设f(x)连续,证明:∫0x[∫0tf(u)du]dt=∫0xf(t)(x-t)dt.

admin2019-09-04  37

问题 设f(x)连续,证明:∫0x[∫0tf(u)du]dt=∫0xf(t)(x-t)dt.

选项

答案今F(x)=∫0xf(t)dt,则F’(x)=f(x),于是∫0x[∫0tf(u)du]dt=∫0xF(t)dt, ∫0xf(t)(x-t)dt=x∫0xf(x)dt-∫0xtf(t)dt=xF(x)-∫0xtdF(t) =xF(x)-tF(t)|0x+∫0xF(t)dt=∫0xF(t)dt. 命题得证. 方法二 因为[*]∫0x[∫0xf(u)du]dt=∫0xf(u)du, [*]∫0xf(t)(x-t)dt=[*][x∫0xf(t)dt-∫0xtf(t)dt]=∫0xf(t)dt, 所以∫0x[∫0xf(u)du]dt-∫0xf(t)(x-t)dt≡C0,取x=0得C0=0,故 ∫0x[∫0tf(u)du]dt=∫0xf(t)(x-t)dt.

解析
转载请注明原文地址:https://jikaoti.com/ti/CZiRFFFM
0

相关试题推荐
最新回复(0)