From Accumulated Amount to Change
An accumulation function records a quantity built up from a rate. Its value at an input depends on the chosen starting value, but the change between two of its values has a simpler description: it depends only on the rate between those two inputs.
For example, if \(A(x)\) is an amount accumulated over time, then \(A(v)-A(u)\) is the change in that amount from time \(u\) to time \(v\). The relevant rate is integrated from \(u\) to \(v\), not necessarily from the original starting time.
Definition and Difference Identity
Let \(f\) be integrable on an interval containing a reference input \(a\) and the inputs under consideration. An accumulation function with initial value \(A(a)\) is defined by
Here \(f(t)\) is the rate being accumulated, and \(A(a)\) is the starting value. The variable \(t\) is a dummy variable of integration; the input of the resulting function is \(x\).
The integral on the right is a signed integral. If \(v<u\), its orientation reverses, so \(\int_u^v f(t)\,dt=-\int_v^u f(t)\,dt\). Thus the identity respects the order of the subtraction \(A(v)-A(u)\).
Why the Identity Works
The result follows from additivity of definite integrals across adjacent intervals. In particular, for any three inputs \(a,u,v\) in the interval, the oriented integrals satisfy
This form of additivity applies regardless of the order of the endpoints, as long as the integrals are defined. Subtracting the part accumulated up to \(u\) leaves exactly the part from \(u\) to \(v\).
By definition, \(A(v)=A(a)+\int_a^v f(t)\,dt\) and \(A(u)=A(a)+\int_a^u f(t)\,dt\).
The common initial value cancels: $$ A(v)-A(u)=\int_a^v f(t)\,dt-\int_a^u f(t)\,dt. $$
Since \(\int_a^v f(t)\,dt=\int_a^u f(t)\,dt+\int_u^v f(t)\,dt\), the remaining term is $$ A(v)-A(u)=\int_u^v f(t)\,dt. $$
This derivation shows why the reference input \(a\) and initial value \(A(a)\) do not appear in the final formula. They describe where the accumulation was started, but not the change between \(u\) and \(v\).
Worked Example: Finding a Difference Algebraically
Worked Example: Finding a Difference Algebraically
Suppose \(A(x)=5+\int_1^x (3t^2-2)\,dt\). Find \(A(3)-A(1)\), and verify the result by calculating both accumulation values.
Here \(f(t)=3t^2-2\), with reference input \(a=1\) and initial value \(A(1)=5\). The difference identity gives
An antiderivative of \(3t^2-2\) is \(t^3-2t\). Therefore,
To verify by evaluating \(A\), first \(A(1)=5+\int_1^1(3t^2-2)\,dt=5+0=5\). Next,
Thus \(A(3)-A(1)=27-5=22\), agreeing with the integral from \(1\) to \(3\).
Interpreting a Difference in Context
When \(f\) describes a rate, the integral gives the net change in the accumulated quantity over the interval. Its units combine the units of the rate with the units of the input. For example, a rate in liters per minute integrated with respect to minutes gives a change in liters.
A positive result means the amount at \(v\) is greater than the amount at \(u\); a negative result means it is smaller. This conclusion concerns the net change. If the rate changes sign during the interval, the integral accounts for both increases and decreases with their signs.
Worked Example: Change in a Volume from a Rate
A tank contains \(30\) liters at time \(t=1\) minute. Its rate of change is \(r(t)=4-t\) liters per minute. Find the change in volume from \(t=1\) to \(t=5\), and then find the volume at \(t=5\).
Let \(V(t)\) be the volume. Since \(V'(t)=r(t)\), the difference identity gives
An antiderivative is \(4t-\frac{t^2}{2}\). Evaluating at both endpoints,
The rate has units of liters per minute, and integration over minutes gives liters. Since \(V(1)=30\) liters,
Check: \(V(5)-V(1)=34-30=4\) liters, which matches the integral.
When the Later Input Comes First
The identity does not require \(u<v\). If \(v<u\), preserve the order in \(A(v)-A(u)\) and in the integral \(\int_u^v\). Reversing the limits changes the sign of the integral, just as reversing the order of a difference changes its sign.
Worked Example: Reversing the Order of the Inputs
Let \(A(x)=7+\int_0^x(t-2)\,dt\). Find \(A(1)-A(4)\) using the difference identity, and check by evaluating the two function values.
Use \(u=4\) and \(v=1\), since the requested difference is \(A(1)-A(4)=A(v)-A(u)\). Then
Now evaluate the accumulation values separately:
Therefore \(A(1)-A(4)=\frac{11}{2}-7=-\frac{3}{2}\), as the oriented integral gives. The negative sign is necessary because the value at \(1\) is less than the value at \(4\).
Common Pitfall: Calculating an Accumulation Value Instead
A frequent error is to calculate the accumulation from the original reference input and report that as the change between two later inputs. The integral for a difference must start at the first input in the difference, not automatically at the accumulation function’s reference input.
Worked Example: Correcting the Wrong Interval
Let \(A(x)=6+\int_0^x t\,dt\). Find \(A(3)-A(1)\).
A wrong calculation is
This computes the accumulated change from \(0\) to \(3\), not from \(1\) to \(3\). The correct calculation is
Check the function values: \(A(3)=6+\frac{9}{2}=\frac{21}{2}\), and \(A(1)=6+\frac{1}{2}=\frac{13}{2}\). Their difference is \(\frac{21}{2}-\frac{13}{2}=\frac{8}{2}=4\). The integral from \(0\) to \(3\) gives \(A(3)-6\), not \(A(3)-A(1)\).
Using the Identity Reliably
When a problem asks for the difference between two accumulation values, first identify the order of the subtraction. If it asks for \(A(v)-A(u)\), the integral is \(\int_u^v f(t)\,dt\). Then evaluate the integral, preserve its sign, and interpret its units if the quantities are contextual.
The identity is useful even when the full formula for \(A\) is complicated or its initial value is unknown. A shared initial value cancels, so the change can be found directly from the rate over the interval of interest. It also provides a check: a separately calculated difference of accumulation values must agree with the corresponding definite integral.
Check Your Understanding
Use the difference identity and pay attention to the order of the inputs.
- If \(A(x)=2+\int_0^x(3t+1)\,dt\), find \(A(2)-A(1)\).
- If \(A(v)-A(u)=\int_u^v f(t)\,dt\), what happens to the sign when \(u\) and \(v\) are exchanged?
- A rate is measured in meters per second and integrated over a time interval in seconds. What are the units of the resulting change?
- A quantity has value \(12\) at \(t=2\), and its rate is \(r(t)=t\). Find its value at \(t=4\).
- In one sentence, explain why the initial value of an accumulation function does not affect the difference between two of its values.
