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#
