设f(x)在[1,2]上连续,证明(∫(2,1)f(x)dx²≦∫(2,1)f²(x)dx
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设f(x)在[1,2]上连续,证明(∫(2,1)f(x)dx²≦∫(2,1)f²(x)dx
![设f(x)在[1,2]上连续,证明(∫(2,1)f(x)dx²≦∫(2,1)f²(x)dx](/uploads/image/z/4075010-26-0.jpg?t=%E8%AE%BEf%28x%29%E5%9C%A8%EF%BC%BB1%2C2%EF%BC%BD%E4%B8%8A%E8%BF%9E%E7%BB%AD%2C%E8%AF%81%E6%98%8E%28%E2%88%AB%282%2C1%29f%28x%29dx%26%23178%3B%E2%89%A6%E2%88%AB%282%2C1%29f%26%23178%3B%28x%29dx)
这是柯西不等式的积分形式
可用二次函数证
区间省略 显然有 0≦∫[ t + f(x) ]²dx
又 ∫[ t + f(x) ]²dx = ∫t²+2 tf(x) +f(x)²dx =t²+2t∫f(x)dx+∫f(x)²dx
所以t²+2t∫f(x)dx+∫f(x)²dx ≥0
所以⊿≤0,即∫f(x)dx²≦∫f²(x)dx
命题得证
可用二次函数证
区间省略 显然有 0≦∫[ t + f(x) ]²dx
又 ∫[ t + f(x) ]²dx = ∫t²+2 tf(x) +f(x)²dx =t²+2t∫f(x)dx+∫f(x)²dx
所以t²+2t∫f(x)dx+∫f(x)²dx ≥0
所以⊿≤0,即∫f(x)dx²≦∫f²(x)dx
命题得证
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