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Net volume change via the Net Change Theorem

A reservoir has water flowing in and out. The net rate of change of volume in the reservoir is given by Q′(t)=6t2−24t+18Q'(t) = 6t^2 - 24t + 18 liters per minute, where tt is measured in minutes.

Use the Net Change Theorem to find the net change in the volume of water in the reservoir over the interval [0,5][0,5] minutes. That is, compute ∫05Q′(t) dt.\int_0^5 Q'(t)\,dt. What is the net change in volume (in liters)?

Hint

Find an antiderivative Q(t)Q(t) of Q′(t)Q'(t), then compute Q(5)−Q(0)Q(5) - Q(0). Remember the Net Change Theorem tells you ∫abQ′(t) dt=Q(b)−Q(a)\int_a^b Q'(t)\,dt = Q(b) - Q(a), so you don't need to worry about the constant of integration.

Step-by-step solution

First, find an antiderivative of Q′(t)=6t2−24t+18Q'(t) = 6t^2 - 24t + 18. Integrating term by term gives Q(t)=2t3−12t2+18tQ(t) = 2t^3 - 12t^2 + 18t (dropping the constant, since it cancels in a definite integral).

By the Net Change Theorem, ∫05Q′(t) dt=Q(5)−Q(0)\int_0^5 Q'(t)\,dt = Q(5) - Q(0). Compute Q(5)=2(125)−12(25)+18(5)=250−300+90=40Q(5) = 2(125) - 12(25) + 18(5) = 250 - 300 + 90 = 40, and Q(0)=0Q(0) = 0. So the net change in volume is 40−0=4040 - 0 = 40 liters.

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