Question

using nonstandard calculus the fundamental theorem of calculus is very straightforward . start with
f'(x) = (f(x+dx)-f(x)) / dx,
so
f'(x)dx = f(x + dx) - f(x)

Answer 1

nonstandard calculus means that infinitesimals exist, so dx exists . also infinitely large, i guess that means 1/dx. but i dont know what that is

Answer 2

ok the proof is straightforward. one sec

Answer 3

Substituting, and assume dx>0 without loss of generality
f'(a)dx = f(a+dx)-f(a),
f'(a+dx)*dx = f((a+dx)+dx) - f(a+dx) =f ( a + 2dx) - f(a+dx)
... in general

Answer 4

f'(a+n*dx) = f(a+ (n+1)*dx) - f( a + n*dx)

Answer 5

ok now consider the interval [a,b] where function is diff. on (a,b)
f'(a)dx + f'(a+dx)*dx + f' ( a +2dx)*dx + ... f'(a+ndx) +... f'(b)=

Answer 6

f'(a)dx + f'(a+dx)*dx + f' ( a +2dx)*dx + ... f'(a+ndx) +... f'(b)= f(a+dx) - f(a) + f(a+2dx) - f(a+dx) + f(a+3dx) - f(a+2dx)+ ...
f ( a + (n+1)dx) - f(a + ndx) + ... + f(b)

Answer 7

the right side cancels, a telescoping sum , the left side is a special sum of infintesimal products ,
dx* ( f'(a) + f'(a+dx) +.... f(b)) = integral f'(x) from a..b

Answer 8

the right side is f(b) - f(a)