设f(x)∈C[0,1],证明∫（π,0）*x*f(sinx)dx =π/2*∫（π,0）*f(sinx)dx
人气：312 ℃ 时间：2019-10-03 08:45:48
解答
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(s...
推荐
- 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）在【0,1】上连续.证明∫（π/2~0)f(cosx)dx=∫（π/2~0)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)
- 如何证明∫[0,π]xf(sinx)dx=π∫[0,π/2]f(sinx)dx
- 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
- To suggest that the student did not do the reading
- -lg5 -lg7 怎么化成1\5 1\7
- at ease,be likely to,in general,lose face,defend aganist,turn one's back to用五句话编一个故事
猜你喜欢
