第三章 一元函数积分学 · 第8题

例题

📝 题目

例 8 设 $a < b$ ,求 $\displaystyle \int \sqrt{\left( {x - a}\right) \left( {b - x}\right) }\mathrm{d}x$ .

💡 答案与解析

解法 1 由配方得到

$$ \left( {x - a}\right) \left( {b - x}\right) = {R}^{2} - {\left( x - \frac{a + b}{2}\right) }^{2}, $$

其中 $R\overset{\text{ 定义 }}{ = }\frac{b - a}{2}$ . 作变量代换 $x = u + \frac{a + b}{2}$ ,则有

$$ \text{ 原式 } = \int \sqrt{{R}^{2} - {u}^{2}}\frac{u = R\sin t}{}{R}^{2}\int {\cos }^{2}t\mathrm{\;d}t $$

$$ = {R}^{2}\int \frac{1 + \cos {2t}}{2}\mathrm{\;d}t = {R}^{2}\left( {\frac{t}{2} + \frac{1}{4}\sin {2t}}\right) + C $$

$$ = \frac{{R}^{2}}{2}t + \frac{{R}^{2}}{2}\sin t\cos t + C $$

$$ = \frac{{R}^{2}}{2}\arcsin \frac{u}{R} + \frac{u}{2}\sqrt{{R}^{2} - {u}^{2}} + C $$

$$ = \frac{1}{4}{\left( b - a\right) }^{2}\arcsin \frac{{2x} - \left( {a + b}\right) }{b - a} $$

$$ + \frac{{2x} - \left( {a + b}\right) }{4}\sqrt{\left( {x - a}\right) \left( {b - x}\right) } + C. $$

解法 2 因为被积函数的定义域为(a, b),所以可设 $x - a =$ $\left( {b - a}\right) {\sin }^{2}t\left( {0 < t < \frac{\pi }{2}}\right)$ . 从而

$$ \sqrt{\left( {x - a}\right) \left( {b - x}\right) } = \left( {b - a}\right) \sin t\cos t, $$

$$ \mathrm{d}x = 2\left( {b - a}\right) \sin t\cos t\mathrm{\;d}t, $$

$$ \int \sqrt{\left( {x - a}\right) \left( {b - x}\right) }\mathrm{d}x = 2{\left( b - a\right) }^{2}\int {\sin }^{2}t{\cos }^{2}t\mathrm{\;d}t $$

$$ = \frac{1}{2}{\left( b - a\right) }^{2}\int {\sin }^{2}{2t}\mathrm{\;d}t = \frac{1}{4}{\left( b - a\right) }^{2}\int \left( {1 - \cos {4t}}\right) \mathrm{d}t $$

$$ = \frac{1}{4}{\left( b - a\right) }^{2}\left( {t - \sin {4t}}\right) + C. \tag{1.1} $$

又注意到

$$ \sin {4t} = 4\sin t\cos t\left( {1 - 2{\sin }^{2}t}\right) $$

$$ = 4\sqrt{\frac{x - a}{b - a}} \cdot \sqrt{1 - \frac{x - a}{b - a}}\left( {1 - 2 \cdot \frac{x - a}{b - a}}\right) $$

$$ = - 4\frac{{2x} - \left( {a + b}\right) }{{\left( b - a\right) }^{2}}\sqrt{\left( {x - a}\right) \left( {b - x}\right) }, $$

故有

$$ \int \sqrt{\left( {x - a}\right) \left( {b - x}\right) }\mathrm{d}x = \frac{1}{4}{\left( b - a\right) }^{2}\arcsin \sqrt{\frac{x - a}{b - a}} $$

$$ + \frac{{2x} - \left( {a + b}\right) }{4}\sqrt{\left( {x - a}\right) \left( {b - x}\right) } + C. $$

\subsubsection{三、联合求解法}