Let #f:[a,b]->RR^+uu{0}# where #int_0^bf(x)dx=0# then #f(x)=0# on #[a,b]# this sentence true or false ?

Let #f(x)# be defined and continuous on #[a,b]# assuming only non negative values and:

#int_a^b f(x) dx = 0#

Consider now any point #xi in (a,b)#. Based on the additivity of the integral:

#(1) int_a^b f(x)dx = int_a^xi f(x)dx + int_xi^b f(x)dx#

Consider now the sum, that is necessarily positive:

#0 <= abs( int_a^xi f(x)dx) + abs(int_xi^b f(x)dx)#

Using the modulus inequality of integrals:

#abs( int_a^xi f(x)dx) + abs(int_xi^b f(x)dx) <= int_a^xi abs(f(x))dx+int_xi^b abs(f(x))dx#

As #f(x)# is non negative, #abs(f(x)) = f(x)#, so:

#abs( int_a^xi f(x)dx) + abs(int_xi^b f(x)dx) <= int_a^xi f(x)dx+int_xi^b f(x)dx#

that is:

#abs( int_a^xi f(x)dx) + abs(int_xi^b f(x)dx) <= int_a^b f(x)dx = 0#

So we have:

# 0 <= abs( int_a^xi f(x)dx) + abs(int_xi^b f(x)dx) <= 0#

or:

#abs( int_a^xi f(x)dx) + abs(int_xi^b f(x)dx) = 0#

but as the two quantities are non-negative, this implies they are both null and then for every #xi in [a,b]#:

#int_a^xi f(x)dx = 0#

Differentiating with respect to #xi# and applying the fundamental theorem of calculus:

#d/(d xi)(int_a^xi f(x)dx) = f(xi) = 0#