Exercise 8C - x轴下方面积练习
以下练习涵盖x轴下方面积计算、跨轴区域面积和综合应用问题。
Question 1 - 基础x轴下方面积
Sketch the following and find the total area of the finite region or regions bounded by the curves and the x-axis:
a) \(y = x(x + 2)\)
b) \(y = (x + 1)(x - 4)\)
c) \(y = (x + 3)x(x - 3)\)
d) \(y = x^2(x - 2)\)
e) \(y = x(x - 2)(x - 5)\)
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提示: 先画图找出与x轴的交点,然后计算积分并取绝对值。
解答:
a) 交点:x = 0, -2
\(\int_{-2}^0 x(x + 2) dx = \int_{-2}^0 (x^2 + 2x) dx = \left[\frac{x^3}{3} + x^2\right]_{-2}^0 = 0 - \left(-\frac{8}{3} + 4\right) = -\frac{4}{3}\)
面积 = \(\frac{4}{3}\)
b) 交点:x = -1, 4
\(\int_{-1}^4 (x + 1)(x - 4) dx = \int_{-1}^4 (x^2 - 3x - 4) dx = \left[\frac{x^3}{3} - \frac{3x^2}{2} - 4x\right]_{-1}^4 = -\frac{125}{6}\)
面积 = \(\frac{125}{6}\)
c) 交点:x = -3, 0, 3
需要分段计算:\(\int_{-3}^0 + \int_0^3\)
面积 = \(\frac{81}{4} + \frac{81}{4} = \frac{81}{2}\)
Question 2 - 跨轴区域面积(配图)
The graph shows a sketch of part of the curve C with equation \(y = x(x + 3)(2 - x)\).
The curve C crosses the x-axis at the origin O and at points A and B.
a) Find the coordinates of point A and point B. (2 marks)
b) Find the area of the finite region bounded by C and the x-axis. (4 marks)
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提示: 先求与x轴的交点,然后分段计算面积(注意跨轴情况)。
解答:
a) 当 y = 0 时:\(x(x + 3)(2 - x) = 0\)
x = 0, -3, 2
所以 A(-3, 0), B(2, 0)
b) 需要分段计算:
\(\int_{-3}^0 x(x + 3)(2 - x) dx + \int_0^2 x(x + 3)(2 - x) dx\)
展开:\(y = -x^3 - x^2 + 6x\)
\(\int_{-3}^0 = \left[-\frac{x^4}{4} - \frac{x^3}{3} + 3x^2\right]_{-3}^0 = \frac{45}{4}\)
\(\int_0^2 = \left[-\frac{x^4}{4} - \frac{x^3}{3} + 3x^2\right]_0^2 = \frac{20}{3}\)
总面积 = \(\frac{45}{4} + \frac{20}{3} = \frac{215}{12}\)
Question 3 - 含未知参数的面积
The shaded area under the graph of the function \(f(x) = 3x^2 - 2x + 2\), bounded by the curve, the x-axis and the lines x = 0 and x = k, is 8. Work out the value of k.
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提示: 建立积分方程 \(\int_0^k (3x^2 - 2x + 2) dx = 8\),然后求解k。
解答:
\(\int_0^k (3x^2 - 2x + 2) dx = 8\)
\(\left[x^3 - x^2 + 2x\right]_0^k = 8\)
\(k^3 - k^2 + 2k = 8\)
\(k^3 - k^2 + 2k - 8 = 0\)
尝试 k = 2:
\(8 - 4 + 4 - 8 = 0\) ✓
所以 k = 2
Question 4 - 综合应用
The finite region R is bounded by the x-axis and the curve with equation \(y = -x^2 + 2x + 3, x \geq 0\).
The curve meets the x-axis at points A and B.
a) Find the coordinates of point A and point B. (2 marks)
b) Find the area of the region R. (4 marks)
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提示: 先求与x轴的交点,然后计算积分面积。
解答:
a) 当 y = 0 时:\(-x^2 + 2x + 3 = 0\)
\(x^2 - 2x - 3 = 0\)
\((x - 3)(x + 1) = 0\)
x = 3, -1
由于 x ≥ 0,所以 A(0, 3), B(3, 0)
b) \(\int_0^3 (-x^2 + 2x + 3) dx\)
\(= \left[-\frac{x^3}{3} + x^2 + 3x\right]_0^3\)
\(= (-9 + 9 + 9) - 0 = 9\)
面积 = 9
Question 5 - 微积分应用
The graph shows part of the curve C with equation \(y = x^2(2 - x)\).
The region R, shown shaded, is bounded by C and the x-axis. Use calculus to find the exact area of R. (5 marks)
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提示: 先找出与x轴的交点,然后计算定积分。
解答:
当 y = 0 时:\(x^2(2 - x) = 0\)
x = 0, 2
\(\int_0^2 x^2(2 - x) dx\)
\(= \int_0^2 (2x^2 - x^3) dx\)
\(= \left[\frac{2x^3}{3} - \frac{x^4}{4}\right]_0^2\)
\(= \left(\frac{16}{3} - 4\right) - 0\)
\(= \frac{16}{3} - \frac{12}{3} = \frac{4}{3}\)
面积 = \(\frac{4}{3}\)