Expert: Paul Klarreich - 12/4/2009

Question
if f is continuous on [a,b]m and there exists numbers α≠ β such that α∫ f(x)dx +β∫ f(x)dx=0  holds for all c∈(a,b) pove that
f(x)=0 for all x∈[a,b]

**note that the first integral goes from a,c and the second integral goes from c,b.  Meaning that that the c is on top and the a is on bottom, and the b is ontop and the c is on the bottom.

I know that I am supposed to use the Fundamental theorem of calculus, but I don't even have a clue to where to start on this problem.

Any help would be great thanks!

Answer
I think this has already been answered.  Check the archives.

Here it is:  I remember it now

{c
| A f(x) dx      << A instead of alpha, and B for beta.72227
}a

PLUS

{b
|  B f(x) dx
}c

= 0

means [Here's your FT of C.]

A(F(c) - F(a)) + B(F(b) - F(c)) = 0

A F(c) - A F(a)) + B F(b) - B F(c) = 0

A F(c)  - B F(c) =   A F(a)) - B F(b)

F(c)(A  - B) =   A F(a)) - B F(b)

F(c) = [A F(a)) - B F(b)]/(A - B)

So F(c) is a constant.  That means  F(x) is a constant on [a,b]

That means F'(x) = 0.
But F'(x) = f(x)

That should do it.