f : [ 0 ; 1 ] → R is continuous and ∫ 0 1 f ( x ) d x = 1 2 Proof that there exists an c ∈ [ 0 , 1 ] such that f ( c ) = c.Idea: So i would define a function g ( x ) = f ( x ) − x and then find points so that g ( x ) would be g ( y ) &lt; 0 and g ( z ) &gt; 0. And proceed with my proof and use the intermediate value theorem.However, i am confused how to do that since i don't have the exact function but only the integral.Can i just define an f ( x ) for which the integral would work like f ( x ) = 1 2 ?

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f  :  [  0  ;  1  ]  →      R   is continuous and      ∫          0              1           f  (  x  )  d  x  =      1    2  Proof that there exists an   c  ∈  [  0  ,  1  ] such that   f  (  c  )  =  c.Idea: So i would define a function   g  (  x  )  =  f  (  x  )  −  x and then find points so that   g  (  x  ) would be   g  (  y  )  &amp;lt;  0 and   g  (  z  )  &amp;gt;  0. And proceed with my proof and use the intermediate value theorem.However, i am confused how to do that since i don&#039;t have the exact function but only the integral.Can i just define an   f  (  x  ) for which the integral would work like   f  (  x  )  =      1    2  ?

darkangel5991we · Accepted Answer

Your idea was just right! If   g  (  x  )  :=  f  (  x  )  −  x then   g  (  x  ) is continuous on   [  0  ,  1  ] and:      ∫    0    1    g  (  x  )        d    x  =      ∫    0    1    f  (  x  )        d    x  −      ∫    0    1    x        d    x  =      1    2    −            1      2        2    =  0You know by the mean value theorem for Riemann integrals of continuous functions that there exists a   c  ∈  [  0  ,  1  ] such that:  0  =      ∫    0    1    g  (  x  )        d    x  =      ∫    0    1    (  f  (  x  )  −  x  )        d    x  =  g  (  c  )  (  1  −  0  )  =  f  (  c  )  −  cThat is,   f  (  c  )  =  c for at least one   c  ∈  [  0  ,  1  ].

Eliza Gregory · Accepted Answer

Let   G  (  x  )  =      ∫    0    x    g  (  x  )     d  x using the definition above (i.e.   g  (  x  )  =  f  (  x  )  −  x)  G  (  0  )  =  0  ,  G  (  1  )  =  0  G  (  x  ) is continuous and differentiable with       G    ′    (  x  )  =  g  (  x  )From the mean value theorem, we know that for all   a  ,  b  ∈  [  0  ,  1  ] with   a  &amp;lt;  b there exists a   a  &amp;lt;  c  &amp;lt;  b such that   g  (  c  )  =            G      (      b      )      −      G      (      a      )              b      −      a      With,   a  =  0 and   b  =  1 there exists a   c  ∈  [  0  ,  1  ] such that   g  (  c  )  =  0

f:[0;1]->R is continuous and int^1_0 f(x)dx=1/2. Proof that there exists an c in [0,1] such that f(c)=c.

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