Chapter Review 8

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Question 1 - 曲线与直线间面积

题目：

The diagram shows the curve with equation \(y = 5 + 2x - x^2\) and the line with equation \(y = 2\). The curve and the line intersect at the points A and B.

a) Find the x-coordinates of A and B.

b) The shaded region R is bounded by the curve and the line. Find the area of R.

Question 1
图1：Question 1 - 曲线与直线交点示意图

解：

  1. a) 求交点：\(5 + 2x - x^2 = 2\)
  2. \(3 + 2x - x^2 = 0\)
  3. \(x^2 - 2x - 3 = 0\)
  4. \((x - 3)(x + 1) = 0\)
  5. \(x = -1\) 或 \(x = 3\)

b) Area = \(\int_{-1}^3 (5 + 2x - x^2) dx - \int_{-1}^3 2 dx\)

= \(\int_{-1}^3 (3 + 2x - x^2) dx\)

= \(\left[3x + x^2 - \frac{x^3}{3}\right]_{-1}^3\)

= \(9 + 9 - 9 - (-3 + 1 + \frac{1}{3}) = \frac{32}{3}\)

Question 3 - 三次函数面积计算

题目：

The diagram shows part of the curve with equation \(y = x^3 - 6x^2 + 9x\). The curve touches the x-axis at A and has a local maximum at B.

a) Show that the equation of the curve may be written as \(y = x(x - 3)^2\), and hence write down the coordinates of A.

b) Find the coordinates of B.

c) The shaded region R is bounded by the curve and the x-axis. Find the area of R.

Question 3
图2：Question 3 - 三次函数曲线图

解：

  1. a) 因式分解：\(y = x^3 - 6x^2 + 9x = x(x^2 - 6x + 9) = x(x - 3)^2\)
  2. A的坐标：(0, 0)
  3. b) 求导：\(\frac{dy}{dx} = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x - 1)(x - 3)\)
  4. 当x = 1时，y = 4，所以B的坐标：(1, 4)
  5. c) 面积计算：Area = \(\int_0^3 x(x - 3)^2 dx\)
  6. = \(\int_0^3 (x^3 - 6x^2 + 9x) dx\)
  7. = \(\left[\frac{x^4}{4} - 2x^3 + \frac{9x^2}{2}\right]_0^3\)
  8. = \(\frac{81}{4} - 54 + \frac{81}{2} = \frac{27}{4}\)

Question 5 - 分数幂函数面积

题目：

The diagram shows a sketch of the curve with equation \(y = 12x^{\frac{1}{2}} - x^{\frac{3}{2}}\) for \(0 \leq x \leq 12\).

a) Show that \(\frac{dy}{dx} = \frac{3}{2}x^{-\frac{1}{2}}(4 - x)\).

b) At the point B on the curve the tangent to the curve is parallel to the x-axis. Find the coordinates of the point B.

c) Find, to 3 significant figures, the area of the finite region bounded by the curve and the x-axis.

Question 5
图3：Question 5 - 分数幂函数曲线图

解：

  1. a) 求导：\(\frac{dy}{dx} = 12 \times \frac{1}{2}x^{-\frac{1}{2}} - \frac{3}{2}x^{\frac{1}{2}} = 6x^{-\frac{1}{2}} - \frac{3}{2}x^{\frac{1}{2}}\)
  2. = \(\frac{3}{2}x^{-\frac{1}{2}}(4 - x)\)
  3. b) 切线平行x轴：\(\frac{dy}{dx} = 0\)，所以x = 4
  4. 当x = 4时，y = 12 × 2 - 8 = 16，所以B的坐标：(4, 16)
  5. c) 面积计算：Area = \(\int_0^{12} (12x^{\frac{1}{2}} - x^{\frac{3}{2}}) dx\)
  6. = \(\left[12 \times \frac{2}{3}x^{\frac{3}{2}} - \frac{2}{5}x^{\frac{5}{2}}\right]_0^{12}\)
  7. = \(\left[8x^{\frac{3}{2}} - \frac{2}{5}x^{\frac{5}{2}}\right]_0^{12}\)
  8. = 8 × 41.57 - \(\frac{2}{5}\) × 497.66 ≈ 332.6 - 199.1 = 133.5

Question 6 - 曲线与水平线间面积

题目：

The diagram shows the curve C with equation \(y = x(8 - x)\) and the line with equation \(y = 12\) which meet at the points L and M.

a) Determine the coordinates of the point M.

b) Given that N is the foot of the perpendicular from M on to the x-axis, calculate the area of the shaded region which is bounded by NM, the curve C and the x-axis.

Question 6
图4：Question 6 - 曲线与水平线交点示意图

解：

  1. a) 求交点：\(x(8 - x) = 12\)
  2. \(8x - x^2 = 12\)
  3. \(x^2 - 8x + 12 = 0\)
  4. \((x - 2)(x - 6) = 0\)
  5. \(x = 2\) 或 \(x = 6\)，所以M的坐标：(6, 12)
  6. b) 面积计算：Area = 矩形面积 - \(\int_0^6 x(8 - x) dx\)
  7. = 6 × 12 - \(\int_0^6 (8x - x^2) dx\)
  8. = 72 - \(\left[4x^2 - \frac{x^3}{3}\right]_0^6\)
  9. = 72 - (144 - 72) = 0

Question 7 - 直线与曲线间面积

题目：

The diagram shows the line \(y = x - 1\) meeting the curve with equation \(y = (x - 1)(x - 5)\) at A and C. The curve meets the x-axis at A and B.

a) Write down the coordinates of A and B and find the coordinates of C.

b) Find the area of the shaded region bounded by the line, the curve and the x-axis.

Question 7
图5：Question 7 - 直线与曲线交点示意图

解：

  1. a) 坐标：A(1, 0)，B(5, 0)
  2. 求C：\(x - 1 = (x - 1)(x - 5)\)
  3. \(x - 1 = (x - 1)(x - 5)\)
  4. \(1 = x - 5\)（因为x ≠ 1）
  5. \(x = 6\)，所以C的坐标：(6, 5)
  6. b) 面积计算：Area = \(\int_1^6 (x - 1) dx - \int_1^6 (x - 1)(x - 5) dx\)
  7. = \(\int_1^6 (x - 1)(1 - (x - 5)) dx\)
  8. = \(\int_1^6 (x - 1)(6 - x) dx\)
  9. = \(\int_1^6 (6x - x^2 - 6 + x) dx\)
  10. = \(\int_1^6 (7x - x^2 - 6) dx\)
  11. = \(\left[\frac{7x^2}{2} - \frac{x^3}{3} - 6x\right]_1^6\)
  12. = 126 - 72 - 36 - (\(\frac{7}{2} - \frac{1}{3} - 6\)) = \(\frac{125}{6}\)

Question 14 - 梯形法则应用

题目：

a) Complete the table below, giving the missing values of y to 3 decimal places.

\(y = \frac{5}{x^2 + 1}\)

b) Use the trapezium rule, with all the values of y from your table, to find an approximate value for the area of R.

c) Use your answer from part b to find an approximate value for \(\int_0^3 (4 + \frac{5}{x^2 + 1}) dx\)

Question 14
图6：Question 14 - 梯形法则应用示意图

解：

  1. a) 完成表格：
  2. x = 1.5时，y = 5/(2.25 + 1) = 5/3.25 ≈ 1.538
  3. x = 2.5时，y = 5/(6.25 + 1) = 5/7.25 ≈ 0.690
  4. b) 梯形法则：h = 0.5
  5. Area ≈ ½ × 0.5 × (5 + 2(4 + 2.5 + 1.538 + 1 + 0.690) + 0.5)
  6. ≈ 0.25 × (5 + 19.456 + 0.5) = 6.239
  7. c) 复合积分：\(\int_0^3 (4 + \frac{5}{x^2 + 1}) dx = 12 + 6.239 = 18.239\)

Question 16 - 两条曲线间面积

题目：

The curve A with equation \(y = 8 + 4x - x^2\) and the curve B with equation \(y = x^2 - 4x + 8\) intersect at two points M and N.

a) Find the coordinates of M and the coordinates of N.

b) Use calculus to find the area of the finite region enclosed by A and B.

解：

  1. a) 求交点：\(8 + 4x - x^2 = x^2 - 4x + 8\)
  2. \(4x - x^2 = x^2 - 4x\)
  3. \(8x - 2x^2 = 0\)
  4. \(2x(4 - x) = 0\)
  5. \(x = 0\) 或 \(x = 4\)
  6. M(0, 8)，N(4, 8)
  7. b) 面积计算：Area = \(\int_0^4 (8 + 4x - x^2) - (x^2 - 4x + 8) dx\)
  8. = \(\int_0^4 (8x - 2x^2) dx\)
  9. = \(\left[4x^2 - \frac{2x^3}{3}\right]_0^4\)
  10. = 64 - \(\frac{128}{3} = \frac{64}{3}\)