Exercise 8B - 曲线下面积计算练习
以下练习涵盖基础面积计算、图形分析和含参数的面积问题。
Question 1 - 基础面积计算
Find the area between the curve with equation \(y = f(x)\), the x-axis and the lines \(x = a\) and \(x = b\) in each of the following cases:
a) \(f(x) = -3x^2 + 17x - 10\); \(a = 1, b = 3\)
b) \(f(x) = 2x^3 + 7x^2 - 4x\); \(a = -3, b = -1\)
c) \(f(x) = -x^4 + 7x^3 - 11x^2 + 5x\); \(a = 0, b = 4\)
d) \(f(x) = \frac{8}{x^2}\); \(a = -4, b = -1\)
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提示: 直接使用面积公式 \(\text{Area} = \int_a^b f(x) dx\),注意正负性。
解答:
a) \(\text{Area} = \int_1^3 (-3x^2 + 17x - 10) dx = \left[-x^3 + \frac{17x^2}{2} - 10x\right]_1^3 = \frac{20}{3}\)
b) \(\text{Area} = \int_{-3}^{-1} (2x^3 + 7x^2 - 4x) dx = \left[\frac{x^4}{2} + \frac{7x^3}{3} - 2x^2\right]_{-3}^{-1} = \frac{32}{3}\)
c) \(\text{Area} = \int_0^4 (-x^4 + 7x^3 - 11x^2 + 5x) dx = \left[-\frac{x^5}{5} + \frac{7x^4}{4} - \frac{11x^3}{3} + \frac{5x^2}{2}\right]_0^4 = \frac{32}{15}\)
d) \(\text{Area} = \int_{-4}^{-1} \frac{8}{x^2} dx = \left[-\frac{8}{x}\right]_{-4}^{-1} = 8 - 2 = 6\)
Question 2 - 图形分析(配图)
The sketch shows part of the curve with equation \(y = x(x^2 - 4)\). Find the area of the shaded region.
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提示: 从图形确定积分区间,注意曲线在x轴下方时面积为负值。
解答:
从图形可以看出,阴影区域在 \(x = 0\) 到 \(x = 2\) 之间。
\(\text{Area} = \int_0^2 x(x^2 - 4) dx = \int_0^2 (x^3 - 4x) dx\)
\(= \left[\frac{x^4}{4} - 2x^2\right]_0^2 = (4 - 8) - 0 = -4\)
由于曲线在x轴下方,面积为 \(| -4 | = 4\)
Question 3 - 有界区域(配图)
The diagram shows a sketch of the curve with equation \(y = 3x + \frac{6}{x^2} - 5, x > 0\).
The region R is bounded by the curve, the x-axis and the lines \(x = 1\) and \(x = 3\). Find the area of R.
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提示: 直接计算给定区间内的定积分。
解答:
\(\text{Area} = \int_1^3 \left(3x + \frac{6}{x^2} - 5\right) dx\)
\(= \left[\frac{3x^2}{2} - \frac{6}{x} - 5x\right]_1^3\)
\(= \left(\frac{27}{2} - 2 - 15\right) - \left(\frac{3}{2} - 6 - 5\right)\)
\(= \frac{27}{2} - 17 - \frac{3}{2} + 11 = 12 - 6 = 6\)
Question 4 - 有限区域面积
Find the area of the finite region between the curve with equation \(y = (3 - x)(1 + x)\) and the x-axis.
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提示: 先因式分解找交点,然后计算面积。
解答:
\(y = (3 - x)(1 + x) = 3 + 2x - x^2\)
与x轴的交点:\(3 + 2x - x^2 = 0\)
\((3 - x)(1 + x) = 0\),所以 \(x = 3\) 或 \(x = -1\)
\(\text{Area} = \int_{-1}^3 (3 + 2x - x^2) dx\)
\(= \left[3x + x^2 - \frac{x^3}{3}\right]_{-1}^3\)
\(= (9 + 9 - 9) - (-3 + 1 + \frac{1}{3}) = 9 - (-\frac{5}{3}) = \frac{32}{3}\)
Question 5 - 三次函数面积
Find the area of the finite region between the curve with equation \(y = x(x - 4)^2\) and the x-axis.
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提示: 展开函数找交点,注意曲线可能穿过x轴。
解答:
\(y = x(x - 4)^2 = x(x^2 - 8x + 16) = x^3 - 8x^2 + 16x\)
与x轴的交点:\(x(x - 4)^2 = 0\)
所以 \(x = 0\) 或 \(x = 4\)(重根)
\(\text{Area} = \int_0^4 (x^3 - 8x^2 + 16x) dx\)
\(= \left[\frac{x^4}{4} - \frac{8x^3}{3} + 8x^2\right]_0^4\)
\(= (64 - \frac{512}{3} + 128) - 0 = 192 - \frac{512}{3} = \frac{64}{3}\)
Question 7 - 含未知参数(配图)
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: 建立含k的积分方程,然后求解k的值。
解答:
\(\text{Area} = \int_0^k (3x^2 - 2x + 2) dx = 8\)
\(\left[x^3 - x^2 + 2x\right]_0^k = 8\)
\(k^3 - k^2 + 2k = 8\)
\(k^3 - k^2 + 2k - 8 = 0\)
尝试 \(k = 2\):\(8 - 4 + 4 - 8 = 0\) ✓
因此 \(k = 2\)