来自胡俊的问题
证明:若函数f(x)在[0,1]上连续,则∫xf(sinx)dx=π/2∫f(sinx)dx(上限π,下限0)
证明:若函数f(x)在[0,1]上连续,则∫xf(sinx)dx=π/2∫f(sinx)dx(上限π,下限0)
1回答
2020-03-13 10:56
证明:若函数f(x)在[0,1]上连续,则∫xf(sinx)dx=π/2∫f(sinx)dx(上限π,下限0)
证明:若函数f(x)在[0,1]上连续,则∫xf(sinx)dx=π/2∫f(sinx)dx(上限π,下限0)
令u=π-x,du=-dx,u:π--->0,则
∫[0--->π]xf(sinx)dx
=-∫[π--->0](π-u)f(sin(π-u))du
=∫[0--->π](π-u)f(sinu)du
=π∫[0--->π]f(sinu)du-∫[0--->π]uf(sinu)du
积分变量可随便换字母
=π∫[0--->π]f(sinx)dx-∫[0--->π]xf(sinx)dx
将-∫[0--->π]xf(sinx)dx移到等式左边与左边合并,然后除去系数
∫[0--->π]xf(sinx)dx=π/2∫[0--->π]f(sinx)dx