
Dayv O. answered 08/07/22
Caring Super Enthusiastic Knowledgeable Calculus Tutor
start with area function F(x)=∫f(t)dt from a to x,,,,,for some continuous f(x),,,,a is a constant
can prove F'(x)=f(x),,,,,,F'(x) is slope of area function as it grows or reduces at x,,,,Fundamental Theorem of Calculus first part --prove with Mean Value Theorem for Integrals.
otherwise it is that
F(x+dx)-F(x)=f(x)dx,,,,,definition of derivative where dx=h and h->0,,,,,F(x+dx)-F(x)=dF= differential of F(x)
integration both sidesˆ
F(x)=∫f(t)dt from a to x,,,,,,same as ∫f(t)dt +C integrated from a to x
The Intuitive part of your query-----draw f(x)=x3 and mark area under curve to x-axis
from x=1 to x=3. The increment in area to be added is 27*dx,,,
x3 at x=3 is 27.
x4/4 is the area function(which can be determined with sums of tiny rectangles and formulas for sums) & the growth rate of area function is the value of f(x) at x=3, which 27 per unit x. That is the derivative of x4/4 is x3.
It is true F(x) is integral of f(x), and f(x) is the derivative of F(x) ,,for example
If it is known the integral of sin(x) = -cos(x) then derivative of -cos(x)=sin(x),,,, same as derivative of cos(x)=(-)sin(x).
If it is known the derivative of -sin(x)=--cos(x) then integral of -cos(x)=-sin(x),,,,,,,same as integral of cos(x)=sin(x).
Which means the one operation reverses the other, but this is not an inverse relationship between functions
If F(x) and f(x) are inverse functions then F[f(x)]=x and f[F(x)]=x