图中所示的是曲线 $y=-x^3+9x^2-15x-3$ 和一条直线。该直线经过曲线的两个驻点。

(a) 证明该直线的方程为

(b) 验证当 $x=3$ 时，曲线也与该直线相交。

(c) 阴影区域由该曲线和直线围成。

求阴影区域的面积。

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。

用导数求两个驻点，再用两点式求过它们的直线。

The derivative is $-3x^2+18x-15=-3(x-1)(x-5)$, so the stationary points have $x=1,5$. Their coordinates are $(1,-10)$ and $(5,22)$. The gradient of the line through them is $(22-(-10))/(5-1)=8$, and its equation is $y+10=8(x-1)$, or $y=8x-18$.

分别代入 $x=3$，确认直线与曲线有同一个纵坐标。

At $x=3$ the line gives $8(3)-18=6$ and the curve gives $-27+81-45-3=6$. Thus both pass through $(3,6)$.

图中只对右侧区域着色。用 $x=3$ 和 $x=5$ 作边界，计算曲线减直线的定积分。

On $3<x<5$, the curve lies above the line since their difference is $-(x-1)(x-3)(x-5)>0$. The shaded area is

也可以分别求曲线下方的积分和直线下方的梯形面积，再相减。

The straight edge joins $(3,6)$ and $(5,22)$. Its trapezium above the $x$-axis has area $(6+22)(5-3)/2=28$. The area under the curve on the same interval is

Subtracting gives the shaded area $32-28=4$.