I am trying to understand the relationship between differentiation and integration. Differentiation has been introduced to me by this diagram:
Which displays that the derivative of a point $x$ on a continuous function $f(x)$ is the gradient of a line which is a tangent to that particular point, as shown in the diagram.
This can be written as
$\lim_{h \to 0} \left(\frac{f\left(x+h\right)-f\left(x\right)}{h}\right)=f'\left(x\right)$
and this gives the gradient of the point $(x, f(x))$.
Next examine this diagram:
The area under the continuous function from $f(x)$ from $a$ to $x$ is $A(x)$, shaded pink. The area of $A(a)$ is $0$ and the area from $a$ to $b$ is $A(b)$. Following this convention the area of $x$ to $x+h$ is
$A\left(x+h\right)-A\left(x\right)$
This section has with $h$ and the area is close to that of a rectangle so it can be said that
$A\left(x+h\right)-A\left(x\right)\approx hf(x)$
If you divide this through by $h$ you get an equation which shows that the derivative of $f(x)$ is $A'(x)$.
$\lim_{h \to 0} \left(\frac{A\left(x+h\right)-A\left(x\right)}{h}\right)=f\left(x\right)$
and hence
$A'(x)=f(x)$
This is where I have a problem, earlier it was stated that the derivative is the gradient of a line which is a tangent to a point on a continuous function. Here however, we have an area, which encloses many points, which are not a line which is a tangent to a continuous function. So I do not see how you can find the derivative of an area because it is not a point that you can find the gradient of a line which is a tangent to it.?
(Images taken from course texts provided by the Open University, Chapters C1 & C2 of the course MST121).