Because of the Fundamental Theorem of Calculus, Which has two important results:
Part I
If #F(x)# is any anti-derivative of #f(x)#, then:
#int_a^b f(x) dx=F(b)-F(a)#,
Part II
# d/dx \ int_a^x \ f(t) \ dt = f(x) # for any constant #a#
(ie the derivative of an integral gives us the original function back).
Derivation
Consider the above image, where #A(x)=F(x)# is definite integral of #f(x)# between #0# and #x#.
If we want to consider the area of the red strip, then we can say that #"A"=F(x+h)-F(x)#. We can also say that #"A"# is approx. equal to a rectangle with height #f(x)# and width #(x+h)-x=h#.
#"A" approx f(x) *h#
If we add on the small difference between the top of the rectangle and the curve, calling it the #"Red excess"#, we can say that:
#"A"=F(x+h)-F(x)=f(x)*h+"Red excess"#
Rearranging to get #f(x)# as the subject yields:
#f(x)=(F(x+h)-F(x))/h-("Red excess")/h#
By looking at the image, it is clear that as #(x+h) -> x#, meaning that as the difference between #(x+h)# and #x#, ie., #h#, becomes smaller and smaller, the area of the whole red strip also becomes smaller and smaller and tends to zero.
#thereforelim_(h->0)# #f(x)=lim_(h->0)# #(F(x+h)-F(x))/h-("Red excess")/h#
#=0#
#lim_(h->0) # #f(x)=f(x)# and #lim_(h->0)# # ("Red excess")/h=0 #
#thereforef(x)=lim_(h->0)# #(F(x+h)-F(x))/h#
This suggests that #f(x)=F'(x)# meaning that differentiating the integral of #f(x)# yields #f(x)#.
