第三章 一元函数积分学及其应用

§3.1 不定积分

1. 计算下列不定积分:

   (2). \(\displaystyle \int \left( \dfrac{1}{\sqrt{x}} - \dfrac{2}{\sqrt{1-x^2}} + 3e^x \right) \mathrm{d} x\);

   (4). \(\displaystyle \int \dfrac{e^{2x} - 1}{e^x + 1} \mathrm{d} x\);

   (6). \(\displaystyle \int (2^xe^x + 1) \mathrm{d} x\);

   (8). \(\displaystyle \int \dfrac{1 - x + x^2}{x + x^3} \mathrm{d} x\);

   (10). \(\displaystyle \int \dfrac{\cos 2x}{\cos^2 x \sin^2 x} \mathrm{d} x\).

解

(2).

\[\begin{split}\int \left( \dfrac{1}{\sqrt{x}} - \dfrac{2}{\sqrt{1-x^2}} + 3e^x \right) \mathrm{d} x & = \int x^{-\frac{1}{2}} \mathrm{d} x - 2\int (1-x^2)^{-\frac{1}{2}} \mathrm{d} x + 3 \int e^x \mathrm{d} x \\ & = 2\sqrt{x} - 2\arcsin x + 3e^x + C.\end{split}\]

(4).

\[\begin{split}\int \dfrac{e^{2x} - 1}{e^x + 1} \mathrm{d} x & = \int \dfrac{(e^x + 1)(e^x - 1)}{e^x + 1} \mathrm{d} x \\ & = \int (e^x - 1) \mathrm{d} x \\ & = e^x - x + C.\end{split}\]

(6).

\[\begin{split}\int (2^xe^x + 1) \mathrm{d} x & = \int (2e)^x \mathrm{d} x + \int \mathrm{d} x \\ & = \dfrac{(2e)^x}{\ln (2e)} + x + C \\ & = \dfrac{2^xe^x}{\ln 2 + 1} + x + C.\end{split}\]

(8).

\[\begin{split}\int \dfrac{1 - x + x^2}{x + x^3} \mathrm{d} x & = \int \dfrac{1 + x^2}{x + x^3} \mathrm{d} x - \int \dfrac{x}{x + x^3} \mathrm{d} x \\ & = \int \dfrac{1}{x} \mathrm{d} x - \int \dfrac{1}{x^2 + 1} \mathrm{d} x \\ & = \ln |x| - \arctan x + C.\end{split}\]

(10).

\[\begin{split}\int \dfrac{\cos 2x}{\cos^2 x \sin^2 x} \mathrm{d} x & = \int \dfrac{\cos 2x}{(\frac{1}{2} \sin 2x)^2} \mathrm{d} x \\ & = 2 \int \dfrac{\mathrm{d} \sin 2x}{\sin^2 2x} \\ & = 2 \cdot \dfrac{-1}{\sin 2x} + C \\ & = -\dfrac{2}{\sin 2x} + C.\end{split}\]

2. 若某曲线过点 \((1, 1)\), 且在任一点 \(x\) 处的切线的斜率为 \(\dfrac{2}{x}\), 求此曲线方程.

解

设曲线方程为 \(y = f(x)\), 则 \(f'(x) = \dfrac{2}{x}\), 从而 \(\displaystyle f(x) = \int \dfrac{2}{x} \mathrm{d} x = 2\ln x + C\), 由 \(f(1) = 1\) 知 \(C = 1\), 因此曲线方程为 \(y = 2\ln x + 1\).

3. 设 \(f'(e^x) = 1 + e^{3x}\), 且 \(f(0) = 1\), 求 \(f(x)\).

解

由题设知 \(f'(e^x) = 1 + e^{3x}\), 从而 \(f'(x) = 1 + x^3\), 那么

\[f(x) = \int (1 + x^3) \mathrm{d} x = x + \dfrac{x^4}{4} + C\]

由 \(f(0) = 1\) 知 \(C = 1\), 因此 \(f(x) = x + \dfrac{x^4}{4} + 1\).

4. 计算下列不定积分:

   (1). \(\displaystyle \int \dfrac{1}{(2x - 5)^{10}} \mathrm{d} x\);

   (3). \(\displaystyle \int \dfrac{x}{\sqrt{1 + x^2}} \mathrm{d} x\);

   (5). \(\displaystyle \int x^2 e^{2x^3} \mathrm{d} x\);

   (7). \(\displaystyle \int \dfrac{\sqrt{1 + 3\ln x}}{x} \mathrm{d} x\);

   (9). \(\displaystyle \int \dfrac{2x - 1}{\sqrt{1 - x^2}} \mathrm{d} x\);

   (11). \(\displaystyle \int \dfrac{1}{4 + 9x^2} \mathrm{d} x\);

   (13). \(\displaystyle \int \sin^2 x \cos^2 x \mathrm{d} x\);

   (15). \(\displaystyle \int x (2x - 3)^{10} \mathrm{d} x\);

   (17). \(\displaystyle \int \dfrac{1}{x^2 \sqrt{1 + x^2}} \mathrm{d} x\).

解

(1). 令 \(u = 2x - 5\), 则 \(\mathrm{d} u = 2 \mathrm{d} x\), 从而有

\[\begin{split}\int \dfrac{1}{(2x - 5)^{10}} \mathrm{d} x & = \dfrac{1}{2} \int u^{-10} \mathrm{d} u = \dfrac{1}{2} \cdot \dfrac{u^{-9}}{-9} + C \\ & = -\dfrac{1}{18(2x - 5)^9} + C.\end{split}\]

接下来, 中间变量 \(u\) 就不再写出了.

(3).

\[\int \dfrac{x}{\sqrt{1 + x^2}} \mathrm{d} x = \int \dfrac{\sqrt{1 + x^2}}{2} \mathrm{d} (1 + x^2) = \sqrt{1 + x^2} + C.\]

(5).

\[\int x^2 e^{2x^3} \mathrm{d} x = \dfrac{1}{6} \int e^{2x^3} \mathrm{d} (2x^3) = \dfrac{1}{6} e^{2x^3} + C.\]

(7).

\[\int \dfrac{\sqrt{1 + 3\ln x}}{x} \mathrm{d} x = \int \sqrt{1 + 3\ln x} \mathrm{d} (\ln x) = \dfrac{2}{9} (1 + 3\ln x)^{\frac{3}{2}} + C.\]

(9).

\[\begin{split}\int \dfrac{2x - 1}{\sqrt{1 - x^2}} \mathrm{d} x & = \int \dfrac{2x}{\sqrt{1 - x^2}} \mathrm{d} x - \int \dfrac{1}{\sqrt{1 - x^2}} \mathrm{d} x \\ & = -\int \dfrac{1}{\sqrt{1 - x^2}} \mathrm{d} (1 - x^2) - \arcsin x + C \\ & = -2 \sqrt{1 - x^2} - \arcsin x + C.\end{split}\]

(11).

\[\int \dfrac{1}{4 + 9x^2} \mathrm{d} x = \dfrac{2}{3} \cdot \dfrac{1}{4} \int \dfrac{1}{1 + \left( \frac{3}{2} x \right)^2} \mathrm{d} \left( \frac{3}{2} x \right) = \dfrac{1}{6} \arctan \dfrac{3}{2} x + C.\]

(13).

\[\begin{split}\int \sin^2 x \cos^2 x \mathrm{d} x & = \dfrac{1}{4} \int \sin^2 2x \mathrm{d} x = \dfrac{1}{8} \int (1 - \cos 4x) \mathrm{d} x \\ & = \dfrac{1}{32} \int (1 - \cos 4x) \mathrm{d} (4x) = \dfrac{1}{32} (4x - \sin 4x) + C.\end{split}\]

(15).

\[\begin{split}\int x (2x - 3)^{10} \mathrm{d} x & = \int \dfrac{1}{2} (2x - 3)^{11} \mathrm{d} x + \int \dfrac{3}{2} (2x - 3)^{10} \mathrm{d} x \\ & = \dfrac{1}{4} \int (2x - 3)^{11} \mathrm{d} (2x - 3) + \dfrac{3}{4} \int (2x - 3)^{10} \mathrm{d} (2x - 3) \\ & = \dfrac{1}{4} \cdot \dfrac{(2x - 3)^{12}}{12} + \dfrac{3}{4} \cdot \dfrac{(2x - 3)^{11}}{11} + C \\ & = \dfrac{1}{48} (2x - 3)^{12} + \dfrac{3}{44} (2x - 3)^{11} + C.\end{split}\]

(17).

\[\begin{split}\int \dfrac{1}{x^2 \sqrt{1 + x^2}} \mathrm{d} x & = -\int \dfrac{1}{\sqrt{1 + x^2}} \mathrm{d} \left( \dfrac{1}{x} \right) = -\int \dfrac{1}{x} \cdot \dfrac{1}{\sqrt{1 + \left(\frac{1}{x}\right)^2}} \mathrm{d} \left( \dfrac{1}{x} \right) \\ & = -\dfrac{1}{2} \int \dfrac{1}{\sqrt{1 + \left(\frac{1}{x}\right)^2}} \mathrm{d} \left( \frac{1}{x} \right)^2 \\ & = -\sqrt{1 + \left(\frac{1}{x}\right)^2} + C \\ & = -\dfrac{\sqrt{x^2 + 1}}{x} + C.\end{split}\]

以上假设了 \(x > 0\), 对于 \(x < 0\) 的情况, 从根式中提出 \(x\) 要变 (2次) 号, 最终结果是一样的.

5. 计算下列不定积分:

   (2). \(\displaystyle \int x \cos (5x + 2) \mathrm{d} x\);

   (4). \(\displaystyle \int \dfrac{\ln x}{\sqrt{x}} \mathrm{d} x\);

   (6). \(\displaystyle \int \ln(1 + x^2) \mathrm{d} x\).

   (8). 设 \(f(x)\) 的一个原函数为 \(x \cos x\), 求积分 \(\displaystyle \int x f'(x) \mathrm{d} x\).

解

(2). 采用分部积分法:

\[\begin{split}\int x \cos (5x + 2) \mathrm{d} x & = \dfrac{1}{5} \int x \mathrm{d} \left( \sin (5x + 2) \right) = \dfrac{1}{5} x \sin (5x + 2) - \dfrac{1}{5} \int \sin (5x + 2) \mathrm{d} x \\ & = \dfrac{1}{5} x \sin (5x + 2) + \dfrac{1}{25} \cos (5x + 2) + C.\end{split}\]

(4). 令 \(x = t^2, t > 0\), 则 \(\mathrm{d} x = 2t \mathrm{d} t\), 从而有

\[\begin{split}\int \dfrac{\ln x}{\sqrt{x}} \mathrm{d} x & = \int \dfrac{2t \ln t^2}{t} \mathrm{d} t = 4 \int \ln t \mathrm{d} t \\ & = 4t \ln t - 4 \int t \mathrm{d} (\ln t) = 4t \ln t - 4 \int t \cdot \dfrac{1}{t} \mathrm{d} t \\ & = 4t \ln t - 4t + C = 4 \sqrt{x} \ln \sqrt{x} - 4 \sqrt{x} + C \\ & = 2 \sqrt{x} \ln x - 4 \sqrt{x} + C.\end{split}\]

也可以直接采用分部积分法:

\[\begin{split}\int \dfrac{\ln x}{\sqrt{x}} \mathrm{d} x & = 2 \int \ln x \mathrm{d} \left( \sqrt{x} \right) = 2 \sqrt{x} \ln x - 2 \int \sqrt{x} \mathrm{d} (\ln x) \\ & = 2 \sqrt{x} \ln x - 2 \int \sqrt{x} \cdot \dfrac{1}{x} \mathrm{d} x \\ & = 2 \sqrt{x} \ln x - 2 \int \dfrac{1}{\sqrt{x}} \mathrm{d} x \\ & = 2 \sqrt{x} \ln x - 4 \sqrt{x} + C.\end{split}\]

(6). 采用分部积分法:

\[\begin{split}\int \ln(1 + x^2) \mathrm{d} x & = x \ln(1 + x^2) - \int x \mathrm{d} (\ln(1 + x^2)) = x \ln(1 + x^2) - \int x \cdot \dfrac{2x}{1 + x^2} \mathrm{d} x \\ & = x \ln(1 + x^2) - 2 \int \dfrac{x^2}{1 + x^2} \mathrm{d} x = x \ln(1 + x^2) - 2 \int \left( 1 - \dfrac{1}{1 + x^2} \right) \mathrm{d} x \\ & = x \ln(1 + x^2) - 2x + 2 \arctan x + C.\end{split}\]

(8). 采用分部积分法:

\[\begin{split}\int x f'(x) \mathrm{d} x & = \int x \mathrm{d} f(x) = x f(x) - \int f(x) \mathrm{d} x \\ & = x (x \cos x)' - x \cos x + C = x \cos x - x^2 \sin x - x \cos x + C \\ &= -x^2 \sin x + C.\end{split}\]

6. 计算下列不定积分:

   (1). \(\displaystyle \int \dfrac{1}{3 + \sin^2 x} \mathrm{d} x\);

   (3). \(\displaystyle \int \cos x \cos 5x \mathrm{d} x\);

   (5). \(\displaystyle \int \dfrac{2x + 5}{x^2 + 4x + 8} \mathrm{d} x\);

   (7). \(\displaystyle \int \dfrac{x}{\sqrt{3 + 4x}} \mathrm{d} x\).

解

(1).

\[\begin{split}\int \dfrac{1}{3 + \sin^2 x} \mathrm{d} x & = \int \dfrac{1}{3\cos^2 x + 4\sin^2 x} \mathrm{d} x = \int \dfrac{\sec^2x \mathrm{d} x}{3 + 4\tan^2 x} \\ & = \int \dfrac{\mathrm{d} \tan x}{3 + 4\tan^2 x} = \dfrac{1}{2\sqrt{3}} \int \dfrac{\mathrm{d} \left( \frac{2}{\sqrt{3}} \tan x \right)}{1 + \left( \frac{2}{\sqrt{3}} \tan x \right)^2} \\ & = \dfrac{1}{2\sqrt{3}} \arctan \left( \dfrac{2}{\sqrt{3}} \tan x \right) + C.\end{split}\]

(3). 利用和差化积公式 \(\cos x \cos 5x = \dfrac{1}{2} (\cos 4x + \cos 6x)\), 从而有

\[\begin{split}\int \cos x \cos 5x \mathrm{d} x & = \dfrac{1}{2} \int \cos 4x \mathrm{d} x + \dfrac{1}{2} \int \cos 6x \mathrm{d} x \\ & = \dfrac{1}{8} \sin 4x + \dfrac{1}{12} \sin 6x + C.\end{split}\]

(5).

\[\begin{split}\int \dfrac{2x + 5}{x^2 + 4x + 8} \mathrm{d} x & = \int \dfrac{2(x + 2) + 1}{(x + 2)^2 + 4} \mathrm{d} (x + 2) \\ & = 2 \int \dfrac{x + 2}{(x + 2)^2 + 4} \mathrm{d} (x + 2) + \int \dfrac{1}{(x + 2)^2 + 4} \mathrm{d} (x + 2) \\ & = \int \dfrac{1}{(x + 2)^2 + 4} \mathrm{d} (x + 2)^2 + \dfrac{1}{2} \int \dfrac{1}{(\frac{x + 2}{2})^2 + 1} \mathrm{d} \left(\dfrac{x + 2}{2}\right) \\ & = \ln \left\lvert (x + 2)^2 + 4 \right\rvert + \dfrac{1}{2} \arctan \dfrac{x + 2}{2} + C \\ & = \ln (x^2 + 4x + 8) + \dfrac{1}{2} \arctan \dfrac{x + 2}{2} + C.\end{split}\]

(7). 令 \(u = \sqrt{3 + 4x}\), 那么 \(\mathrm{d} x = \dfrac{u \mathrm{d} u}{2}\), 从而有

\[\begin{split}\int \dfrac{x}{\sqrt{3 + 4x}} \mathrm{d} x & = \int \dfrac{u^2 - 3}{4u} \cdot \dfrac{u \mathrm{d} u}{2} = \dfrac{1}{8} \int (u^2 - 3) \mathrm{d} u \\ & = \dfrac{1}{8} \cdot \dfrac{u^3}{3} - \dfrac{3}{8} u + C \\ & = \dfrac{1}{24} (3 + 4x)^{\frac{3}{2}} - \dfrac{3}{8} \sqrt{3 + 4x} + C \\ & = \sqrt{3 + 4x} \left( \dfrac{1}{24} (3 + 4x) - \dfrac{3}{8} \right) + C \\ & = \dfrac{4x - 6}{24} \sqrt{3 + 4x} + C \\ & = \dfrac{2x - 3}{12} \sqrt{3 + 4x} + C.\end{split}\]

§3.2 定积分

2. 设 \(x\) 轴上有一根细棒, 位于 \(x = a\) 到 \(x = b\) 的区间上, 这棒在 \(x\) 处的线密度为 \(\rho(x)\), 试用定积分表示这细棒的质量.

解

设细棒的质量为 \(m\), 则有

\[m = \int_a^b \rho(x) \mathrm{d} x.\]

3. 利用定积分的几何意义, 给出下列定积分的值:

   (1). \(\displaystyle \int_a^b x \mathrm{d} x\);

   (3). \(\displaystyle \int_{-\pi}^{\pi} \sin x \mathrm{d} x\);

   (5). \(\displaystyle \int_0^4 (2 - x) \mathrm{d} x\).

解

(1). 假设 \(a < b\).

定积分 \(\displaystyle \int_a^b x \mathrm{d} x\) 表示 \(x\) 从 \(a\) 到 \(b\) 曲线 \(y = x\) 与 \(x\) 轴之间 (带正负号) 的面积. 当 \(a, b\) 同号时, 这是一个底边长 \(|a|, |b|\), 高为 \(|a - b|\) 的梯形, 面积为 \(\dfrac{|a| + |b|}{2} |a - b|\). 当 \(a, b > 0\) 时, 面积为正的, 当 \(a, b < 0\) 时, 面积为负的. 值为 \(\dfrac{b^2 - a^2}{2}\).

当 \(a \leqslant 0 \leqslant b\), 定积分 \(\displaystyle \int_a^b x \mathrm{d} x\) 表示两个三角形的面积之差 (包括等于 \(0\) 时退化的情况). 这是两个等腰直角三角形, 直角边长分别为 \(-a, b\), 面积之差为 \(\dfrac{b^2 - a^2}{2}\).

(3). \(\sin x\) 在 \((-\pi, 0)\) 取值为负, \((0, \pi)\) 取值为正, 因此定积分 \(\displaystyle \int_{-\pi}^{\pi} \sin x \mathrm{d} x\) 表示 这两部分曲线与 \(x\) 轴围成 (带正负号) 的面积之和. 正两部分面积正好绝对值相等, 符号相反, 因此定积分的值为 \(0\).

(5). \(\displaystyle \int_0^4 (2 - x) \mathrm{d} x\) 表示 \(x\) 从 \(0\) 到 \(4\) 曲线 \(y = 2 - x\) 与 \(x\) 轴之间 (带正负号) 的面积. \(x\) 从 \(0\) 到 \(2\) 时, \(y = 2 - x\) 在 \(x\) 轴上方, 面积为正, \(x\) 从 \(2\) 到 \(4\) 时, \(y = 2 - x\) 在 \(x\) 轴下方, 面积为负. 这两部分面积绝对值相等, 符号相反, 因此定积分的值为 \(0\).

4. 利用定积分的性质, 比较下列各组积分值的大小:

   (2). \(\displaystyle \int_0^1 e^x \mathrm{d} x\) 与 \(\displaystyle \int_0^1 (1 + x) \mathrm{d} x\).

解

由于在区间 \((0, 1)\) 上有不等式 \(e^x > 1 + x\), 因此有 \(\displaystyle \int_0^1 e^x \mathrm{d} x > \int_0^1 (1 + x) \mathrm{d} x\).

5. 证明下列不等式:

   (2). \(\displaystyle 2 e^{-\frac{1}{4}} < \int_0^2 e^{x^2 - x} \mathrm{d} x < 2 e^2\).

证明

由于 \(e^{x^2 - x} = e^{\left( x - \frac{1}{2} \right)^2 - \frac{1}{4}}\) 在区间 \([0, 2]\) 上的最小值为 \(e^{-\frac{1}{4}}\), 最大值为 \(e^2\), 因此有

\[2 e^{-\frac{1}{4}} = \int_0^2 e^{-\frac{1}{4}} \mathrm{d} x < \int_0^2 e^{x^2 - x} \mathrm{d} x < \int_0^2 e^2 \mathrm{d} x = 2 e^2.\]

6. 设函数 \(f(x)\) 在区间 \([1, 3]\) 上的平均值为 \(6\), 求定积分 \(\displaystyle \int_1^3 f(x) \mathrm{d} x\).

解

函数 \(f(x)\) 在区间 \([1, 3]\) 上的平均值为 \(6\), 也就是说有

\[\dfrac{\int_1^3 f(x) \mathrm{d} x}{3 - 1} = 6,\]

从而有 \(\displaystyle \int_1^3 f(x) \mathrm{d} x = 12\).

§3.3 定积分的计算

1. 计算下列各题:

   (2). 设 \(\displaystyle f(x) = \int_0^x e^{-t^2} \mathrm{d} t\), 求 \(f''(1)\);

   (4). 求 \(\displaystyle \dfrac{\mathrm{d}}{\mathrm{d} x} \int_{x^2}^{x^3} \dfrac{1}{\sqrt{1 + u^4}} \mathrm{d} u\);

   (6). 求极限 \(\displaystyle \lim_{x \to 0} \dfrac{\int_0^x t(t + \sin t) \mathrm{d} t}{\int_x^0 \ln (1 + t^2) \mathrm{d} t}\).

解

(1). \(f'(x) = e^{-x^2}\), \(f''(x) = -2x e^{-x^2}\), 因此 \(f''(1) = -2e^{-1}\).

(3). \(\displaystyle \dfrac{\mathrm{d}}{\mathrm{d} x} \int_{x^2}^{x^3} \dfrac{1}{\sqrt{1 + u^4}} \mathrm{d} u = \dfrac{1}{\sqrt{1 + x^{12}}} \cdot 3x^2 - \dfrac{1}{\sqrt{1 + x^8}} \cdot 2x = \dfrac{3x^2}{\sqrt{1 + x^{12}}} - \dfrac{2x}{\sqrt{1 + x^8}}\).

(5).

\[\begin{split}\displaystyle \lim_{x \to 0} \dfrac{\int_0^x t(t + \sin t) \mathrm{d} t}{\int_x^0 \ln (1 + t^2) \mathrm{d} t} & = \lim_{x \to 0} \dfrac{\int_0^x t(t + \sin t) \mathrm{d} t}{-\int_0^x \ln (1 + t^2) \mathrm{d} t} = -\lim_{x \to 0} \dfrac{x(x + \sin x)}{\ln (1 + x^2)} \\ & = -\lim_{x \to 0} \dfrac{2x + x \cos x + \sin x}{\frac{2x}{1 + x^2}} \\ & = -\lim_{x \to 0} (1 + x^2) \dfrac{2x + x \cos x + \sin x}{2x} \\ & = -2.\end{split}\]

备注

一般地, 如果 \(\displaystyle f(x) = \int_{\varphi(x)}^{\psi(x)} g(t) \mathrm{d} t\), 那么

\[f'(x) = g(\psi(x)) \psi'(x) - g(\varphi(x)) \varphi'(x).\]

2. 设 \(y = f(x)\) 是由方程 \(\displaystyle x^2 y = \int_0^y \sqrt{1 + t^2} \mathrm{d} t\) 所确定的隐函数, 试求 \(y = f(x)\) 的微分 \(\mathrm{d} y\).

解

对方程两边求微分, 有

\[2x y \mathrm{d} x + x^2 \mathrm{d} y = \sqrt{1 + y^2} \mathrm{d} y,\]

移项之后有

\[\mathrm{d} y = \dfrac{2x y}{\sqrt{1 + y^2} - x^2} \mathrm{d} x.\]

3. 设函数 \(f(x)\) 在区间 \([a, b]\) 上连续且单调增加, 令

   \[F(x) = \dfrac{1}{x - a} \int_a^x f(t) \mathrm{d} t \quad (a < x \leqslant b),\]

   试证明在区间 \((a, b]\) 上恒有 \(F'(x) \geqslant 0\).

证明

由于 \(f(x)\) 在区间 \([a, b]\) 上连续且单调增加, 所以有

\[F'(x) = \dfrac{1}{x - a} \cdot f(x) - \dfrac{1}{(x - a)^2} \int_a^x f(t) \mathrm{d} t.\]

进一步由积分中值定理, 存在 \(\xi \in (a, x)\) 使得 \(\displaystyle \int_a^x f(t) \mathrm{d} t = f(\xi) (x - a)\), 因此有

\[F'(x) = \dfrac{1}{x - a} \cdot f(x) - \dfrac{f(\xi) (x - a)}{(x - a)^2} = \dfrac{1}{x - a} \cdot \left( f(x) - f(\xi) \right).\]

由于 \(f(x)\) 在区间 \([a, b]\) 上连续且单调增加, 因此有 \(f(x) \geqslant f(\xi)\), 从而有 \(F'(x) \geqslant 0\).

4. 计算下列定积分:

   (1). \(\displaystyle \int_0^4 (2 - \sqrt{x})^2 \mathrm{d} x\);

   (3). \(\displaystyle \int_0^1 \dfrac{1}{\sqrt{4-u^2}} \mathrm{d} u\);

   (5). 设 \(\displaystyle f(x) = \begin{cases} \frac{x}{2} + 1, & 0 \leqslant x \leqslant 2 \\ x, & 2 < x \leqslant 3 \end{cases}\), 求 \(\displaystyle \int_0^3 f(x) \mathrm{d} x\).

   (7). \(\displaystyle \int_0^2 (2 - x)^2 (2 + x) \mathrm{d} x\);

   (9). \(\displaystyle \int_0^{\pi} (1 - \sin^3 \varphi) \mathrm{d} \varphi\).

解

(1). 令 \(t = \sqrt{x}\), 那么 \(x = t^2, \mathrm{d} x = 2t \mathrm{d} t\), 从而有

\[\begin{split}\int_0^4 (2 - \sqrt{x})^2 \mathrm{d} x & = \int_0^2 (2 - t)^2 \cdot 2t \mathrm{d} t = 2 \int_0^2 (4 - 4t + t^2) t \mathrm{d} t \\ & = 2 \int_0^2 (4t - 4t^2 + t^3) \mathrm{d} t = 2 \left. \left[ 2t^2 - \dfrac{4}{3} t^3 + \dfrac{1}{4} t^4 \right] \right|_0^2 \\ & = 2 \left( 8 - \dfrac{32}{3} + 4 \right) = \dfrac{8}{3}.\end{split}\]

(3). 令 \(u = 2 \sin \varphi\), 那么 \(\mathrm{d} u = 2 \cos \varphi \mathrm{d} \varphi\), 从而有

\[\begin{split}\int_0^1 \dfrac{1}{\sqrt{4-u^2}} \mathrm{d} u & = \int_0^{\frac{\pi}{6}} \dfrac{1}{\sqrt{4 - 4 \sin^2 \varphi}} \cdot 2 \cos \varphi \mathrm{d} \varphi \\ & = \int_0^{\frac{\pi}{6}} \dfrac{1}{\sqrt{4 \cos^2 \varphi}} \cdot 2 \cos \varphi \mathrm{d} \varphi = \int_0^{\frac{\pi}{6}} \dfrac{1}{2 \cos \varphi}\ \cdot 2 \cos \varphi \mathrm{d} \varphi \\ & = \int_0^{\frac{\pi}{6}} \mathrm{d} \varphi = \dfrac{\pi}{6}.\end{split}\]

(5). 根据定积分对积分区间的可加性, 有

\[\begin{split}\int_0^3 f(x) \mathrm{d} x & = \int_0^2 f(x) \mathrm{d} x + \int_2^3 f(x) \mathrm{d} x = \int_0^2 \left( \dfrac{x}{2} + 1 \right) \mathrm{d} x + \int_2^3 x \mathrm{d} x \\ & = \left. \left( \dfrac{x^2}{4} + x \right) \right|_0^2 + \left. \dfrac{x^2}{2} \right|_2^3 = 3 + \dfrac{9}{2} - 2 = \dfrac{11}{2}.\end{split}\]

(7).

\[\begin{split}\int_0^2 (2 - x)^2 (2 + x) \mathrm{d} x & = \int_2^0 x^2 (4 - x) \mathrm{d} (2-x) = \int_0^2 x^2 (4 - x) \mathrm{d} x \\ & = \int_0^2 (4x^2 - x^3) \mathrm{d} x = \left. \left( \dfrac{4}{3} x^3 - \dfrac{1}{4} x^4 \right) \right|_0^2 \\ & = \dfrac{32}{3} - 4 = \dfrac{20}{3}.\end{split}\]

(9). 由于 \(\sin^3 \varphi = \dfrac{3}{4} \sin \varphi - \dfrac{1}{4} \sin 3\varphi\), 因此有

\[\begin{split}\int_0^{\pi} (1 - \sin^3 \varphi) \mathrm{d} \varphi & = \int_0^{\pi} \left( 1 - \dfrac{3}{4} \sin \varphi + \dfrac{1}{4} \sin 3\varphi \right) \mathrm{d} \varphi \\ & = \left. \left( \varphi + \dfrac{3}{4} \cos \varphi - \dfrac{1}{12} \cos 3\varphi \right) \right|_0^{\pi} \\ & = \pi - \dfrac{3}{4} + \dfrac{1}{12} - (0 + \dfrac{3}{4} - \dfrac{1}{12}) \\ & = \pi - \dfrac{4}{3}.\end{split}\]

5. 计算下列定积分:

   (2). \(\displaystyle \int_0^{\pi} \dfrac{\sin x}{1 + \cos^2 x} \mathrm{d} x\);

   (4). \(\displaystyle \int_0^1 x^2 \sqrt{1 - x^2} \mathrm{d} x\);

   (6). \(\displaystyle \int_1^2 \dfrac{\sqrt{x^2 - 1}}{x} \mathrm{d} x\);

   (8). \(\displaystyle \int_{-1}^1 \dfrac{x}{\sqrt{5 - 4x}} \mathrm{d} x\).

解

(2).

\[\begin{split}\int_0^{\pi} \dfrac{\sin x}{1 + \cos^2 x} \mathrm{d} x & = - \int_0^{\pi} \dfrac{\mathrm{d} \cos x}{1 + \cos^2 x} = - \left. \arctan \cos x \right|_0^{\pi} \\ & = - \left( \arctan (-1) - \arctan 1 \right) = - \left( -\dfrac{\pi}{4} - \dfrac{\pi}{4} \right) = \dfrac{\pi}{2}.\end{split}\]

(4).

\[\begin{split}\int_0^1 x^2 \sqrt{1 - x^2} \mathrm{d} x & = \dfrac{1}{2} \int_0^1 \sqrt{x^2 (1 - x^2)} \mathrm{d} x^2 = \dfrac{1}{2} \int_0^1 \sqrt{x (1 - x)} \mathrm{d} x \\ & = \dfrac{1}{2} \int_0^1 \sqrt{\dfrac{1}{4} - \left( x - \dfrac{1}{2} \right)^2} \mathrm{d} \left( x - \dfrac{1}{2} \right) \\ & = \dfrac{1}{8} \int_0^1 \sqrt{1 - \left( 2x - 1 \right)^2} \mathrm{d} \left( 2x - 1 \right) \\ & = \dfrac{1}{8} \int_{-1}^1 \sqrt{1 - x^2} \mathrm{d} x \\ & = \dfrac{1}{4} \int_{0}^1 \sqrt{1 - x^2} \mathrm{d} x \\ & = \dfrac{1}{4} \int_{0}^{\frac{\pi}{2}} \sqrt{1 - \sin^2 \varphi} \mathrm{d} \sin \varphi \\ & = \dfrac{1}{4} \int_{0}^{\frac{\pi}{2}} \cos^2 \varphi \mathrm{d} \varphi \\ & = \dfrac{1}{4} \int_{0}^{\frac{\pi}{2}} \dfrac{1 + \cos 2\varphi}{2} \mathrm{d} \varphi \\ & = \dfrac{1}{8} \left. \left( \varphi + \dfrac{1}{2} \sin 2\varphi \right) \right|_0^{\frac{\pi}{2}} \\ & = \dfrac{\pi}{16}.\end{split}\]

另解: 令 \(x = \sin t\), 积分区域变为 \([0, \frac{\pi}{2}]\), 从而有

\[\begin{split}\int_0^1 x^2 \sqrt{1 - x^2} \mathrm{d} x & = \int_0^{\frac{\pi}{2}} \sin^2 t \cos t \mathrm{d} \sin t = \int_0^{\frac{\pi}{2}} \sin^2 t \cos^2 t \mathrm{d} t \\ & = \dfrac{1}{4} \int_0^{\frac{\pi}{2}} \sin^2 2t \mathrm{d} t \\ & = \dfrac{1}{4} \int_0^{\frac{\pi}{2}} \dfrac{1 - \cos 4t}{2} \mathrm{d} t \\ & = \dfrac{1}{8} \int_0^{\frac{\pi}{2}} \left( 1 - \cos 4t \right) \mathrm{d} t \\ & = \dfrac{1}{8} \int_0^{\frac{\pi}{2}} \mathrm{d} t - \dfrac{1}{8} \int_0^{\frac{\pi}{2}} \cos 4t \mathrm{d} t \\ & = \dfrac{\pi}{16}.\end{split}\]

(6). 令 \(x = \sec \varphi\), 积分区域变为 \([0, \frac{\pi}{3}]\), 从而有

\[\begin{split}\int_1^2 \dfrac{\sqrt{x^2 - 1}}{x} \mathrm{d} x & = \int_{0}^{\frac{\pi}{3}} \dfrac{\tan \varphi}{\sec \varphi} \cdot \sec \varphi \tan \varphi \mathrm{d} \varphi \\ & = \int_{0}^{\frac{\pi}{3}} \tan^2 \varphi \mathrm{d} \varphi \\ & = \int_{0}^{\frac{\pi}{3}} \sec^2 \varphi \mathrm{d} \varphi - \int_{0}^{\frac{\pi}{3}} \mathrm{d} \varphi \\ & = \left. \tan \varphi \right|_0^{\frac{\pi}{3}} - \left. \varphi \right|_0^{\frac{\pi}{3}} \\ & = \sqrt{3} - \dfrac{\pi}{3}.\end{split}\]

(8). 令 \(t = \sqrt{5 - 4x}\), 那么 \(x = \dfrac{5 - t^2}{4}\), \(\mathrm{d} x = -\dfrac{t}{2} \mathrm{d} t\), 从而有

\[\begin{split}\int_{-1}^1 \dfrac{x}{\sqrt{5 - 4x}} \mathrm{d} x & = \int_{3}^1 \dfrac{\frac{5 - t^2}{4}}{t} \cdot \left( -\dfrac{t}{2} \right) \mathrm{d} t = \dfrac{1}{8} \int_1^{3} \left( 5 - t^2 \right) \mathrm{d} t \\ & = \dfrac{1}{8} \left. \left( 5t - \dfrac{t^3}{3} \right) \right|_1^{3} = \dfrac{1}{8} \left( 15 - \dfrac{27}{3} - 5 + \dfrac{1}{3} \right) \\ & = \dfrac{1}{6}.\end{split}\]

6. 计算下列定积分:

   (1). \(\displaystyle \int_0^1 x \ln(1 + x) \mathrm{d} x\);

   (3). \(\displaystyle \int_0^{\sqrt{3}} \ln \left( x + \sqrt{1 + x^2} \right) \mathrm{d} x\);

   (5). \(\displaystyle \int_0^{\frac{\sqrt{2}}{2}} \arccos x \mathrm{d} x\);

   (7). \(\displaystyle \int_{-1}^1 \dfrac{x^2 \sin^5 x + 1}{1 + x^2} \mathrm{d} x\).

解

(1).

\[\begin{split}\int_0^1 x \ln(1 + x) \mathrm{d} x & = \dfrac{1}{2} \int_0^1 \ln(1 + x) \mathrm{d} x^2 = \left. \dfrac{1}{2} \ln(1 + x) \cdot x^2 \right|_0^1 - \dfrac{1}{2} \int_0^1 \dfrac{x^2}{1 + x} \mathrm{d} x \\ & = \dfrac{1}{2} \ln 2 - \dfrac{1}{2} \int_0^1 \left( x - 1 + \dfrac{1}{1 + x} \right) \mathrm{d} x \\ & = \dfrac{1}{2} \ln 2 - \dfrac{1}{2} \left. \left( \dfrac{x^2}{2} - x + \ln(1 + x) \right) \right|_0^1 \\ & = \dfrac{1}{2} \ln 2 - \dfrac{1}{2} \left( \dfrac{1}{2} - 1 + \ln 2 \right) \\ & = \dfrac{1}{4}.\end{split}\]

(3).

\[\begin{split}\int_0^{\sqrt{3}} \ln \left( x + \sqrt{1 + x^2} \right) \mathrm{d} x & = \left. x \ln \left( x + \sqrt{1 + x^2} \right) \right|_0^{\sqrt{3}} - \int_0^{\sqrt{3}} x \dfrac{1 + \dfrac{x}{\sqrt{1 + x^2}}}{x + \sqrt{1 + x^2}} \mathrm{d} x \\ & = \sqrt{3} \ln \left( \sqrt{3} + 2 \right) - \int_0^{\sqrt{3}} \dfrac{x}{\sqrt{1 + x^2}} \mathrm{d} x \\ & = \sqrt{3} \ln \left( \sqrt{3} + 2 \right) - \dfrac{1}{2} \int_0^{\sqrt{3}} \dfrac{\mathrm{d} x^2}{\sqrt{1 + x^2}} \\ & = \sqrt{3} \ln \left( \sqrt{3} + 2 \right) - \dfrac{1}{2} \int_0^{\sqrt{3}} \dfrac{\mathrm{d} \left( 1 + x^2 \right)}{\sqrt{1 + x^2}} \\ & = \sqrt{3} \ln \left( \sqrt{3} + 2 \right) - \left. \sqrt{1 + x^2} \right|_0^{\sqrt{3}} \\ & = \sqrt{3} \ln \left( \sqrt{3} + 2 \right) - 1.\end{split}\]

(5).

\[\begin{split}\int_0^{\frac{\sqrt{2}}{2}} \arccos x \mathrm{d} x & = \left. x \arccos x \right|_0^{\frac{\sqrt{2}}{2}} - \int_0^{\frac{\sqrt{2}}{2}} \dfrac{x}{-\sqrt{1 - x^2}} \mathrm{d} x \\ & = \dfrac{\pi}{4} \cdot \dfrac{\sqrt{2}}{2} - \left. \sqrt{1 - x^2} \right|_0^{\frac{\sqrt{2}}{2}} \\ & = \dfrac{\pi}{8} - \dfrac{\sqrt{2}}{2} + 1.\end{split}\]

(7). 因为 \(\dfrac{x^2 \sin^5 x}{1 + x^2}\) 是奇函数, 所以 \(\displaystyle \int_{-1}^1 \dfrac{x^2 \sin^5 x}{1 + x^2} \mathrm{d} x = 0\), 因此有

\[\begin{split}\int_{-1}^1 \dfrac{x^2 \sin^5 x + 1}{1 + x^2} \mathrm{d} x & = \int_{-1}^1 \dfrac{1}{1 + x^2} \mathrm{d} x = \left. \arctan x \right|_{-1}^1 \\ & = \arctan 1 - \arctan (-1) = \dfrac{\pi}{2}.\end{split}\]

7. 设 \(f(x)\) 在区间 \([a, b]\) 上连续, 证明 \(\displaystyle \int_a^b f(x) \mathrm{d} x = \int_a^b f(a + b - x) \mathrm{d} x\).

证明

令 \(t = a + b - x\), 那么 \(x = a + b - t, \mathrm{d} t = -\mathrm{d} x\), 积分区间变为 \([a + b - b, a + b - a] = [a, b]\), 从而有

\[\begin{split}\int_a^b f(a + b - x) \mathrm{d} x & = -\int_{a + b - a}^{a + b - b} f(t) \mathrm{d} t \\ & = -\int_b^a f(t) \mathrm{d} t = \int_a^b f(t) \mathrm{d} t \\ & = \int_a^b f(x) \mathrm{d} x.\end{split}\]

8. 设 \(a > 0\), 试证明: \(\displaystyle \int_0^a x^3 f(x^2) \mathrm{d} x = \dfrac{1}{2} \int_0^{a^2} x f(x) \mathrm{d} x\).

证明

\(\displaystyle \int_0^a x^3 f(x^2) \mathrm{d} x = \dfrac{1}{2} \int_0^a x^2 f(x^2) \mathrm{d} (x^2) = \dfrac{1}{2} \int_0^{a^2} x f(x) \mathrm{d} x\).

9. 证明: \(\displaystyle \int_0^{\pi} \sin^n x \mathrm{d} x = 2 \int_0^{\frac{\pi}{2}} \sin^n x \mathrm{d} x\).

证明

令 \(t = x - \dfrac{\pi}{2}\), 那么 \(x = t + \dfrac{\pi}{2}\), \(\mathrm{d} t = \mathrm{d} x\), 积分区间变为 \([-\dfrac{\pi}{2}, \dfrac{\pi}{2}]\), 从而有

\[\int_0^{\pi} \sin^n x \mathrm{d} x = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^n \left( t + \dfrac{\pi}{2} \right) \mathrm{d} t = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^n t \mathrm{d} t.\]

由于 \(\cos^n t\) 是偶函数, 因此有

\[\int_0^{\pi} \sin^n x \mathrm{d} x = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^n t \mathrm{d} t = 2 \int_0^{\frac{\pi}{2}} \cos^n t \mathrm{d} t = 2 \int_0^{\frac{\pi}{2}} \sin^n x \mathrm{d} x.\]

§3.4 定积分的应用

1. 求下列各曲线所围成的图形的面积:

   (1). \(y = 9 - x^2, y = 0\);

   (3). \(y = x^3, x = 0, y = 1\);

   (5). \(y = \sin x, x = -\pi, x = \dfrac{\pi}{2}, y = 0\);

   (7). \(r = 2a (2 + \cos \theta)\).

解

(1). \(y = 9 - x^2\) 与 \(y = 0\) 的交点为 \(x = \pm 3\), 因此所围成的图形的面积 \(S\) 为

\[S = \int_{-3}^3 (9 - x^2) \mathrm{d} x = \left. \left( 9x - \dfrac{x^3}{3} \right) \right|_{-3}^3 = 36.\]

(3). \(y = x^3, x = 0, y = 1\) 所围成的图形为正方形 \([0, 1] \times [0, 1]\) 内, 位于曲线 \(y = x^3\) 之上的部分, 因此所围成的图形的面积 \(S\) 为

\[S = \int_0^1 (1 - x^3) \mathrm{d} x = \left. \left( x - \dfrac{x^4}{4} \right) \right|_0^1 = \dfrac{3}{4}.\]

(5). \(y = \sin x, x = -\pi, x = \dfrac{\pi}{2}, y = 0\) 所围成的图形分为两部分, 一部分为 \([-\pi, 0] \times [0, 1]\) 内在曲线 \(y = \sin x\) 之上的部分; 另一部分为 \([0, \frac{\pi}{2}] \times [0, 1]\) 内在曲线 \(y = \sin x\) 之下的部分, 因此所围成的图形的面积 \(S\) 为

\[S = \int_{-\pi}^0 (0 - \sin x) \mathrm{d} x + \int_0^{\frac{\pi}{2}} (\sin x - 0) \mathrm{d} x = \left. \left( \cos x \right) \right|_{-\pi}^0 - \left. \cos x \right|_0^{\frac{\pi}{2}} = 3.\]

(7). \(r = 2a (2 + \cos \theta)\) 所围成的图形为 \(\theta\) 从 \(0\) 增加到 \(2\pi\) 形成的闭合曲线所围成的图形, 因此所围成的图形的面积 \(S\) 为

\[\begin{split}S & = \int_0^{2\pi} \dfrac{1}{2} r^2 \mathrm{d} \theta = \int_0^{2\pi} \dfrac{1}{2} \cdot 4a^2 (2 + \cos \theta)^2 \mathrm{d} \theta \\ & = 2a^2 \int_0^{2\pi} \left( 4 + 4 \cos \theta + \cos^2 \theta \right) \mathrm{d} \theta \\ & = 2a^2 \int_0^{2\pi} \left( 4 + 4 \cos \theta + \dfrac{1 + \cos 2\theta}{2} \right) \mathrm{d} \theta \\ & = 2a^2 \left. \left( 4\theta + 4 \sin \theta + \dfrac{\theta}{2} + \dfrac{\sin 2\theta}{4} \right) \right|_0^{2\pi} \\ & = 2a^2 \left( 8\pi + 0 + \pi + 0 \right) = 18 \pi a^2.\end{split}\]

2. 求抛物线 \(y = -x^2 + 4x - 3\) 与其在点 \((0, -3)\) 和 \((3, 0)\) 处的切线所围成的平面图形的面积.

解

抛物线 \(y = -x^2 + 4x - 3\) 的导函数为 \(y' = -2x + 4\), 因此在点 \(A = (0, -3)\) 处的切线方程为 \(y = 4x - 3\), 在点 \(B = (3, 0)\) 处的切线方程为 \(y = -2x + 6\), 两条切线的交点为 \(C = \left( \frac{3}{2}, 3 \right)\). 因此所围成的图形的为三角形 \(\triangle ABC\) 内位于抛物线 \(y = -x^2 + 4x - 3\) 之上的部分. 因此所围成的图形的面积 \(S\) 为

\[\begin{split}S & = \int_0^{3/2} (4x - 3 - (-x^2 + 4x - 3)) \mathrm{d} x + \int_{3/2}^3 (-2x + 6 - (-x^2 + 4x - 3)) \mathrm{d} x \\ & = \int_0^{3/2} x^2 \mathrm{d} x + \int_{3/2}^3 (x^2 - 6x + 9) \mathrm{d} x \\ & = \left. \left( \dfrac{x^3}{3} \right) \right|_0^{3/2} + \left. \left( \dfrac{x^3}{3} - 3x^2 + 9x \right) \right|_{3/2}^3 \\ & = \dfrac{9}{8} - 0 + \left( 9 - 27 + 27 - \dfrac{9}{8} + \dfrac{27}{4} - \dfrac{27}{2} \right) \\ & = \dfrac{9}{8} + \dfrac{9}{8} = \dfrac{9}{4}.\end{split}\]

4. 求摆线 \(x = a(t - \sin t), y = a(1 - \cos t)\) 的一拱 \((0 \leqslant t \leqslant 2\pi)\) 的长度.

解

摆线长 \(\displaystyle \ell = \int_0^{2\pi} \sqrt{\left( \dfrac{\mathrm{d} x}{\mathrm{d} t} \right)^2 + \left( \dfrac{\mathrm{d} y}{\mathrm{d} t} \right)^2} \mathrm{d} t\), 因此有

\[\begin{split}\ell & = \int_0^{2\pi} \sqrt{a^2 \left( 1 - \cos t \right)^2 + a^2 \sin^2 t} \mathrm{d} t = \int_0^{2\pi} a \sqrt{2 - 2 \cos t} \mathrm{d} t \\ & = \int_0^{2\pi} a \sqrt{4 \sin^2 \frac{t}{2}} \mathrm{d} t = 2a \int_0^{2\pi} \sin \frac{t}{2} \mathrm{d} t = -4a \left. \cos \frac{t}{2} \right|_0^{2\pi} \\ & = 8a.\end{split}\]

6. 设抛物线 \(y^2 = 2x\) 与直线 \(y = x - 4\) 围成的平面区域为 \(D\),

   (1). 求 \(D\) 的面积；

   (2). 求 \(D\) 绕 \(x\) 轴旋转一周所生成的旋转体体积.

解

(1). 抛物线 \(y^2 = 2x\) 与直线 \(y = x - 4\) 的交点为 \(A = (8, 4)\), \(B = (2, -2)\), 因此所围成的图形为三角形 \(\triangle OAB\) 内位于抛物线 \(y^2 = 2x\) 以及直线 \(y = x - 4\) 之间的部分. 以 \(y\) 为自变量, 那么所围成的图形的面积 \(S\) 为直线 \(x = y + 4\) 之下, 抛物线 \(x = \dfrac{y^2}{2}\) 之上的部分:

\[\begin{split}S_D & = \int_{-2}^4 \left( y + 4 - \dfrac{y^2}{2} \right) \mathrm{d} y = \left. \left( \dfrac{y^2}{2} + 4y - \dfrac{y^3}{6} \right) \right|_{-2}^4 \\ & = 8 + 16 - \dfrac{64}{6} - \left( 2 - 8 + \dfrac{8}{6} \right) = 18.\end{split}\]

(2). 令点 \(E = (4, 0), F = (8, 0)\), 那么旋转体的体积等于曲线 \(y = \sqrt{2x}\), 直线 \(x = 8\) 与 \(x\) 轴所围成的图形绕 \(x\) 轴旋转一周所形成的旋转体的体积, 减去以 \(EF\) 为高的圆锥的体积, 即

\[\begin{split}V & = \pi \int_0^8 \left( \sqrt{2x} \right)^2 \mathrm{d} x - \dfrac{1}{3} \pi \cdot 4^2 \cdot 4 \\ & = 2 \pi \int_0^8 x \mathrm{d} x - \dfrac{64}{3} \pi = \left. \pi x^2 \right|_0^8 - \dfrac{64}{3} \pi \\ & = 64 \pi - \dfrac{64}{3} \pi = \dfrac{128}{3} \pi.\end{split}\]

8. 求曲线 \(xy = 1\) 与直线 \(x = 1, x = 2, y = 0\) 所围成的平面区域绕 \(y\) 轴旋转一周所形成的旋转体体积.

解

曲线 \(xy = 1\) 与直线 \(x = 1, x = 2, y = 0\) 所围成的平面区域绕 \(y\) 轴旋转一周所形成的旋转体可以分为两部分. 第一部分为曲线 \(x = \dfrac{1}{y}\), 直线 \(y = 1, y = \dfrac{1}{2}\) 与 \(y\) 轴所围成的曲边梯形绕 \(y\) 轴旋转一周所形成的旋转体减去矩形 \([0, 1] \times [\frac{1}{2}, 1]\) 绕 \(y\) 轴旋转一周所形成的旋转体, 其体积为

\[\begin{split}S_1 & = \pi \int_{\frac{1}{2}}^1 \left( \dfrac{1}{y} \right)^2 \mathrm{d} y - \left( 1 - \dfrac{1}{2} \right) \cdot \pi \cdot 1^2 \\ & = \pi \int_{\frac{1}{2}}^1 \dfrac{1}{y^2} \mathrm{d} y - \dfrac{\pi}{2} = \left. -\dfrac{\pi}{y} \right|_{\frac{1}{2}}^1 - \dfrac{\pi}{2} \\ & = -\pi + 2 \pi - \dfrac{\pi}{2} \\ & = \dfrac{\pi}{2}.\end{split}\]

第二部分为矩形 \([1, 2] \times [0, \frac{1}{2}]\) 绕 \(y\) 轴旋转一周所形成的旋转体, 其体积为

\[S_2 = \dfrac{1}{2} \cdot \pi \cdot 2^2 - \dfrac{1}{2} \cdot \pi \cdot 1^2 = \dfrac{3\pi}{2}.\]

所以所围成的图形的面积 \(S = S_1 + S_2 = \dfrac{\pi}{2} + \dfrac{3\pi}{2} = 2\pi\).

10. 设某水库的闸门为一等腰梯形, 下底为 2m, 上底为 6m, 高为 10m. 当水库水齐闸门顶时, 求闸门所受的水压力.

解

水深 \(h\) 处的压强为 \(\rho g h\), 其中 \(\rho\) 为水的密度, \(g\) 为重力加速度. 水深 \(h\) 处闸门宽 \(w\) 为 \(w = 6 - \dfrac{4}{10} h\), 因此闸门所受的水压力

\[\begin{split}F & = \int_0^{10} \rho g h \cdot \left( 6 - \dfrac{4}{10} h \right) \mathrm{d} h = \rho g \int_0^{10} \left( 6h - \dfrac{4}{10} h^2 \right) \mathrm{d} h \\ & = \rho g \left. \left( 3h^2 - \dfrac{4}{30} h^3 \right) \right|_0^{10} = \rho g \left( 300 - \dfrac{400}{3} \right) \\ & = \dfrac{500}{3} \rho g.\end{split}\]

§3.5 广义积分

1. 计算下列广义积分:

   (2). \(\displaystyle \int_2^{+\infty} \dfrac{x}{\sqrt{1 + x^2}} \mathrm{d} x\);

   (4). \(\displaystyle \int_1^{+\infty} \dfrac{1}{\sqrt{x}(1 + x)} \mathrm{d} x\);

   (6). \(\displaystyle \int_0^2 \dfrac{1}{(1 - x)^2} \mathrm{d} x\).

解

(2).

\[\begin{split}\int_2^{+\infty} \dfrac{x}{\sqrt{1 + x^2}} \mathrm{d} x & = \dfrac{1}{2} \int_2^{+\infty} \dfrac{\mathrm{d} (1 + x^2)}{\sqrt{1 + x^2}} \\ & = \left. \sqrt{1 + x^2} \right|_2^{+\infty} = +\infty.\end{split}\]

该广义积分发散.

(4).

\[\begin{split}\int_1^{+\infty} \dfrac{1}{\sqrt{x}(1 + x)} \mathrm{d} x & = 2 \int_1^{+\infty} \dfrac{\mathrm{d} \sqrt{x}}{1 + \left( \sqrt{x} \right)^2} = 2 \cdot \left. \arctan \sqrt{x} \right|_1^{+\infty} \\ & = 2 \cdot \left( \dfrac{\pi}{2} - \dfrac{\pi}{4} \right) = \dfrac{\pi}{2}.\end{split}\]

(6).

\[\begin{split}\int_0^2 \dfrac{1}{(1 - x)^2} \mathrm{d} x & = \int_0^1 \dfrac{1}{(1 - x)^2} \mathrm{d} x + \int_1^2 \dfrac{1}{(1 - x)^2} \mathrm{d} x \\ & = \left. \dfrac{1}{1 - x} \right|_0^1 + \left. \dfrac{1}{1 - x} \right|_1^2.\end{split}\]

该广义积分发散.

2. 讨论广义积分 \(\displaystyle \int_2^{+\infty} \dfrac{1}{x (\ln x)^k} \mathrm{d} x\) 的敛散性, 若收敛, 求其值. 又当 \(k\) 为何值时, 该广义积分取得最小值.

解

由于

\[\begin{split}\int_2^{+\infty} \dfrac{1}{x (\ln x)^k} \mathrm{d} x = \int_2^{+\infty} \dfrac{\mathrm{d} (\ln x)}{(\ln x)^k} = \begin{cases} \left. \dfrac{1}{(1 - k)(\ln x)^{k - 1}} \right|_2^{+\infty}, & k \neq 1 \\ \left. \dfrac{1}{\ln x} \right|_2^{+\infty}, & k = 1 \end{cases}\end{split}\]

所以当 \(k > 1\) 时, 该广义积分收敛, 值为 \(\dfrac{1}{(k - 1)(\ln 2)^{k - 1}}\); 当 \(k \leqslant 1\) 时, 该广义积分发散.

令 \(f(k) = (k - 1)(\ln 2)^{k - 1}, k > 1\), 那么

\[f'(k) = (\ln 2)^{k - 1} + (k - 1)(\ln 2)^{k - 1} \cdot \ln \ln 2 = (\ln 2)^{k - 1} \left( 1 + (k - 1) \ln \ln 2 \right).\]

由于 \(\ln 2 \in (0, 1)\), \(\ln \ln 2 < 0\), 令 \(f'(k) = 0\) 解得 \(k = 1 - \dfrac{1}{\ln \ln 2}\). 当 \(1 < k < 1 - \dfrac{1}{\ln \ln 2}\) 时, \(f'(k) > 0\); 当 \(k > 1 - \dfrac{1}{\ln \ln 2}\) 时, \(f'(k) < 0\), 因此当 \(k = 1 - \dfrac{1}{\ln \ln 2}\) 时, \(f(k)\) 取得极大值. 它是 \(f(k)\) 唯一的极大值点, 因此是其最大值点, 从而是该广义积分的最小值点.

3. 设 \(\displaystyle f(x) = \begin{cases} \lambda e^{-\lambda x}, & x \geqslant 0 \\ 0, & x < 0 \end{cases}\), 其中 \(\lambda > 0\), 试求 \(\displaystyle \int_{-\infty}^{+\infty} xf(x) \mathrm{d} x\) 与 \(\displaystyle \int_{-\infty}^{+\infty} x^2 f(x) \mathrm{d} x\).

解

\[\begin{split}\int_{-\infty}^{+\infty} xf(x) \mathrm{d} x & = \int_0^{+\infty} x \cdot \lambda e^{-\lambda x} \mathrm{d} x = - \int_0^{+\infty} x \mathrm{d} e^{-\lambda x} \\ & = - \left. x e^{-\lambda x} \right|_0^{+\infty} + \int_0^{+\infty} e^{-\lambda x} \mathrm{d} x \\ & = \left. - \dfrac{1}{\lambda} e^{-\lambda x} \right|_0^{+\infty} = \dfrac{1}{\lambda}.\end{split}\]

\[\begin{split}\int_{-\infty}^{+\infty} x^2 f(x) \mathrm{d} x & = \int_0^{+\infty} x^2 \cdot \lambda e^{-\lambda x} \mathrm{d} x = - \int_0^{+\infty} x^2 \mathrm{d} e^{-\lambda x} \\ & = - \left. x^2 e^{-\lambda x} \right|_0^{+\infty} + \int_0^{+\infty} 2x e^{-\lambda x} \mathrm{d} x \\ & = \dfrac{2}{\lambda} \int_{-\infty}^{+\infty} xf(x) \mathrm{d} x \\ & = \dfrac{2}{\lambda^2}.\end{split}\]