I am trying to understand or find a proof for the following statement:
$$ \frac{ A(2.1)-A(2) }{0.1} \approx f(2) \approx1 $$
f is defined in this graph:
And A is the full rectangular area bounded by x axis on the lower part and by y= f(t) in the upper part like the entire pink-shaded region.
I believe $$\frac{ A(2.1)-A(2) }{0.1}$$ is the slope of the line tangent to the graph of A at the given interval.
But I do not understand why this is approximately f(2).
It seems that A(x) represent the area under f(x) from 0 to x, thus for the FTC
$$A(x)=\int_0^x f(t) dt \implies A’(x)=f(x)$$
then
$$ A’(2)\approx \frac{ A(2.1)-A(2) }{0.1} \approx f(2) \approx1 $$