Exercise 8A - 定积分基础练习
以下练习涵盖定积分的基本计算、含参数积分和实际应用问题。
Question 1 - 基础定积分计算
计算以下定积分：
a) \(\int_2^5 x^3 dx\)
b) \(\int_1^3 x^4 dx\)
c) \(\int_0^4 \sqrt{x} dx\)
d) \(\int_1^3 \frac{3}{x^2} dx\)
提示：使用基本积分公式，注意定积分不需要常数项。
解答：
a) \(\int_2^5 x^3 dx = \left[\frac{x^4}{4}\right]_2^5 = \frac{625}{4} - \frac{16}{4} = \frac{609}{4}\)
b) \(\int_1^3 x^4 dx = \left[\frac{x^5}{5}\right]_1^3 = \frac{243}{5} - \frac{1}{5} = \frac{242}{5}\)
c) \(\int_0^4 \sqrt{x} dx = \left[\frac{2}{3}x^{\frac{3}{2}}\right]_0^4 = \frac{2}{3} \cdot 8 = \frac{16}{3}\)
d) \(\int_1^3 \frac{3}{x^2} dx = \left[-\frac{3}{x}\right]_1^3 = -1 - (-3) = 2\)
Question 2 - 复合函数积分
计算以下定积分：
a) \(\int_1^2 \left(\frac{2}{x^3} + 3x\right) dx\)
b) \(\int_0^2 (2x^3 - 4x + 5) dx\)
c) \(\int_4^9 \left(\sqrt{x} - \frac{6}{x^2}\right) dx\)
d) \(\int_1^8 (x^{-\frac{1}{3}} + 2x - 1) dx\)
提示：分别对每一项积分，注意负指数和分数指数的处理。
解答：
a) \(\int_1^2 \left(\frac{2}{x^3} + 3x\right) dx = \left[-\frac{1}{x^2} + \frac{3x^2}{2}\right]_1^2 = \left(-\frac{1}{4} + 6\right) - \left(-1 + \frac{3}{2}\right) = \frac{19}{4}\)
b) \(\int_0^2 (2x^3 - 4x + 5) dx = \left[\frac{x^4}{2} - 2x^2 + 5x\right]_0^2 = (8 - 8 + 10) - 0 = 10\)
c) \(\int_4^9 \left(\sqrt{x} - \frac{6}{x^2}\right) dx = \left[\frac{2}{3}x^{\frac{3}{2}} + \frac{6}{x}\right]_4^9 = (18 + \frac{2}{3}) - (\frac{16}{3} + \frac{3}{2}) = \frac{35}{6}\)
d) \(\int_1^8 (x^{-\frac{1}{3}} + 2x - 1) dx = \left[\frac{3}{2}x^{\frac{2}{3}} + x^2 - x\right]_1^8 = (6 + 64 - 8) - (\frac{3}{2} + 1 - 1) = \frac{119}{2}\)
Question 4 - 含参数积分
Given that \(A\) is a constant and \(\int_1^4 (6\sqrt{x} - A) dx = A^2\), show that there are two possible values for \(A\) and find these values.
提示：先计算定积分得到关于A的表达式，然后建立方程求解。
解答：
\(\int_1^4 (6\sqrt{x} - A) dx = \left[4x^{\frac{3}{2}} - Ax\right]_1^4\)
\(= (32 - 4A) - (4 - A)\)
\(= 28 - 3A\)
根据题意：\(28 - 3A = A^2\)
\(A^2 + 3A - 28 = 0\)
\((A + 7)(A - 4) = 0\)
因此：\(A = -7\) 或 \(A = 4\)
Question 6 - 分数形式答案
Evaluate \(\int_4^{12} \frac{2}{\sqrt{x}} dx\), giving your answer in the form \(a + b\sqrt{3}\), where \(a\) and \(b\) are integers. (4 marks)
提示：注意 \(\frac{1}{\sqrt{x}} = x^{-\frac{1}{2}}\)，积分后需要化简根式。
解答：
\(\int_4^{12} \frac{2}{\sqrt{x}} dx = \int_4^{12} 2x^{-\frac{1}{2}} dx\)
\(= \left[4x^{\frac{1}{2}}\right]_4^{12}\)
\(= 4\sqrt{12} - 4\sqrt{4}\)
\(= 4 \cdot 2\sqrt{3} - 4 \cdot 2\)
\(= 8\sqrt{3} - 8\)
因此：\(a = -8, b = 8\)
Question 7 - 含未知上限的积分
Given that \(\int_1^k \frac{1}{\sqrt{x}} dx = 3\), calculate the value of \(k\). (4 marks)
提示：先计算含k的积分表达式，然后令其等于3求解k。
解答：
\(\int_1^k \frac{1}{\sqrt{x}} dx = \left[2\sqrt{x}\right]_1^k\)
\(= 2\sqrt{k} - 2\sqrt{1}\)
\(= 2\sqrt{k} - 2\)
根据题意：\(2\sqrt{k} - 2 = 3\)
\(2\sqrt{k} = 5\)
\(\sqrt{k} = \frac{5}{2}\)
\(k = \frac{25}{4}\)
Question 8 - 实际应用问题
The speed, \(v \text{ m/s}\), of a train at time \(t\) seconds is given by \(v = 20 + 5t, 0 \leq t \leq 10\).
The distance, \(s\) metres, travelled by the train in 10 seconds is given by \(s = \int_0^{10} (20 + 5t) dt\). Find the value of \(s\).
提示：这是物理应用问题，速度的积分等于位移。
解答：
\(s = \int_0^{10} (20 + 5t) dt\)
\(= \left[20t + \frac{5t^2}{2}\right]_0^{10}\)
\(= (200 + 250) - (0 + 0)\)
\(= 450 \text{ 米}\)