Why the Basepoint Matters
A definite integral can describe how much a rate accumulates between two inputs. When we define an accumulation function, one input is fixed as the starting point, or basepoint, while the other input varies. Choosing a different basepoint changes the accumulation values, but it does so in a particularly simple way: every value changes by the same constant.
This matters when comparing two descriptions of the same rate. Their graphs may have different vertical positions because they measure accumulation from different starting points. The difference between those positions is determined exactly by the integral between the basepoints.
Accumulation Functions With Fixed Basepoints
Let \(f\) be integrable on an interval \(I\). For any fixed basepoint \(a\) in \(I\), define an accumulation function by integrating \(f\) from \(a\) to a variable input \(x\) in \(I\).
The letter \(t\) is the integration variable; \(x\) is the input of the function. We could use another letter in place of \(t\) without changing the value. The basepoint \(a\), by contrast, is fixed in this definition.
We use the oriented-integral convention: reversing the endpoints changes the sign, so \(\int_u^v f(t)\,dt=-\int_v^u f(t)\,dt\), and an integral with identical endpoints is zero. These conventions let the same accumulation definition apply whether \(x\) is to the left or to the right of \(a\).
To compare basepoints, take \(a,b\in I\) and use the same integrand \(f\) for both functions. Their common input \(x\) must also lie in \(I\), so all the integrals involved are defined.
The Constant-Offset Formula
The additivity property of definite integrals says that, for an integrable function and endpoints in its interval of integration, an integral can be split at an intermediate point. In particular, splitting the integral from \(a\) to \(x\) at \(b\) gives
This relation holds regardless of the order of \(a\), \(b\), and \(x\), provided the integrals use the oriented-integral convention. Rearranging isolates the accumulation from \(b\) to \(x\).
The term \(\int_a^b f(t)\,dt\) depends on the two fixed basepoints and on \(f\), but it does not depend on \(x\). Therefore, changing the basepoint adds a constant vertical offset to the accumulation function.
The formula gives the signed offset directly. If \(\int_a^b f(t)\,dt\) is positive, then \(A_b\) lies below \(A_a\) by that amount. If the integral is negative, then \(A_b\) lies above \(A_a\). If it is zero, the two accumulation functions agree at every input.
Why the Formula Works
The proof uses only the defining integrals and additivity. It does not require a special formula for \(f\), or a separate calculation at every input. The same portion from \(a\) to \(b\) is removed from the accumulation measured from \(a\), leaving the accumulation measured from \(b\).
By definition, \(A_a(x)=\int_a^x f(t)\,dt\) and \(A_b(x)=\int_b^x f(t)\,dt\).
Additivity gives \(\int_a^x f(t)\,dt=\int_a^b f(t)\,dt+\int_b^x f(t)\,dt\).
Subtract \(\int_a^b f(t)\,dt\) from both sides to obtain \(A_b(x)=A_a(x)-\int_a^b f(t)\,dt\).
The subtracted integral has fixed endpoints, so its value is the same for every \(x\). The change is a constant vertical shift.
Because the argument applies to every allowed \(x\), it proves the comparison for the entire domain, not merely at a sample input. Continuity of \(f\) on the interval is a common sufficient condition for the required integrals to exist; more generally, the result applies whenever the relevant definite integrals are defined and additive.
Worked Example: A Polynomial Rate
Worked Example: A Polynomial Rate
Let \(f(t)=2t+1\), and compare the accumulation functions with basepoints \(0\) and \(2\). We calculate the offset from \(0\) to \(2\):
Here \(a=0\) and \(b=2\), so the constant-offset formula gives
To see the functions explicitly, evaluate each defining integral:
Subtracting confirms the claimed difference: \(A_2(x)-A_0(x)=(x^2+x-6)-(x^2+x)=-6\), independent of \(x\). As a direct check at \(x=2\), \(A_2(2)=0\) and \(A_0(2)=2^2+2=6\), so \(A_2(2)-A_0(2)=-6\), as required by the defining integrals.
Worked Example: A Piecewise Rate
Worked Example: A Piecewise Rate
On the interval \([-1,3]\), define \(f(t)=t+2\) for \(-1\le t\le0\) and \(f(t)=2\) for \(0\le t\le3\). Compare basepoints \(a=-1\) and \(b=2\). First calculate the integral between them by splitting at the breakpoint \(0\):
The first piece is
The second piece is
Therefore, the offset integral is \(\frac{3}{2}+4=\frac{11}{2}\), and the comparison formula is
Check at \(x=3\). From \(-1\) to \(3\), the integral is \(\frac{3}{2}+\int_0^3 2\,dt=\frac{3}{2}+6=\frac{15}{2}\), so \(A_{-1}(3)=\frac{15}{2}\). From \(2\) to \(3\), \(A_2(3)=\int_2^3 2\,dt=2\). Their difference is \(2-\frac{15}{2}=-\frac{11}{2}\), which matches the formula.
Worked Example: Reversing the Basepoint Order
Worked Example: Reversing the Basepoint Order
Let \(f(t)=3-t\) on \([0,4]\). Compare \(A_3\) and \(A_1\). This time the first basepoint, \(3\), is to the right of the second, \(1\). The oriented integral from \(3\) to \(1\) is negative:
Evaluating the bracket gives
Use \(a=3\) and \(b=1\) in the formula. Since \(\int_3^1 f(t)\,dt=-2\),
Check at \(x=0\). First, \(A_3(0)=\int_3^0(3-t)\,dt=-\int_0^3(3-t)\,dt=-\frac{9}{2}\). Also, \(A_1(0)=\int_1^0(3-t)\,dt=-\int_0^1(3-t)\,dt=-\left(3-\frac{1}{2}\right)=-\frac{5}{2}\). Thus \(A_1(0)-A_3(0)=-\frac{5}{2}+\frac{9}{2}=2\), confirming the upward shift.
Interpreting the Offset
The offset is the net accumulation between the two basepoints, with its sign determined by the order of integration and the values of \(f\). It is not generally the geometric area between the graph of \(f\) and the horizontal axis: where \(f\) is negative, the definite integral counts signed contribution. If \(f\) changes sign between the basepoints, positive and negative contributions can cancel.
Geometrically, the graphs of \(A_a\) and \(A_b\) have the same shape and are translated vertically relative to one another. The formula identifies exactly how far: \(A_b-A_a=\int_b^a f(t)\,dt\). This lets us compare their values without recalculating the variable-limit integral at every input.
In an application, units also clarify the meaning. If \(f\) is measured in gallons per minute and the input is measured in minutes, then an integral between basepoints is measured in gallons. The vertical offset between the two accumulation functions is therefore a fixed amount of accumulated volume, not a change in the rate.
A Common Sign Mistake
A frequent error is to subtract the basepoint integral in the wrong order. For the polynomial example, a mistaken calculation might say
This has the wrong sign. At \(x=2\), the definition requires \(A_2(2)=\int_2^2(2t+1)\,dt=0\), whereas the incorrect expression gives \(A_0(2)+6=6+6=12\). The correct relation is \(A_2(x)=A_0(x)-6\), and it gives \(A_2(2)=6-6=0\).
A reliable check is to substitute \(x=b\): by definition \(A_b(b)=0\). Any proposed comparison formula should agree with that value. For reversed endpoints, keep the orientation convention in view rather than replacing an integral by its unsigned area.
Check Your Understanding
Use the constant-offset formula and the oriented-integral convention to answer these questions.
- If \(A_a(x)=\int_a^x f(t)\,dt\), write \(A_b(x)\) in terms of \(A_a(x)\) and an integral with fixed endpoints.
- Suppose \(\int_a^b f(t)\,dt=4\). Is \(A_b\) above or below \(A_a\), and by how much?
- For \(f(t)=t\), compare basepoints \(0\) and \(2\) by calculating the integral between them.
- Why does the relation between two accumulation functions hold for every allowed \(x\), rather than only at the basepoints?
- If \(\int_a^b f(t)\,dt=0\), what can you conclude about \(A_a\) and \(A_b\)?
