Given a function #f(x)# continuous in the interval #[a,b]# consider the function:
#F(x) = int_a^x f(t)dt#
Evaluate now the incremental ratio of #F(x)#:
#(F(x+h)-F(x))/h = 1/h(int_a^(x+h) f(t)dt -int_a^x f(t)dt)#
#(F(x+h)-F(x))/h = 1/h int_x^(x+h) f(t)dt #
Based on the mean value theorem there must be a point #c_h in (x,x+h)# such that:
#(F(x+h)-F(x))/h = f(c_h)#
Clearly when #h->0# also #c_h -> x# and because #f(x)# is continuous #f(c_h) -> f(x)#. Then:
#lim_(h->0) (F(x+h)-F(x))/h = lim_(h->0) f(c_h) = f(x)#
which means:
#F'(x) = f(x)#
Given now any two points #a_0, b_0 in [a,b]# we have:
#int_(a_0)^(b_0) f(t) dt = int_(a_0)^a f(t) dt + int_a^(b_0) f(t) dt #
#int_(a_0)^(b_0) f(t) dt = - int_a^(a_0) f(t) dt + int_a^(b_0) f(t) dt #
#int_(a_0)^(b_0) f(t) dt = F(b_0) - F(a_0) #
Besides if #Phi(x)# is a primitive of #f(x)# that is if:
#Phi'(x) = f(x)#
from the linearity of the differentiation it follows that:
#d/dx (Phi(x) - F(x)) = d/dx Phi(x) - d/dx F(x) = f(x) -f(x) = 0#
So:
#Phi(x) = F(x) + C#
But then:
#Phi(b_0) - Phi(a_0) = F(b_0) - F(a_0)#
and we can conclude that:
#Phi'(x) = f(x) => int_a^b f(t)dt = Phi(b)-Phi(a)#
