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
$$ \angle {POQ} = \pi - \left( {\frac{\pi }{3} + \frac{\pi }{3}}\right) = \frac{\pi }{3}. $$