\sqrt x-2x-2)\,\mathrm{d}x\\ =&\,\Big[\frac{10}{3}x^{3/2}-x^2-2x\Big]_{1/4}^4\\ =&\,\frac83+\frac7{48}=\frac{45}{16}. \end{align*}

解法二

思路

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也可以取水平截面。把两条边界都写成 xx 关于 yy 的式子,用右端减左端。

答题过程

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The root curve is x=y2/25x=y^2/25 and the line is x=(y−2)/2x=(y-2)/2. Their intersections have y=5/2y=5/2 and y=10y=10. On this interval the line is the right boundary, so the area is

 ∫5/210(y−22−y225) dy= [y24−y−y375]5/210= 4516.\begin{align*} &\,\int_{5/2}^{10}\Big(\frac{y-2}{2}-\frac{y^2}{25}\Big)\,\mathrm{d}y\\ =&\,\Big[\frac{y^2}{4}-y-\frac{y^3}{75}\Big]_{5/2}^{10}\\ =&\,\frac{45}{16}. \end{align*}

解法三

思路

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也可先求曲线下方的面积,再扣掉直线下方的梯形面积。梯形的两条平行边是交点的纵坐标。

答题过程

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The intersections are (1/4,5/2)(1/4,5/2) and (4,10)(4,10). The area under the root curve between them is

[103x3/2]1/44=103(8−18)=1054.\Big[\frac{10}{3}x^{3/2}\Big]_{1/4}^4=\frac{10}{3}\Big(8-\frac18\Big)=\frac{105}{4}.

The area under the straight line is a trapezium of area

12(52+10)(4−14)=37516.\frac12\Big(\frac52+10\Big)\Big(4-\frac14\Big)=\frac{375}{16}.

Hence the shaded area is 105/4−375/16=45/16105/4-375/16=45/16.