Introduction
An accumulation function assigns a value to each input by integrating an integrand from a fixed basepoint to that input. This value can be understood geometrically: it records the signed area between the graph of the integrand and the horizontal axis.
The word signed matters. A region where the graph lies above the axis contributes positively; a region where it lies below contributes negatively. When both occur, the contributions combine with their signs. So the accumulation value need not equal the total geometric area of all the regions.
Accumulation Values and Signed Area
Let \(f\) be continuous on an interval containing a basepoint \(a\), and define the accumulation function by integrating \(f\) from \(a\) to an input \(x\). This tutorial focuses on inputs \(x\ge a\), so the interval being considered is \([a,x]\).
The letter \(t\) is the integration variable: it ranges across the interval being accumulated. The input \(x\) specifies the right endpoint. In this interpretation, \(A(x)\) is positive when the area above the axis exceeds the area below it, negative when the area below exceeds the area above, and zero when the two contributions are equal.
More precisely, let \(P\) be the geometric area of the part of the region above the horizontal axis, and let \(N\) be the geometric area of the part below the axis. Then the accumulation value is \(P-N\). If the graph touches or crosses the axis more than once, divide the interval into the corresponding regions and combine each region according to its sign.
When \(f(t)\ge 0\) throughout \([a,x]\), there is no below-axis contribution, so the accumulation value is the ordinary area \(P\). When \(f(t)\le 0\) throughout the interval, the accumulation value is \(-N\): the magnitude equals the geometric area, but the value is negative.
Why the Signs Give the Correct Value
The signed-area interpretation follows from how a definite integral is built. In a Riemann sum, each thin rectangle has signed height approximately \(f(t)\). Its contribution is approximately \(f(t)\,\Delta t\): positive for a rectangle above the axis and negative for one below. Adding these contributions and taking the limit gives the definite integral.
We can also make the relationship precise by separating the positive and negative parts of the integrand. Define \(f_+(t)=\max(f(t),0)\) and \(f_-(t)=\max(-f(t),0)\). At every point, \(f_+\) records the height above the axis, while \(f_-\) records the magnitude of the height below it. Therefore \(f=f_+-f_-\).
This also explains why crossings of the axis matter: the sign of the height changes, so the corresponding contribution changes from positive to negative, or from negative to positive. The integral combines all these contributions into a single value.
Worked Examples
Worked Example: A Larger Region Above the Axis
Let \(f(t)=3-t\), with basepoint \(a=0\). Find and interpret \(A(4)=\int_0^4(3-t)\,dt\).
The graph crosses the horizontal axis where \(3-t=0\), so \(t=3\). It is above the axis from \(0\) to \(3\), and below it from \(3\) to \(4\).
The region above the axis is a triangle with base \(3-0=3\) and height \(f(0)=3\). Its area is \(\frac12(3)(3)=\frac92\).
The region below the axis is a triangle with base \(4-3=1\) and height \(|f(4)|=|3-4|=1\). Its geometric area is \(\frac12(1)(1)=\frac12\), and its signed contribution is \(-\frac12\).
Thus the accumulation value is \(\frac92-\frac12=\frac82=4\).
Check by evaluating the integral. An antiderivative is \(3t-\frac{t^2}{2}\), so $$ \int_0^4(3-t)\,dt = \left[3t-\frac{t^2}{2}\right]_0^4 = \left(12-\frac{16}{2}\right)-\left(0-\frac{0}{2}\right) = 12-8 = 4. $$ The integral agrees with the signed-area calculation.
Worked Example: Equal Contributions Cancel
Let \(f(t)=t-2\), with basepoint \(a=0\). Find \(A(4)=\int_0^4(t-2)\,dt\) by interpreting the regions.
The graph crosses the axis when \(t-2=0\), so \(t=2\). On \([0,2]\), it lies below the axis. The region is a triangle with base \(2\) and height \(|f(0)|=2\), giving geometric area \(\frac12(2)(2)=2\). Its signed contribution is \(-2\).
On \([2,4]\), the graph lies above the axis. This triangle has base \(4-2=2\) and height \(f(4)=4-2=2\), so its area and signed contribution are both \(\frac12(2)(2)=2\).
The contributions sum to \(-2+2=0\). Check directly: $$ \int_0^4(t-2)\,dt = \left[\frac{t^2}{2}-2t\right]_0^4 = \left(\frac{16}{2}-8\right)-\left(\frac{0}{2}-0\right) = (8-8)-0 = 0. $$ The two equal regions have opposite signs, so the accumulation value is zero.
Worked Example: Curved Regions on Both Sides
Let \(f(t)=t^2-1\), with basepoint \(a=0\). Find and interpret \(A(2)=\int_0^2(t^2-1)\,dt\).
The graph crosses the axis in this interval when \(t^2-1=0\), which gives \(t=1\). It is below the axis on \([0,1]\) and above the axis on \([1,2]\). The regions are curved, so calculate their areas using integrals.
The geometric area below the axis is $$ \int_0^1(1-t^2)\,dt = \left[t-\frac{t^3}{3}\right]_0^1 = \left(1-\frac13\right)-(0-0) = \frac23. $$ Its signed contribution to \(A(2)\) is \(-\frac23\).
The area above the axis is $$ \int_1^2(t^2-1)\,dt = \left[\frac{t^3}{3}-t\right]_1^2 = \left(\frac83-2\right)-\left(\frac13-1\right) = \frac23-\left(-\frac23\right) = \frac43. $$ Therefore the signed-area calculation gives \(-\frac23+\frac43=\frac23\).
Check by evaluating the integral over the full interval: $$ \int_0^2(t^2-1)\,dt = \left[\frac{t^3}{3}-t\right]_0^2 = \left(\frac83-2\right)-0 = \frac83-\frac63 = \frac23. $$ This matches the result from the two separate regions.
A Common Mistake: Making Every Region Positive
A frequent error is to add the geometric areas on both sides of the axis when calculating an accumulation value. That produces the total geometric area, not the signed area.
For example, in the second worked example, each triangle has geometric area \(2\). Adding them gives \(2+2=4\). But this is not \(A(4)\), because the first triangle is below the axis and contributes \(-2\), not \(+2\). The correct calculation is \(-2+2=0\), which agrees with evaluating the definite integral.
Using the Interpretation
The signed-area viewpoint lets you check whether a computed accumulation value is plausible. If the graph stays above the axis, its accumulation value from the basepoint cannot be negative. If it stays below, the value cannot be positive. If the graph lies on both sides, compare the sizes of the positive and negative contributions.
The interpretation is also useful when a graph is easier to understand than an antiderivative. Simple regions, such as triangles, can be evaluated from their geometric dimensions. For curved regions, the definite integral provides the exact signed value. Either way, the sign is determined by the graph's position relative to the horizontal axis.
Keep the distinction clear: an accumulation value is a signed quantity, not automatically the sum of all geometric areas. Each region contributes according to whether its graph lies above or below the axis on the interval from the basepoint to the input.
Check Your Understanding
Answer each question using the signed-area interpretation.
- If \(f(t)\ge 0\) throughout \([a,x]\), how is \(A(x)\) related to the geometric area between the graph and the axis?
- A graph encloses a geometric area of \(5\) above the axis and an area of \(3\) below it between \(a\) and \(x\). What is \(A(x)\)?
- For \(f(t)=t-1\) on \([0,2]\), identify the intervals where the graph is below and above the axis. What are the signed contributions of the two triangular regions?
- Why does adding the geometric areas on both sides of the axis generally fail to give an accumulation value?
- If \(A(x)=0\), must there be no area between the graph and the axis? Explain.
