设f(x)在x∈[0,1]上连续,且 ∫(0,1)f(x)dx=1,求I=∫(0,1)dx∫(x,1) f (x)f(y
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设f(x)在x∈[0,1]上连续,且 ∫(0,1)f(x)dx=1,求I=∫(0,1)dx∫(x,1) f (x)f(y)dy
1均为上限~
1均为上限~
![设f(x)在x∈[0,1]上连续,且 ∫(0,1)f(x)dx=1,求I=∫(0,1)dx∫(x,1) f (x)f(y](/uploads/image/z/145714-58-4.jpg?t=%E8%AE%BEf%28x%29%E5%9C%A8x%E2%88%88%5B0%2C1%5D%E4%B8%8A%E8%BF%9E%E7%BB%AD%2C%E4%B8%94+%E2%88%AB%280%2C1%29f%28x%29dx%3D1%2C%E6%B1%82I%3D%E2%88%AB%280%2C1%29dx%E2%88%AB%28x%2C1%29+f+%28x%29f%28y)
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再答:
![](http://img.wesiedu.com/upload/9/b5/9b5853e23a60372c9e81910a10220456.jpg)
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