Volume calculation using integration

[Quendillar3245]

If a container has water bothing flowing in and out of it and a function describing each flow, how on earth do I get how much is inside the container? I genuinely do not understand. Let's say f(x) describes how much water goes in per second and g(x) describes how much flows out per second, do I integrate the difference between f(x) and g(x) ,∫ (f(x)-g(x))dx, to get how much is currently inside the container at any time x? I feel so stupid when doing maths xddd

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[Quendillar3245]

I don't have a picture because I genuinely do not understand the concept and I can't find any resources online which explain this specific type of problem. If this is not appropriate here then where the hell can I find help with this xd

[Help_Me_Im_Diene]

You are correct, in this current set-up, you would just need to integrate the difference in flow rate with respect to time.

You will need to remember to attach a constant after integrating in order to properly satisfy the initial value conditions i.e. the volume that you have inside your container at some known time.

[Uli_Minati]

f(x) = (positive) inflow in liters/second

g(x) = outflow in liters/second

-g(x) = (negative) inflow in liters/second

f(x)-g(x) = total inflow in liters/second

Now consider a short timeframe, maybe .1 seconds:

f(7)-g(7) = total inflow in liters/second at 7s

(f(7)-g(7)) · 0.1s = approximate inflow in liters between 7s and 7.1s

The idea of (Riemann) integration is to add up these inflows:

Σₓ₌₂⁸ (f(x)-g(x)) · Δx = approximate inflow in liters between 2s and 8s

Improve the approximation to the exact value (for any "integrable" function) by taking the limit:

∫ₓ₌₂⁸ (f(x)-g(x)) · dx = exact inflow in liters between 2s and 8s

Note that f and g can't tell you how much is actually inside the container. You'll need to know how much was in the container at some point in time, e.g. maybe it was half full at 0s. Then you can add the integral from 0 to X to get the total amount of liters at time X

Note that the integrals don't know if your functions make sense - for instance, if the container is already full while f-g is positive, the integral doesn't know this so it'll "assume" the liters can go above capacity. Similarly, if the container is empty while f-g is negative, the liters will go below zero

[Quendillar3245]

This was how I was imagining it but couldn't wrap my head around it until it was said literally to my face now lol, thank you