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CIE 9709 2026 June Paper 15 Q10

A Level / CIE / P1

CIE 9709 2026 June Paper 15 Paper · Question 10

题目

Problem

The diagram shows the curve with equation y=x2−12x+37y=x^2-12x+37 and the line y=−4x+30y=-4x+30.

Find the area of the shaded region.

(7)
题目中文翻译

图中所示的是曲线 y=x2−12x+37y=x^2-12x+37 和直线 y=−4x+30y=-4x+30。

求阴影区域的面积。

(7)
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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联立直线和曲线求积分端点。图中直线在上,面积是两者纵坐标之差的定积分。

答题过程

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The intersection equation is x2−8x+7=(x−1)(x−7)=0x^2-8x+7=(x-1)(x-7)=0, giving x=1,7x=1,7. Between these values the line lies above the curve. The shaded area is

 ∫17(−x2+8x−7) dx= [−x33+4x2−7x]17= 36.\begin{align*} &\,\int_1^7(-x^2+8x-7)\,\mathrm{d}x\\ =&\,\Big[-\frac{x^3}{3}+4x^2-7x\Big]_1^7\\ =&\,36. \end{align*}

解法二

思路

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也可以先用梯形公式求直线下方的面积,再减去同一区间内曲线下方的面积。

答题过程

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The endpoints of the straight edge are (1,26)(1,26) and (7,2)(7,2). The area under the line on this interval is 12(26+2)(7−1)=84\frac12(26+2)(7-1)=84. The parabola is (x−6)2+1>0(x-6)^2+1>0, and the area under it is

[(x−6)33+x]17=48.\Big[\frac{(x-6)^3}{3}+x\Big]_1^7=48.

The difference is 84−48=3684-48=36.