作业帮f(x)在〔0,1〕内连续,证明：∫{0,1}∫{x,1}∫{x,y}f(x)f(y)f(z)dxdydz={[∫{0,1}f(x)dx]^3}/6
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/09 09:01:54
![f(x)在〔0,1〕内连续,证明：∫{0,1}∫{x,1}∫{x,y}f(x)f(y)f(z)dxdydz={[∫{0,1}f(x)dx]^3}/6](/uploads/image/z/6398086-22-6.jpg?t=f%28x%29%E5%9C%A8%E3%80%940%2C1%E3%80%95%E5%86%85%E8%BF%9E%E7%BB%AD%2C%E8%AF%81%E6%98%8E%EF%BC%9A%E2%88%AB%7B0%2C1%7D%E2%88%AB%7Bx%2C1%7D%E2%88%AB%7Bx%2Cy%7Df%28x%29f%28y%29f%28z%29dxdydz%3D%7B%5B%E2%88%AB%7B0%2C1%7Df%28x%29dx%5D%5E3%7D%2F6)
f(x)在〔0,1〕内连续,证明：∫{0,1}∫{x,1}∫{x,y}f(x)f(y)f(z)dxdydz={[∫{0,1}f(x)dx]^3}/6
f(x)在〔0,1〕内连续,证明：∫{0,1}∫{x,1}∫{x,y}f(x)f(y)f(z)dxdydz={[∫{0,1}f(x)dx]^3}/6
f(x)在〔0,1〕内连续,证明：∫{0,1}∫{x,1}∫{x,y}f(x)f(y)f(z)dxdydz={[∫{0,1}f(x)dx]^3}/6
把积分顺序的六种可能都写一遍,用变量替换证明这六个积分都一样,并且和就是右边……