📝 题目
例 9 设曲线 ${x}^{\frac{1}{n}} + {y}^{\frac{1}{n}} = 1\left( {n > 1}\right)$ 在第一象限与坐标轴围成的面积为 $I\left( n\right)$ . 证明:
(1) $I\left( n\right) = {2n}{\int }_{0}^{1}{\left( 1 - {t}^{2}\right) }^{n}{t}^{{2n} - 1}\mathrm{\;d}t$ ;
(2) $\mathop{\sum }\limits_{{n = 1}}^{{+\infty }}I\left( n\right) < 4$ .
💡 答案与解析
解 (1) $I\left( n\right) = {\int }_{0}^{1}{\left( 1 - {x}^{\frac{1}{n}}\right) }^{n}\mathrm{\;d}x$ ,作变量替换 $x = {t}^{2n}$ ,即得
$$ I\left( n\right) = {2n}{\int }_{0}^{1}{\left( 1 - {t}^{2}\right) }^{n}{t}^{{2n} - 1}\mathrm{\;d}t. $$
(2)因为
$$ 0 \leq I\left( n\right) = {2n}{\int }_{0}^{1}{\left( 1 - {t}^{2}\right) }^{n}{t}^{{2n} - 1}\mathrm{\;d}t $$
$$ = {2n}{\int }_{0}^{1}\left( {1 - {t}^{2}}\right) {\left( 1 - {t}^{2}\right) }^{n - 1}{t}^{{2n} - 2}t\mathrm{\;d}t $$
$$ \leq {2n}{\int }_{0}^{1}{\left( 1 - {t}^{2}\right) }^{n - 1}{t}^{{2n} - 2}\mathrm{\;d}t $$
$$ = {2n}{\int }_{0}^{1}{\left\lbrack \left( 1 - {t}^{2}\right) {t}^{2}\right\rbrack }^{n - 1}\mathrm{\;d}t $$
$$ \leq {2n}{\int }_{0}^{1}{\left( \frac{1}{4}\right) }^{n - 1}\mathrm{\;d}t = \frac{2n}{{4}^{n - 1}}, $$
注意到 $\mathop{\sum }\limits_{{n = 1}}^{{+\infty }}{x}^{n} = \frac{x}{1 - x}\left( {\left| x\right| < 1}\right)$ ,逐项求导,得
$$ \mathop{\sum }\limits_{{n = 1}}^{{+\infty }}n{x}^{n - 1} = \frac{1}{{\left( 1 - x\right) }^{2}}\;\left( {\left| x\right| < 1}\right) . $$
当 $x = \frac{1}{4}$ 时,上式给出:
$$ \mathop{\sum }\limits_{{n = 1}}^{{+\infty }}I\left( n\right) \leq \mathop{\sum }\limits_{{n = 1}}^{{+\infty }}\frac{2n}{{4}^{n - 1}} = \frac{32}{9} < 4. $$