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.

ajanlr 2022-10-15 Answered
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 ) < 0 and g ( z ) > 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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Answers (2)

darkangel5991we
Answered 2022-10-16 Author has 10 answers
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 = 0
You 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 ) c
That is, f ( c ) = c for at least one c [ 0 , 1 ].
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Eliza Gregory
Answered 2022-10-17 Author has 3 answers
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 < b there exists a a < c < 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
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