Introduction
An accumulation function records how much net change has built up by a given input, relative to a chosen reference point. Its value at one input describes the accumulated amount up to that point. The difference between two of its values describes the change accumulated between the corresponding inputs.
The Fundamental Theorem of Calculus makes this relationship precise. If an accumulation function changes at rate \(f\), then the definite integral of \(f\) over an interval is the change in the accumulation function across that interval. This lets us compute a definite integral from endpoint values, even when we do not calculate the rate's integral directly.
Accumulation Values and Changes
Suppose \(f\) is a rate defined on an interval, and choose a fixed reference input \(c\). The accumulation function based at \(c\) is defined by integrating the rate from \(c\) to a variable input \(x\).
The variable \(t\) is a dummy variable of integration: it marks the input over which the rate is accumulated. The endpoint \(x\) determines the value of the accumulation function. If \(f\) is continuous, the first part of the Fundamental Theorem of Calculus gives \(A'(x)=f(x)\).
In particular, \(A(x)\) is the accumulated net change from \(c\) to \(x\), not necessarily the change from some other input \(u\) to \(x\). To find the change between \(u\) and \(v\), subtract the accumulation value at \(u\) from the value at \(v\).
The Endpoint Difference Formula
The second part of the Fundamental Theorem of Calculus states that if \(f\) is continuous on an interval and \(F\) is an antiderivative of \(f\) there, then integrating \(f\) from \(u\) to \(v\) gives the difference between the antiderivative's endpoint values. An accumulation function \(A\) with \(A'=f\) is such an antiderivative.
This formula uses the endpoint values in a specific order: the value at the upper endpoint minus the value at the lower endpoint. The order matters. If the bounds are reversed, the integral changes sign, and the endpoint difference reverses as well.
Why the Formula Works for an Accumulation Function
When \(A(x)=\int_c^x f(t)\,dt\), the relationship between the change in \(A\) and the integral over a subinterval follows directly from the additivity of definite integrals. This also explains why the chosen reference point \(c\) does not affect the change between \(u\) and \(v\).
By definition, \(A(v)=\int_c^v f(t)\,dt\) and \(A(u)=\int_c^u f(t)\,dt\).
The change from \(u\) to \(v\) is \(A(v)-A(u)\).
Splitting the integral at \(u\) gives \(\int_c^v f(t)\,dt=\int_c^u f(t)\,dt+\int_u^v f(t)\,dt\), so subtracting the first term leaves \(\int_u^v f(t)\,dt\).
Thus the accumulation values measured from \(c\) contain enough information to recover the net accumulation between any two endpoints. Equivalently, if \(A'=f\), the Fundamental Theorem gives the same endpoint-difference formula directly.
Worked Examples
Worked Example: Finding a Rate Integral from Accumulation Values
Suppose \(A(x)\) is the total net quantity accumulated from a fixed reference input to \(x\), and its rate is \(f(x)=A'(x)\). Given \(A(2)=5.5\) and \(A(6)=13.25\), find \(\int_2^6 f(t)\,dt\).
By the endpoint difference formula, the integral is the value at the upper endpoint minus the value at the lower endpoint:
Substitute the supplied values:
Therefore the net quantity accumulated from input \(2\) to input \(6\) is \(7.75\). The calculation uses only the change in the accumulation function; it does not require finding a formula for \(f\).
Check: The endpoint difference is \(A(6)-A(2)=13.25-5.5=7.75\), which matches the stated integral value.
Worked Example: Using an Explicit Accumulation Function
Let \(A(x)=x^3-2x\), and let \(f(x)=A'(x)\). Find \(\int_{-1}^3 f(t)\,dt\).
First differentiate \(A\) to identify its rate:
Now evaluate the accumulation function at the two endpoints. At \(x=3\),
At \(x=-1\),
Apply the endpoint difference formula in the order upper endpoint minus lower endpoint:
Check: Since \(A'(x)=3x^2-2=f(x)\), \(A\) is an antiderivative of the stated rate. Its endpoint difference is \(21-1=20\), confirming the integral value.
Worked Example: Reversing the Endpoints
Suppose an accumulation function has \(A(1)=4\) and \(A(5)=-2\), and \(A'=f\). Find \(\int_5^1 f(t)\,dt\).
The lower endpoint is \(5\), and the upper endpoint is \(1\). Use the same upper-minus-lower rule, keeping that order:
Substitute the values and calculate:
Check: In the forward direction, the integral is \(\int_1^5 f(t)\,dt=A(5)-A(1)=-2-4=-6\). Reversing the bounds negates it, so \(\int_5^1 f(t)\,dt=6\), as found.
Worked Example: Recovering an Unknown Accumulation Value
Suppose \(A'=f\), \(A(2)=9\), and \(\int_2^7 f(t)\,dt=-3\). Find \(A(7)\).
Use the endpoint difference formula, keeping the integral's order of bounds:
Substitute the known integral and endpoint value:
Add \(9\) to both sides:
Check: The endpoint difference is \(A(7)-A(2)=6-9=-3\), which equals the given integral. The negative integral means the accumulated value decreased by \(3\) from input \(2\) to input \(7\).
Interpreting the Result
A definite integral obtained this way is a net change. It is the final accumulation value minus the initial accumulation value. If the result is positive, the accumulation function increased overall on the interval; if it is negative, it decreased overall. A result of zero means the endpoint values are equal, even if the function rose or fell at intermediate inputs.
For example, if \(A(4)=12\) and \(A(9)=12\), then \(\int_4^9 f(t)\,dt=0\). This says that the net change over the interval is zero. It does not say that \(f\) was zero at every input, or that there was no change at intermediate points.
In applications, \(A\) might represent accumulated distance with direction, net fluid volume added, or total change in a measured quantity. The same rule applies: the definite integral of the rate over an interval is the difference between the final and initial accumulated values. Units are consistent too: if \(A\) is measured in litres, then the integral of its rate over the corresponding time interval is also measured in litres.
A Common Endpoint-Order Mistake
A frequent error is to subtract the lower-endpoint value from the upper-endpoint value in the wrong order. The integral \(\int_u^v f(t)\,dt\) corresponds to \(A(v)-A(u)\), not \(A(u)-A(v)\).
For instance, if \(A(2)=5\) and \(A(6)=11\), an incorrect calculation for \(\int_2^6 f(t)\,dt\) would be:
The correction is to match the order of the endpoints in the integral:
This also provides a useful check: if the bounds are reversed, both the integral and the endpoint subtraction must reverse. Writing the bounds and endpoint values in corresponding order before calculating helps prevent sign errors.
Summary
If \(A\) is an accumulation function whose derivative is the rate \(f\), then the definite integral of the rate between two inputs is the change in \(A\) between those inputs:
Use the endpoint at the top of the integral first, then subtract the endpoint at the bottom. The result is signed net change. The formula can be used to calculate an integral from known accumulation values or to find a missing accumulation value from an integral and one known endpoint value.
Check Your Understanding
Use the endpoint difference formula and show the subtraction order in each answer.
- If \(A'=f\), \(A(3)=8\), and \(A(7)=14\), what is \(\int_3^7 f(t)\,dt\)?
- If \(A(1)=-2\) and \(A(4)=5\), what is \(\int_4^1 A'(t)\,dt\)?
- If \(A(2)=10\) and \(\int_2^6 f(t)\,dt=4\), where \(A'=f\), find \(A(6)\).
- What does \(\int_u^v f(t)\,dt=0\) tell you about \(A(u)\) and \(A(v)\), if \(A'=f\)? Does it require \(f\) to be zero throughout the interval?
- For an accumulation function \(A(x)=\int_c^x f(t)\,dt\), explain why \(A(v)-A(u)\) does not depend on the choice of reference input \(c\).
