Using Accumulation Functions and Definite Integrals in Applied Contexts
Introduced| English |
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| definite integral/ˈdefɪnət ˈɪntɪɡrəl/ |
Start here, add the change, land there
- Many applied problems give a rate and a starting amount, and ask for a later total.
- The key relationship: final value = initial value + accumulated change.
- The accumulated change is the definite integral of the rate over the interval.
- This one idea powers tanks, populations, accounts, and more.
The accumulation formula
- If $R(t)$ is the rate and $Q$ the quantity, then:
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$$Q(b)=Q(a)+\int_a^b R(t)\,dt$$
- Start with $Q(a)$, add the net accumulated change $\displaystyle\int_a^b R(t)\,dt$, arrive at $Q(b)$.
- It's the FTC in applied clothing: the integral of a rate is the net change of its quantity.
Accumulated inflow is the area
y = 4t (inflow rate)
The area under the inflow rate $R(t)$ is the water added; the total is the initial amount plus that area.
The applied accumulation relationship is: final value =...
Final = initial + accumulated change.
To find the final amount, add the accumulated change to the ____ value.
Final = initial + integral of the rate.
Reading the context and units
- A definite integral 定积分 of a rate has units of (rate units)$\times$(time) = the quantity's units.
- $\int$ of litres/min over minutes gives litres; $\int$ of people/year over years gives people.
- The sign matters: a negative rate (draining, cooling) subtracts from the total.
- Always interpret the integral in the problem's own words and units.
A tank has $50$ L; water flows in at $R(t)=4t$ for $0\le t\le 3$ ($\int_0^3 4t\,dt=18$). Find the total at $t=3$.
$50+18=68$ L.
If $R$ is in litres/min and $t$ in minutes, $\int R\,dt$ has units of...
(litres/min)×(min) = litres.
Net vs. gross change
- The integral $\int_a^b R\,dt$ is the net change — inflow minus outflow.
- If a tank is filled and drained, the net integral accounts for both automatically.
- To find total added or total removed separately, integrate only the positive or negative parts.
- But "final = initial + net accumulated change" always uses the plain signed integral.
The definite integral of the rate gives the change, not the final total by itself.
You must add the initial value.
For an outflow (draining) rate, the integral over the interval is...
An outflow rate is negative → subtracts.
Don't report the integral alone as the final amount — it's only the change. You must add the initial value: final $=$ initial $+\int_a^b R\,dt$. And keep the sign of the rate: an outflow rate is negative, so its integral correctly subtracts from the starting total.
A tank holds $50$ L at $t=0$. Water flows in at $R(t)=4t$ L/min for $0\le t\le 3$. How much at $t=3$?
- Accumulated change: $\displaystyle\int_0^3 4t\,dt=\Big[2t^2\Big]_0^3=18$ L.
- Final $=$ initial $+$ change $=50+18=68$ L.
- (The integral $18$ is the added water; $68$ is the total.)
In applied contexts, final value = initial value + accumulated change, where the change is the definite integral of the rate: $Q(b)=Q(a)+\int_a^b R(t)\,dt$. The integral carries the quantity's units and its sign (inflow adds, outflow subtracts). Never forget to add the initial value.