Practice question · Multiple choice

Water flows into a tank at rate $r(t) > 0$ , then out at rate $|r(t)|$ when $r(t) < 0$ . What does $\int_0^T r(t)\,dt$ report?

Hints

Where $r$ dips negative, the integral counts that area negatively.
"Final volume" would also need the starting volume, the integral only measures the change.

Show the answer

A. The net change in water volume

Why

Signed area = net change: outflow intervals subtract. To get the final amount you must add the initial state, $V(T) = V(0) + \int_0^T r\,dt$ , the accumulation form of the fundamental theorem.

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Practise Integrals as accumulation

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Water flows into a tank at rate r(t)>0r(t) > 0r(t)>0, then out at rate ∣r(t)∣|r(t)|∣r(t)∣ when r(t)<0r(t) < 0r(t)<0. What does ∫0Tr(t) dt\int_0^T r(t)\,dt∫0T​r(t)dt report?

Practise Integrals as accumulation

More questions on Integrals as accumulation

Water flows into a tank at rate $r(t) > 0$ , then out at rate $|r(t)|$ when $r(t) < 0$ . What does $\int_0^T r(t)\,dt$ report?