Is this example a correct illustration of the Fundamental Theorem Of Calculus ( Part I)?

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  • Suppose function $f(x)$ is a function giving back as output the increment of my bank account at every moment $x$ ( $x$ starting at $0$) . Suppose also ( maybe) that $f$ cannot take negative values.

The area under $C_f$ is the total amount of my bank account .

That is, if $A$ is the total amount function , $A(x) = \int_{0}^{x} f(t)dt$.

  • Now , what is the ( instantaneous) rate of growth of my total amount?

By FTC, the instantaneous rate of growth of my total amount at each moment $x$ is simply $f(x)$ : $A'(x)=f(x)$.

in other words , my gain at each moment is also the ( instantaneous) rate at which my total amount is growing.

Is this correct?

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Assume the total amount of your bank account changes continuously through your continuous labor and pauses. The "increment at any moment $x$" then is $0$ (otherwise you had $\infty$ in less than a minute). What can be $\ne0$ however is the $${\rm instantaneous\ increment\ intensity} \quad f(t) \qquad{\rm [dollars\ per\ minute]}\ .$$ When you have, e.g., $f(t_0)=0.5$ this means that in the tiny time interval $[t_0, \>t_0+dt]$ you earn $0.5\ dt$ dollars. Note that doubling this tiny $dt$ lets your "interval salary" double as well, which is reasonable.

Given this interpretation, and a starting total amount $a(0)$ we then have a total amount $a(T)$ at time $T>0$ given by $$a(T)=a(0)+\int_0^T f(t)\>dt\ .\tag{1}$$ When your wage (dollars per minute) is doubled for night time work then the instantaneous increment intensity $f$ is doubled during such times, and this will find its effect during the integration $(1)$.