📝 题目
解( 1 )因为
$$ I = {\int }_{0}^{a}f\left( x\right) g\left( x\right) \mathrm{d}x = \frac{u - x}{}{\int }_{0}^{a}f\left( {a - u}\right) g\left( {a - u}\right) \mathrm{d}u $$
$$ \overset{\left( {3.4}\right) }{ = }k{\int }_{0}^{a}f\left( u\right) \mathrm{d}u - I, $$
所以
$$ I = \frac{1}{2}k{\int }_{0}^{a}f\left( x\right) \mathrm{d}x. $$
(2)令 $f\left( x\right) = \frac{\sin x}{1 + {\cos }^{2}x},g\left( x\right) = x$ ,显然, $f\left( x\right) ,g\left( x\right)$ 满足 (3.4) 式,所以
$$ {\int }_{0}^{\pi }\frac{x\sin x}{1 + {\cos }^{2}x}\mathrm{\;d}x = \frac{\pi }{2}{\int }_{0}^{\pi }\frac{\sin x}{1 + {\cos }^{2}x}\mathrm{\;d}x $$
$$ \overset{x = - \cos x}{ = }\frac{\pi }{2}{\int }_{-1}^{1}\frac{1}{1 + {u}^{2}}\mathrm{\;d}u = \frac{{\pi }^{2}}{4}. $$
💡 答案与解析
解( 1 )因为
$$ I = {\int }_{0}^{a}f\left( x\right) g\left( x\right) \mathrm{d}x = \frac{u - x}{}{\int }_{0}^{a}f\left( {a - u}\right) g\left( {a - u}\right) \mathrm{d}u $$
$$ \overset{\left( {3.4}\right) }{ = }k{\int }_{0}^{a}f\left( u\right) \mathrm{d}u - I, $$
所以
$$ I = \frac{1}{2}k{\int }_{0}^{a}f\left( x\right) \mathrm{d}x. $$
(2)令 $f\left( x\right) = \frac{\sin x}{1 + {\cos }^{2}x},g\left( x\right) = x$ ,显然, $f\left( x\right) ,g\left( x\right)$ 满足 (3.4) 式,所以
$$ {\int }_{0}^{\pi }\frac{x\sin x}{1 + {\cos }^{2}x}\mathrm{\;d}x = \frac{\pi }{2}{\int }_{0}^{\pi }\frac{\sin x}{1 + {\cos }^{2}x}\mathrm{\;d}x $$
$$ \overset{x = - \cos x}{ = }\frac{\pi }{2}{\int }_{-1}^{1}\frac{1}{1 + {u}^{2}}\mathrm{\;d}u = \frac{{\pi }^{2}}{4}. $$