📝 题目
解( 1 )由题设条件, 有
$$ {\int }_{a}^{b}f\left( x\right) \mathrm{d}x\frac{x = \frac{a + b}{2} + u}{2}{\int }_{-\frac{b - a}{2}}^{\frac{b - a}{2}}f\left( {\frac{a + b}{2} + u}\right) \mathrm{d}u $$
$$ = {\int }_{-\frac{b - a}{2}}^{0}f\left( {\frac{a + b}{2} + u}\right) \mathrm{d}u + {\int }_{0}^{\frac{b - a}{2}}f\left( {\frac{a + b}{2} + u}\right) \mathrm{d}u $$
$$ = {\int }_{0}^{\frac{b - a}{2}}f\left( {\frac{a + b}{2} - u}\right) \mathrm{d}u + {\int }_{0}^{\frac{b - a}{2}}f\left( {\frac{a + b}{2} + u}\right) \mathrm{d}u $$
$$ \overset{\left( {3.3}\right) \text{ 式 }}{ = }2{\int }_{0}^{\frac{b - a}{2}}f\left( {\frac{a + b}{2} + u}\right) \mathrm{d}u, $$
$$ {\int }_{a}^{b}g\left( x\right) \mathrm{d}x = \frac{x + b + u}{2}{\int }_{-\frac{b - a}{2}}^{\frac{b - a}{2}}g\left( {\frac{a + b}{2} + u}\right) \mathrm{d}u $$
$$ = {\int }_{-\frac{b - a}{2}}^{0}g\left( {\frac{a + b}{2} + u}\right) \mathrm{d}u + {\int }_{0}^{\frac{b - a}{2}}g\left( {\frac{a + b}{2} + u}\right) \mathrm{d}u $$
$$ = {\int }_{0}^{\frac{b - a}{2}}g\left( {\frac{a + b}{2} - u}\right) \mathrm{d}u + {\int }_{0}^{\frac{b - a}{2}}g\left( {\frac{a + b}{2} + u}\right) \mathrm{d}u $$
$$ = {\int }_{0}^{\frac{b - a}{2}}\left\{ {g\left( {\frac{a + b}{2} - u}\right) + g\left( {\frac{a + b}{2} + u}\right) }\right\} \mathrm{d}u $$
$$ \frac{\left( {3.3}\right) }{n} = 0\text{ . } $$
(2) $\displaystyle{\int }_{0}^{\pi }\frac{x}{1 + {\cos }^{2}x}\mathrm{\;d}x = {\int }_{0}^{\pi }\frac{x - \frac{\pi }{2}}{1 + {\cos }^{2}x}\mathrm{\;d}x + \frac{\pi }{2}{\int }_{0}^{\pi }\frac{1}{1 + {\cos }^{2}x}\mathrm{\;d}x}$
$$ \overset{\text{ 由(1) }}{ = }\pi {\int }_{0}^{\frac{\pi }{2}}\frac{1}{1 + {\cos }^{2}x}\mathrm{\;d}x $$
$$ = \pi {\int }_{0}^{\frac{\pi }{2}}\frac{1}{{\cos }^{2}x\left( {1 + {\sec }^{2}x}\right) }\mathrm{d}x $$
$$ \frac{u = \tan x}{}\pi {\int }_{0}^{\infty }\frac{1}{2 + {u}^{2}}\mathrm{\;d}u = \frac{{\pi }^{2}}{2\sqrt{2}}. $$
💡 答案与解析
解( 1 )由题设条件, 有
$$ {\int }_{a}^{b}f\left( x\right) \mathrm{d}x\frac{x = \frac{a + b}{2} + u}{2}{\int }_{-\frac{b - a}{2}}^{\frac{b - a}{2}}f\left( {\frac{a + b}{2} + u}\right) \mathrm{d}u $$
$$ = {\int }_{-\frac{b - a}{2}}^{0}f\left( {\frac{a + b}{2} + u}\right) \mathrm{d}u + {\int }_{0}^{\frac{b - a}{2}}f\left( {\frac{a + b}{2} + u}\right) \mathrm{d}u $$
$$ = {\int }_{0}^{\frac{b - a}{2}}f\left( {\frac{a + b}{2} - u}\right) \mathrm{d}u + {\int }_{0}^{\frac{b - a}{2}}f\left( {\frac{a + b}{2} + u}\right) \mathrm{d}u $$
$$ \overset{\left( {3.3}\right) \text{ 式 }}{ = }2{\int }_{0}^{\frac{b - a}{2}}f\left( {\frac{a + b}{2} + u}\right) \mathrm{d}u, $$
$$ {\int }_{a}^{b}g\left( x\right) \mathrm{d}x = \frac{x + b + u}{2}{\int }_{-\frac{b - a}{2}}^{\frac{b - a}{2}}g\left( {\frac{a + b}{2} + u}\right) \mathrm{d}u $$
$$ = {\int }_{-\frac{b - a}{2}}^{0}g\left( {\frac{a + b}{2} + u}\right) \mathrm{d}u + {\int }_{0}^{\frac{b - a}{2}}g\left( {\frac{a + b}{2} + u}\right) \mathrm{d}u $$
$$ = {\int }_{0}^{\frac{b - a}{2}}g\left( {\frac{a + b}{2} - u}\right) \mathrm{d}u + {\int }_{0}^{\frac{b - a}{2}}g\left( {\frac{a + b}{2} + u}\right) \mathrm{d}u $$
$$ = {\int }_{0}^{\frac{b - a}{2}}\left\{ {g\left( {\frac{a + b}{2} - u}\right) + g\left( {\frac{a + b}{2} + u}\right) }\right\} \mathrm{d}u $$
$$ \frac{\left( {3.3}\right) }{n} = 0\text{ . } $$
(2) $\displaystyle{\int }_{0}^{\pi }\frac{x}{1 + {\cos }^{2}x}\mathrm{\;d}x = {\int }_{0}^{\pi }\frac{x - \frac{\pi }{2}}{1 + {\cos }^{2}x}\mathrm{\;d}x + \frac{\pi }{2}{\int }_{0}^{\pi }\frac{1}{1 + {\cos }^{2}x}\mathrm{\;d}x}$
$$ \overset{\text{ 由(1) }}{ = }\pi {\int }_{0}^{\frac{\pi }{2}}\frac{1}{1 + {\cos }^{2}x}\mathrm{\;d}x $$
$$ = \pi {\int }_{0}^{\frac{\pi }{2}}\frac{1}{{\cos }^{2}x\left( {1 + {\sec }^{2}x}\right) }\mathrm{d}x $$
$$ \frac{u = \tan x}{}\pi {\int }_{0}^{\infty }\frac{1}{2 + {u}^{2}}\mathrm{\;d}u = \frac{{\pi }^{2}}{2\sqrt{2}}. $$