Net Change from a Rate Function

If you know the rate at which something is changing, the integral of that rate gives you the total change. This is sometimes called the net change theorem — and it's just the Fundamental Theorem in disguise. For a tank filling with water:

$\text{total water} = \int_{0}^{T} r(t)\,dt$

The rate $r(t)$ has units of liters per minute, and integrating with respect to time (in minutes) gives liters.

Step-by-Step Solution

Given: $r(t) = 5t - t^{2}$ liters/min, time from $t = 0$ to $t = 5$.

Find: The total volume of water that enters the tank.

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Step 1 — Set up the integral.

$V = \int_{0}^{5} (5t - t^{2})\,dt$

Step 2 — Find the antiderivative term by term.

$\int 5t\,dt = \dfrac{5t^{2}}{2}, \qquad \int t^{2}\,dt = \dfrac{t^{3}}{3}$

So:

$\int (5t - t^{2})\,dt = \dfrac{5t^{2}}{2} - \dfrac{t^{3}}{3} + C$

Step 3 — Apply the Fundamental Theorem of Calculus.

$V = \left[\dfrac{5t^{2}}{2} - \dfrac{t^{3}}{3}\right]_{0}^{5}$

Step 4 — Evaluate at $t = 5$.

$\dfrac{5(25)}{2} - \dfrac{125}{3} = \dfrac{125}{2} - \dfrac{125}{3}$

Common denominator $6$:

$= \dfrac{375}{6} - \dfrac{250}{6} = \dfrac{125}{6}$

Step 5 — Evaluate at $t = 0$.

$\dfrac{5(0)^{2}}{2} - \dfrac{0^{3}}{3} = 0$

Step 6 — Subtract.

$V = \dfrac{125}{6} - 0 = \dfrac{125}{6} \approx 20.83\;\text{liters}$

Step 7 — Sanity check the rate function.

Notice that $r(t) = t(5 - t)$, so the rate is zero at $t = 0$ and $t = 5$, and maximum at $t = 2.5$ where $r(2.5) = 6.25\;\text{L/min}$. The total integral is the area under this downward parabola from 0 to 5 — symmetric around $t = 2.5$, peaking at 6.25 L/min, with a base of 5 minutes. The area roughly looks like a $5 \times 6.25 \times 2/3 = 20.83\;\text{L}$ (using the parabola-area-as-2/3-of-bounding-rectangle rule). ✓

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Answer: The total volume of water that enters the tank from $t = 0$ to $t = 5$ is

$\boxed{\;V = \dfrac{125}{6} \approx 20.83\;\text{liters}\;}$

The rate is zero at both endpoints and peaks at $r(2.5) = 6.25\;\text{L/min}$ in the middle. The integral computes the area under the rate curve, which is the accumulated volume.

Try It

• The animation runs from $t = 0$ to $t = 5$ and shows the tank filling in real time.
• The rate curve $r(t)$ is plotted on the right; the shaded area under it grows as $t$ advances and equals the current volume.
• At $t = 5$, the total area = $125/6 \approx 20.83$ L exactly.
• Adjust the upper time bound to see partial accumulations.