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

ajanlr

Answered question

2022-10-15

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 ?

Answer & Explanation

darkangel5991we

darkangel5991we

Beginner2022-10-16Added 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 ].
Eliza Gregory

Eliza Gregory

Beginner2022-10-17Added 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

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?