方企勤 第三章 一元函数积分学 第4题
📝 题目
例 4 设 $a < c < d < b$ ,求 $\displaystyle{\int }_{c}^{d}\frac{\mathrm{d}x}{\sqrt{\left( {x - a}\right) \left( {b - x}\right) }}$ .
💡 答案解析
解 先计算不定积分.
$$
\int \frac{\mathrm{d}x}{\sqrt{\left( {x - a}\right) \left( {b - x}\right) }} = \int \frac{{2d}\sqrt{x - a}}{\sqrt{b - x}}
$$
$$
= 2\int \frac{d\sqrt{x - a}}{\sqrt{{\left( \sqrt{b - a}\right) }^{2} - {\left( \sqrt{x - a}\right) }^{2}}}
$$
$$
\frac{u = \sqrt{x - a}}{c = \sqrt{b - a}}2\int \frac{\mathrm{d}u}{\sqrt{{c}^{2} - {u}^{2}}} = 2\arcsin \frac{u}{c} + C
$$
$$
= 2\arcsin \sqrt{\frac{x - a}{b - a}} + C.
$$
再应用微积分基本公式, 得
$$
{\int }_{c}^{d}\frac{\mathrm{d}x}{\sqrt{\left( {x - a}\right) \left( {b - x}\right) }} = {\left. 2\arcsin \sqrt{\frac{x - a}{b - a}}\right| }_{c}^{d}
$$
$$
= 2\arcsin \sqrt{\frac{d - a}{b - a}} - 2\arcsin \sqrt{\frac{c - a}{b - a}}.
$$
评注 为了说明本例定积分的几何意义,让我们来求由 $x = c$ 至 $x = d$ 曲线 $y = \sqrt{\left( {x - a}\right) \left( {b - x}\right) }$ 这一段弧长. 设
$$
f\left( x\right) = \sqrt{\left( {x - a}\right) \left( {b - x}\right) },
$$
及
$$
A = \left( {a,0}\right) ,\;C = \left( {c,0}\right) ,\;D = \left( {d,0}\right) ,\;B = \left( {b,0}\right) ,
$$
$$
P = \left( {c,f\left( c\right) }\right) ,\;Q = \left( {d,f\left( d\right) }\right) ,
$$
那么以线段 ${AB}$ 为直径的上半圆正是曲线 $y = \sqrt{\left( {x - a}\right) \left( {b - x}\right) }$ ,其半径 $R = \frac{b - a}{2}$ (见图 3.9). 因为
$$
{y}^{2} = \left( {x - a}\right) \left( {b - x}\right) ,
$$
所以
$$
{\left( y{y}^{\prime }\right) }^{2} = {\left( \frac{a + b}{2} - x\right) }^{2} = {R}^{2} - {y}^{2},
$$
\begin{center}
\includegraphics[max width=0.2\textwidth]{images/030.jpg}
\end{center}
\hspace*{3em}
图 3.9
即得 $1 + {y}^{\prime 2} = \frac{{R}^{2}}{{y}^{2}}$ . 因此,由 $x = c$ 至 $x = d$ 曲线 $y = \sqrt{\left( {x - a}\right) \left( {b - x}\right) }$ 这一段弧 $\overset{⏜}{PQ}$ 的长度为
$$
{\int }_{c}^{d}\sqrt{1 + {y}^{\prime 2}}\mathrm{\;d}x = {\int }_{c}^{d}\frac{R}{y}\mathrm{\;d}x = {\int }_{c}^{d}\frac{R}{\sqrt{\left( {x - a}\right) \left( {b - x}\right) }}\mathrm{d}x.
$$
于是
$$
{\int }_{c}^{d}\frac{1}{\sqrt{\left( {x - a}\right) \left( {b - x}\right) }}\mathrm{d}x = \frac{1}{R}{\int }_{c}^{d}\sqrt{1 + {y}^{\prime 2}}\mathrm{\;d}x = \frac{\text{ 弧长. }}{\text{ 半径 }}
$$
由此可见本例定积分 $\displaystyle{\int }_{c}^{d}\frac{1}{\sqrt{\left( {x - a}\right) \left( {b - x}\right) }}\mathrm{d}x$ 的几何意义正是弧 $\overset{⏜}{PQ}$
所对的圆心角 $\angle {POQ}$ 的弧度数. 注意到
$$
\angle {POQ} = \left\{ \begin{array}{ll} \pi - \left( {\angle {POC} + \angle {QOD}}\right) & \left( {c < \frac{a + b}{2} < d}\right) ; \\ \angle {POC} - \angle {QOD} & \left( {\frac{a + b}{2} < c < d}\right) ; \\ \angle {QOD} - \angle {POC} & \left( {c < d < \frac{a + b}{2}}\right) . \end{array}\right.
$$
(4.1)
因为 (4.1) 式右端容易计算,所以用 (4.1) 式右端计算 $\angle {POQ}$ ,有时甚至可以直接写出答案. 请看下面一道填空题:
$$
{\int }_{\frac{1}{4}}^{\frac{3}{4}}\frac{\mathrm{d}x}{\sqrt{x\left( {1 - x}\right) }}\mathrm{d}x = \text{ \_\_\_\_\_. }
$$
此题答案应填 $\frac{\pi }{3}$ . 因为这时 $c = \frac{1}{4} < \frac{1}{2} < \frac{3}{4} = d$ ,并且 $\angle {POC} =$ $\angle {QOD} = \frac{\pi }{3}$ ,所以
$$
\angle {POQ} = \pi - \left( {\frac{\pi }{3} + \frac{\pi }{3}}\right) = \frac{\pi }{3}.
$$