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If oil is leaking from a pipe at a rate of r(t) = 2t gallons per min at t minutes then you can find the total amount of oil leaked in the first 4 minutes by finding the area under the graph of r(t) from 0 to 4. This is the definite integral of 2t from 0 to 4 which equals t^2 evaluated at 4 minus t^2 evaluated at 0 or 16 minus 0 equals 16.

What meaning if any can be given to the area under the curve of the total oil leaked function R(t) = t^2 from 0 to 4?

Basically what meaning can be given to the area under a function's graph without considering the function as a rate of change function?

Feel free to use different units. For example maybe start with a rate of change function of P'(x) = 5x + 1 representing marginal profit in dollars per unit sold if it helps give meaning to the area under the graph of the antiderivative of the function.

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Here is a possible scenario:

If $r(t)=2t$ $(t\geq0)$ is the intensity of leaking in gallons/min then$$R(t):=\int_0^t r(\tau)\>d\tau=t^2\qquad(t\geq0)$$ is the amount of oil (in gallons) lost in the first $t$ minutes.

This oil is missing in some production plant downstream, and they have to borrow oil elsewhere, for an interest. To borrow $R(t)$ gallons of oil during the time interval $[t,t+\Delta t]$ typically costs $\gamma\> R(t)\>\Delta t$ dollars interest; here $\gamma$ is a small constant. It follows that the total amount of interest $C(t)$ accrued up to time $t$ is given by $$C(t)=\gamma\>\int_0^t R(\tau)\>d\tau\ .$$

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Answering to the question:
Basically what meaning can be given to the area under a function's graph without considering the function as a rate of change function?
you can consider that the area under $y(x)$, divided by the length of the considered interval in $x$, gives the average value of the $y$ over that interval.

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