From Accumulated Value to Change
An accumulation function records a signed total from a fixed starting input to a variable input. Often, however, the quantity of interest is not the total from that fixed starting point. It is the change between two inputs: how much accumulated value was added or removed as the input moved from \(u\) to \(v\).
The key fact is that this change depends only on the rate over the interval from \(u\) to \(v\). The fixed basepoint used to define the accumulation cancels out. The resulting identity connects the difference of two function values directly to a definite integral.
The basepoint \(c\) is held fixed while \(x\) varies. The integral is oriented: reversing its bounds reverses its sign. These conventions let the same definition apply whether \(x\) is to the left or to the right of \(c\).
The Interval-Increment Identity
The result is an application of additivity of definite integrals: splitting an interval at an interior point splits the integral into the sum over its pieces. With oriented bounds, the corresponding additivity relation holds for any three endpoints in the interval:
Because \(A(v)=\int_c^v f(t)\,dt\) and \(A(u)=\int_c^u f(t)\,dt\), subtracting \(A(u)\) from both sides gives the identity. This is the reason the fixed basepoint disappears: both accumulated values include the same segment from \(c\), and the difference isolates the segment from \(u\) to \(v\).
\(A(v)=\int_c^v f(t)\,dt\) and \(A(u)=\int_c^u f(t)\,dt\).
Split the oriented integral from \(c\) to \(v\) at \(u\), obtaining \(\int_c^v f=\int_c^u f+\int_u^v f\).
Subtract \(\int_c^u f(t)\,dt=A(u)\) from each side. The result is \(A(v)-A(u)=\int_u^v f(t)\,dt\).
This proof does not require that \(u\), \(v\), and \(c\) occur in any particular order. The additivity relation uses oriented integrals, so it already accounts for reversed bounds. If \(u=v\), both sides are zero: \(A(u)-A(u)=0\), and \(\int_u^u f(t)\,dt=0\).
Computing a Change from a Polynomial Rate
Worked Example: Change in a Polynomial Accumulation
Let \(A(x)=\int_0^x (2t+1)\,dt\). Find the change in accumulation from \(u=1\) to \(v=3\), first by evaluating \(A(3)-A(1)\) and then by integrating directly over \([1,3]\).
An antiderivative of \(2t+1\) is \(t^2+t\), since \(\frac{d}{dt}(t^2+t)=2t+1\). Therefore, for any input \(x\),
At \(x=3\), \(A(3)=3^2+3=9+3=12\). At \(x=1\), \(A(1)=1^2+1=1+1=2\). Hence
Directly,
The two calculations agree, as the identity requires. The antiderivative check is \(\frac{d}{dt}(t^2+t)=2t+1\), the given integrand. Thus the change in accumulated value from input \(1\) to input \(3\) is \(10\).
Reversed Endpoints and the Sign of the Change
If the input moves from a larger value \(u\) to a smaller value \(v\), the identity remains unchanged in form. The integral \(\int_u^v f(t)\,dt\) has reversed orientation, so it is the negative of \(\int_v^u f(t)\,dt\). The difference \(A(v)-A(u)\) uses the same direction: final value minus initial value.
Worked Example: Change When the Input Moves Backward
Let \(A(x)=\int_0^x t\,dt\). Find \(A(-1)-A(2)\), and verify it by evaluating the integral from \(2\) to \(-1\).
An antiderivative of \(t\) is \(\frac{t^2}{2}\), since \(\frac{d}{dt}\left(\frac{t^2}{2}\right)=t\). Thus
Evaluate both accumulation values:
Therefore the change from \(2\) to \(-1\) is
The interval integral gives the same result:
The antiderivative check is \(\frac{d}{dt}(t^2/2)=t\). The negative result is consistent with both calculations: the accumulation at \(-1\) is \(3/2\) less than the accumulation at \(2\).
Net Change Can Hide Activity
The interval integral measures signed, or net, change. Positive contributions and negative contributions within the interval can cancel. Consequently, a zero increment means the accumulation has the same value at the two endpoints; it does not by itself mean that the rate was zero throughout the interval.
Worked Example: A Piecewise Rate with Zero Net Change
Suppose \(f(t)=2\) for \(0\le t\le2\) and \(f(t)=-1\) for \(2<t\le5\), and define \(A(x)=\int_0^x f(t)\,dt\). Find the change from \(u=1\) to \(v=4\).
First calculate the accumulation at \(1\). On \([0,1]\), the rate is \(2\), so
To calculate \(A(4)\), split at the breakpoint \(t=2\), where the rate changes:
Thus the endpoint difference is
Now compute the interval integral directly, splitting it at the same breakpoint:
The identity is verified: \(A(4)-A(1)=\int_1^4 f(t)\,dt=0\). There is positive accumulation over \([1,2]\) and an equal negative contribution over \([2,4]\). The net change is zero even though the rate is nonzero on both pieces.
Reading the Identity in an Accumulation Model
Suppose \(f(t)\) is a rate measured in units of some quantity per unit of input. Then integrating \(f\) from \(u\) to \(v\) gives a net change in the quantity. The interval-increment identity says that this net change is exactly the difference between the accumulated values at the two endpoints. The units are consistent: a rate multiplied by an interval of input has the units of the accumulated quantity.
This interpretation is useful even when the basepoint is chosen for convenience rather than because it represents a meaningful starting time or position. To find the change between two inputs, one need not separately interpret the accumulation from the basepoint to each endpoint. The interval integral isolates the change of interest.
The identity also clarifies what an integral can and cannot tell you. It determines net change over the interval. If positive and negative parts cancel, the net change can be small or zero even when there has been substantial activity. The total accumulated amount without cancellation is a different quantity; it is not what this signed integral necessarily reports.
A Common Sign Mistake
A frequent error is to reverse the bounds in the integral but forget to reverse its sign. In the second example, one might incorrectly write
This integral runs from \(-1\) to \(2\), the opposite direction from the change being calculated. Its value is \(3/2\), whereas the change from \(2\) to \(-1\) is \(-3/2\). The correction is to keep the endpoints in the same order as “final value minus initial value”:
Equivalently, one may reverse the integral bounds only if one also changes the sign: \(\int_2^{-1}t\,dt=-\int_{-1}^2t\,dt\). Checking the direction of change before evaluating helps prevent this error.
Check Your Understanding
Use the interval-increment identity and keep track of the orientation of each integral.
- If \(A(x)=\int_2^x f(t)\,dt\), write \(A(5)-A(1)\) as a single definite integral.
- If \(A(x)=\int_0^x (t+1)\,dt\), compute \(A(2)-A(-1)\) using the endpoint values and verify the result by integrating over the interval from \(-1\) to \(2\).
- For an integrable \(f\), what does \(A(u)-A(v)\) equal in terms of an integral with bounds \(u\) and \(v\)?
- If \(A(4)-A(1)=0\), must \(f\) equal zero throughout \([1,4]\)? Explain what the identity does imply.
- In a rate model, what does the sign of \(A(v)-A(u)\) say about the net change between the two inputs?
