设f(x)连续,∫0xtf(x-t)dt=1-cosx,求∫0π/2f(x)dx.

admin2022-08-19  40

问题 设f(x)连续,∫0xtf(x-t)dt=1-cosx,求∫0π/2f(x)dx.

选项

答案由∫0xtf(x-t)dt[*]∫x0(x-u)f(u)(-du)=∫0x(x-u)f(u)du =x∫0xf(u)du-∫0xuf(u)du, 得x∫0xf(u)du-∫0xuf(u)du=1-cosx, 两边求导得∫0xf(u)du=sinx, [*]

解析
转载请注明原文地址:https://kaotiyun.com/show/IVR4777K
0

最新回复(0)