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

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

admin2022-08-19  44
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)|f(x)dx-∫abf(x)F(x)dx=F2(b)-∫abF(x)dF(x) =F2(b)-(1/2)F2(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/BtR4777K
0

考研数学三

相关试题推荐
随机试题
最新回复(0)