设f(x)在[a,b]上连续,证明:∫abf(x)dx∫xbf(y)dy=1/2[∫abf(x)dx]2.

admin2018-05-21  22

问题 设f(x)在[a,b]上连续,证明:∫abf(x)dx∫xbf(y)dy=1/2[∫abf(x)dx]2

选项

答案令F(x)=∫axf(t)dt, 则∫abf(x)dx∫xbf(y)dy=∫abf(x)[F(b))-F(x)]dx =F(b)∫abf(x)dx-∫abf(x)F(x)dx=F2(b)-∫abF(x)dF(x) =F2(b)-[*]F(x)|ab=1/2F2(b)=1/2[∫abf(x)dx]2

解析
转载请注明原文地址:https://kaotiyun.com/show/xdr4777K
0

最新回复(0)