How to show $\alpha\int_a^cf(x)dx+\beta \int_c^b f(x)dx =0$ means $f(x)=0$ everywhere $?$ Question :
Let $$f\ :\ [a,b]\rightarrow \mathbb R$$   be  continuous . And  there  exist  constants  $\alpha$  $\beta$  such  that  $\forall c\in [a,b]$  $$\alpha \int_a^c f(x)dx +\beta \int_c^b f(x)dx =0 $$ 
Prove  that  $f=0$  allover  $[a,b]$.
Attempt :(not much useful ) 
Since  the  equation  holds  for  all  $c\in [a,b]$   , if  take  $c=a$  then  we  have $$\beta \int_a^b f(x)dx =0 $$ 
And  taking  $c=b$  we  get $$\alpha \int_a^b f(x)dx=0 $$
See  if  both  $\alpha=0$  and  $\beta=0$  hold  then  there  is  nothing  to  do . If  only  one  of  them  is  $0$   and  the  other  is  non-zero  then  from  the  last  two  equations  we  get $$\int_a^b f(x)dx =0 $$ and  this  does   not  ensure   $f$'s  being  identically  $0$  on  $[a,b]$ (for  example : $f(x)=sin x$  and $a=0$,$b=2\pi $ ) . 
So  our  only  possibility  is  that  both $\alpha \neq 0$  and $\beta \neq 0$  must  hold .
Now  I  cannot  proceed  any  further  from  this  point . Please  give  me  some   hints  as  to  what  to  do  next .
Thaanks.

If $\alpha  \neq \beta$, and we differentiate with respect to $c$, we get
$(\alpha-\beta)f(c) = 0$, from which we get $f(c) = 0$, hence $f = 0$ on $[a,b]$.
If $\alpha = \beta$, all we can conclude is that $\alpha \int_a^b f(x) dx = 0$, and so either $\alpha = 0$, or $f$ has zero average over $[a,b]$.
