设f(x)∈C[0,1],证明∫(π,0)*x*f(sinx)dx =π/2*∫(π,0)*f(sinx)dx
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/04/28 14:02:30
设f(x)∈C[0,1],证明∫(π,0)*x*f(sinx)dx =π/2*∫(π,0)*f(sinx)dx
设x = π - y,dx = - dy
当x = 0,y = π
当x = π,y = 0
∫(0→π) xf(sinx) dx = - ∫(π→0) (π - y)f(sin(π - y)) dy
= π∫(0→π) f(siny) dy - ∫(0→π) yf(siny) dy
= π∫(0→π) f(sinx) dx - ∫(0→π) xf(sinx) dx,这里的y是假变量
2∫(0→π) xf(sinx) dx = π∫(0→π) f(sinx) dx,重复,移项
∴∫(0→π) xf(sinx) dx = (π/2)∫(0→π) f(sinx) dx
当x = 0,y = π
当x = π,y = 0
∫(0→π) xf(sinx) dx = - ∫(π→0) (π - y)f(sin(π - y)) dy
= π∫(0→π) f(siny) dy - ∫(0→π) yf(siny) dy
= π∫(0→π) f(sinx) dx - ∫(0→π) xf(sinx) dx,这里的y是假变量
2∫(0→π) xf(sinx) dx = π∫(0→π) f(sinx) dx,重复,移项
∴∫(0→π) xf(sinx) dx = (π/2)∫(0→π) f(sinx) dx
设f(x)∈C[0,1],证明∫(π,0)*x*f(sinx)dx =π/2*∫(π,0)*f(sinx)dx
设f(x)在【0,1】上连续.证明∫(π/2~0)f(cosx)dx=∫(π/2~0)f(sinx)dx
设f(x)连续,证明(积分区间为0到π)∫xf(sinx)dx=(π/2)∫f(sinx)dx
设f(x)连续,证明(积分区间为0到2π)∫xf(cosx)dx=π∫f(sinx)dx
证明∫(0,π)f(sinx)dx=2∫(0,π/2)f(sinx)dx
设f(x)为连续函数,证明:∫(0,π)f(丨cosx丨)dx=2∫(0,π/2)f(sinx)dx
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)
若f(x)在[0,1]上连续,证明 ∫【上π/2下0】f(sinx)dx= ∫【上π/2下0】f(cosx)dx
∫f(sinx,cosx)dx=∫f(cosx,sinx)dx上下限是[0,π/2]
证明:定积分∫(0到π)f(sinx)dx=2∫(0到π/2)f(sinx)dx,
如何证明∫[0,π]xf(sinx)dx=π∫[0,π/2]f(sinx)dx
证明∫(上π,下0)xf(sinx)dx=π/2∫(上π,下0)f(sinx)dx