设∫(0,1)dx∫(0,1)xf(y)dy=1,则∫(0,1)f(1-x)dx=
人气:339 ℃ 时间:2019-12-10 16:44:21
解答
∫(0,1)dx∫(0,1)xf(y)dy=∫(0->1)xdx∫(0->1) f(y)dy=(1/2)∫(0->1) f(y)dy=1
所以∫(0->1) f(y)dy=2
∫(0,1)f(1-x)dx=-∫(0->1)f(1-x)d(1-x)=∫(0->1)f(y)dy=2应该是-2吧?是2,
中间少写了一步。
-∫(0->1)f(1-x)d(1-x)= -∫(1->0)f(y)dy=∫(0->1)f(y)dy=2谢谢了!
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