Expert: Paul Klarreich - 12/16/2008

Question
I am having trouble with an analysis problem.  
If f is continuous on [a,b] and there exists numbers x,y which are not equal, such that
x integral from a to c f(x)dx + y integral c to b f(x)dx =0
hold for all c in (a,b).  Prove that f(x) =0 for all x in [a,b]

Answer
Questioner:   Gloria
Category:  Advanced Math
 
Subject:  Math Analysis HW
Question:  I am having trouble with an analysis problem.  
If f is continuous on [a,b] and there exists numbers x,y which are not equal, such that
x integral from a to c f(x)dx + y integral c to b f(x)dx =0
hold for all c in (a,b).  Prove that f(x) =0 for all x in [a,b]
......................
Hi, Gloria,

Your notation is not clear, but I think you mean the following:

Given a,b, f(t), there exists a pair of numbers x /= y, such that FOR EVERY c in [a,b],

 {c
x |  f(t) dt PLUS
 }a

 {b
y |  f(t) dt
 }c

= 0

[To avoid confusion between x's, I changed the variable of integration to t.  It's a dummy, anyway.]

Now there exists F, an antiderivative of f(t) such that:

x (F(c) - F(a)) + y (F(b) - F(c)) = 0

x F(c) - x F(a) + y F(b) -  y F(c) = 0


(x - y) F(c) = x F(a) - y F(b)


      x F(a) - y F(b)
F(c) = ---------------
           x - y


But the right side does not depend on c,  So it is a constant.

F(c) = K.

and thus  F'(c) = f(c) = 0 for all c.

Does that do it for you?