Accumulating Signed Change from a Base Point
An integral can describe the net amount accumulated as an input moves from one endpoint to another. Fix a starting input \(a\), and let the other endpoint \(x\) vary. The resulting accumulation function assigns a definite integral to each allowed value of \(x\).
The fixed value \(a\) is called the base point. It determines where accumulation is measured from. The value \(A(x)\) is not generally the total geometric area between the graph and the horizontal axis: it is a signed integral, so regions below the axis count negatively, and reversing the endpoints reverses the sign.
The letter used inside the integral can be changed without changing the value: \(\int_a^x f(t)\,dt\) and \(\int_a^x f(u)\,du\) express the same accumulation. The endpoint \(x\), not the name of the integration variable, is the input to \(A\).
The Value at the Base Point
At \(x=a\), the interval of accumulation has zero length. A definite integral over an interval with identical endpoints is zero, so every accumulation function has a known value at its base point:
This holds regardless of the value of \(f(a)\). For instance, if \(f(a)=7\), that does not make \(A(a)=7\): \(f(a)\) is the integrand’s value at one input, while \(A(a)\) measures accumulation over an interval of zero length.
Reading Accumulation as Signed Area
When a graph or formula makes the relevant regions easy to identify, definite integrals can be evaluated as signed geometric areas. On an interval where \(f\) is above the horizontal axis, the integral contributes positive area. Where \(f\) is below the axis, it contributes negative area. If the interval is split into simpler pieces, add the signed contributions.
If \(x<a\), the integral’s endpoints run in the reverse order from the usual left-to-right direction. The reversal rule gives
Thus a region that has positive geometric area between \(x\) and \(a\) contributes a negative value to \(A(x)\) when the accumulation is taken from \(a\) back to \(x\). The graph’s position and the order of the endpoints both matter.
Worked Example: Evaluate Values from a Formula
Worked Example: Evaluate Values from a Formula
Let \(f(t)=t+2\) on \([0,3]\), and define \(A(x)=\int_1^x (t+2)\,dt\). Find \(A(1)\), \(A(0)\), and \(A(3)\).
At the base point: Since \(x=1\) is the base point,
At \(x=0\): First find the area under the graph from \(0\) to \(1\). The line \(f(t)=t+2\) has endpoint heights \(f(0)=2\) and \(f(1)=3\). The region is a trapezoid with width \(1\), so its area is
The integral runs from \(1\) to \(0\), so reverse the endpoints and change the sign:
At \(x=3\): From \(t=1\) to \(t=3\), the endpoint heights are \(f(1)=3\) and \(f(3)=5\), and the width is \(2\). The trapezoid’s area is
Check: Splitting the last interval at \(t=2\), the trapezoid areas are \(\frac{3+4}{2}\cdot1=\frac{7}{2}\) on \([1,2]\) and \(\frac{4+5}{2}\cdot1=\frac{9}{2}\) on \([2,3]\). Their sum is \(\frac{7}{2}+\frac{9}{2}=8\), as found. The results also give \(A(1)=0\) at the base point and a negative value at \(x=0<1\), as expected from endpoint reversal.
Why Add Areas Across Intervals?
Definite integrals are additive across adjacent intervals. In particular, for a point \(y\) between \(a\) and \(x\), the accumulation from \(a\) to \(x\) can be split at \(y\):
This says that the net accumulation over the whole interval is the sum of the net accumulations over its pieces. The pieces must retain their signs. One way to see why the rule applies is to split a Riemann sum for the whole interval into the subintervals on either side of \(y\); the sum of the two parts is the sum over the whole interval, and the corresponding integral is therefore the sum of the two integrals.
Equivalently, for any two inputs \(x\) and \(y\) in the domain, additivity and reversal of endpoints give
This identity is useful for interpreting changes between accumulation values. It follows by splitting the integral at \(y\): \(\int_a^x f=\int_a^y f+\int_y^x f\), then subtracting \(A(y)=\int_a^y f\). It concerns integral values only; it does not require calculating a derivative.
Worked Example: Read Values from a Graph
Worked Example: Read Values from a Graph
Suppose the graph of \(f\) consists of straight segments connecting \((-2,0)\), \((0,2)\), \((2,0)\), and \((4,-2)\). Define \(A(x)=\int_{-1}^x f(t)\,dt\). Find \(A(-1)\), \(A(1)\), and \(A(3)\).
At the base point,
For \(A(1)\), the graph is above the axis from \(-1\) to \(1\). On \([-1,0]\), its heights are \(f(-1)=1\) and \(f(0)=2\), so the trapezoid area is \(\frac{1+2}{2}\cdot1=\frac{3}{2}\). On \([0,1]\), its heights are \(2\) and \(1\), so the area is \(\frac{2+1}{2}\cdot1=\frac{3}{2}\). Therefore,
For \(A(3)\), include the region from \(1\) to \(3\) as well. From \(1\) to \(2\), the heights are \(1\) and \(0\), so the positive triangular area is \(\frac{1}{2}\cdot1\cdot1=\frac{1}{2}\). From \(2\) to \(3\), the graph is below the axis; its heights in magnitude are \(0\) and \(1\), so its signed triangular area is \(-\frac{1}{2}\cdot1\cdot1=-\frac{1}{2}\). Thus
Check: The two new contributions cancel: the positive area on \([1,2]\) is \(\frac12\), and the negative area on \([2,3]\) is \(-\frac12\). Hence \(A(3)=A(1)=3\), consistent with the signed areas read from the graph.
Worked Example: Use a Piecewise Formula and a Base Point Inside the Domain
Worked Example: Use a Piecewise Formula and a Base Point Inside the Domain
On \([-2,2]\), let
Define \(A(x)=\int_1^x f(t)\,dt\). Find \(A(-1)\), \(A(0)\), and \(A(2)\).
To find \(A(-1)\), reverse the bounds and split at \(0\):
The first integral is a rectangle of height \(2\) and width \(1\), so it is \(2\). The second is a trapezoid with heights \(2\) and \(3\) and width \(1\), so it is \(\frac{2+3}{2}\cdot1=\frac52\). Therefore,
For \(A(0)\), reverse the bounds from \(1\) to \(0\). On \([0,1]\), the area is \(\frac52\), so
For \(A(2)\), the graph on \([1,2]\) has endpoint heights \(3\) and \(4\) and width \(1\). Thus
Check: The values satisfy \(A(0)-A(-1)=-\frac52-(-\frac92)=2\). This equals \(\int_{-1}^0 f(t)\,dt\), the rectangle of height \(2\) and width \(1\). Also, \(A(2)-A(0)=\frac72-(-\frac52)=6\), equal to the trapezoid area \(\frac52\) on \([0,1]\) plus the trapezoid area \(\frac72\) on \([1,2]\). Both differences agree with the signed areas on the corresponding intervals.
A Common Sign Mistake
A frequent error is to treat the region between the graph and the axis as positive area even when the integral’s endpoints are reversed. In the first worked example, the geometric area from \(0\) to \(1\) is \(\frac52\). A tempting but incorrect calculation is
The area has been calculated correctly, but its sign has not. Since the endpoints run from \(1\) back to \(0\), the correct calculation is
Keep two sign sources distinct: a region below the axis contributes negatively, and reversing the integral’s bounds also changes the sign. When both apply, both must be accounted for.
Why the Base Point Matters
The base point determines the reference from which all the values of an accumulation function are measured. If the same integrand is used with a different base point, the new function generally has different values. For example, \(\int_1^x f(t)\,dt\) and \(\int_0^x f(t)\,dt\) differ by the accumulated integral from \(0\) to \(1\), wherever all these integrals are defined.
Accumulation functions are useful for describing how a net quantity changes relative to an initial location. The integrand may represent a rate, a density, or another quantity whose contributions are accumulated over an interval. The signed integral keeps track of both increases and decreases. In this tutorial, the key task is to identify the base point, respect endpoint order, and combine the signed contributions—not to confuse the integrand’s value with the accumulated value.
Check Your Understanding
Use the definition of an accumulation function and signed integral values to answer the following questions.
- If \(A(x)=\int_4^x f(t)\,dt\), what is \(A(4)\), and why?
- If the geometric area under \(f\) from \(2\) to \(5\) is \(7\), what is \(\int_5^2 f(t)\,dt\), assuming \(f\) is above the axis on that interval?
- A graph has signed area \(3\) from \(a\) to \(b\), then signed area \(-1\) from \(b\) to \(c\). What is \(\int_a^c f(t)\,dt\)?
- Why can an accumulation value be negative even when the graph of \(f\) is above the axis between \(x\) and \(a\)?
- In a piecewise formula, why might it be helpful to split the integral at the input where the formula changes?
