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Math Help - if f is continuous on [a,b]....

  1. #1
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    if f is continuous on [a,b]....

    Prove that if f is continuous on [a,b] with f(x) \geq 0 \ , \forall x \in [a,b] then there is c \in [a,b] \ such \ that \ f(c)=[\frac{1}{b-a} \int_a^bf^2]
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    Quote Originally Posted by flower3 View Post
    Prove that if f is continuous on [a,b] with f(x) \geq 0 \ , \forall x \in [a,b] then there is c \in [a,b] \ such \ that \ f(c)=[\frac{1}{b-a} \int_a^bf^2]

    This is false: take f(x)=10\,,\,\,[a,b]=[0,1]\Longrightarrow \frac{1}{1-0}\int\limits_0^1 10^2\,dx=100\neq 10=f(c)\,\,\forall\,c\in [0,1] .

    The claim though is true if instead f^2 we put f in the integral: is just Roll's theorem.

    Tonio
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by flower3 View Post
    Prove that if f is continuous on [a,b] with f(x) \geq 0 \ , \forall x \in [a,b] then there is c \in [a,b] \ such \ that \ f(c)=[\frac{1}{b-a} \int_a^bf^2]
    I agree with tonio, except I think he meant MVT. Let F(x)=\int_a^x f(x)\text{ }dx. By the FTC this is continuous on [a,b] and diff. on (a,b) so there exists some c\in(a,b) such that f(c)=F'(c)=\frac{1}{b-a}\left\{\int_a^b \text{ }f(x)dx-\int_a^a\text{ }f(x)dx\right\}=\frac{1}{b-a}\int_a^b\text{ }f(x)\text{ }dx
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