Introduction
An accumulation function can be built by fixing one endpoint of an integral and allowing the other endpoint to vary. The position of that moving endpoint matters: reversing the order of integration reverses the sign of the integral.
This tutorial focuses on an accumulation whose lower endpoint varies while its upper endpoint remains fixed. The key is to keep the orientation visible, then interpret the value as signed area.
Variable Lower Endpoint, Fixed Upper Endpoint
Let \(f\) be integrable on an interval containing a fixed number \(b\) and a variable number \(x\). Define the accumulation
The variable \(t\) is the integration variable; \(x\) is the input that determines the lower endpoint. The fixed number \(b\) is the upper endpoint. As \(x\) changes, the interval being accumulated changes.
Recall the orientation convention for definite integrals: reversing the endpoints changes the sign. Thus, whenever the integral is defined,
This identity is useful because many calculations or area readings are naturally made from the fixed endpoint \(b\) out to \(x\). Reversing the bounds lets us express that calculation in the order we need, but the minus sign must remain.
How to Read the Sign of the Accumulation
There are two separate sign effects to keep track of. First, the function can be above or below the horizontal axis, which determines the sign of its signed area. Second, the order of the endpoints determines the orientation of the integral.
If \(x<b\), the integral \(\int_x^b f(t)\,dt\) is taken in the usual left-to-right direction. Its value is the signed area from \(x\) to \(b\): regions above the axis contribute positively, and regions below contribute negatively.
If \(x>b\), the integral \(\int_x^b f(t)\,dt\) has reversed orientation. It is the negative of the signed area from \(b\) to \(x\). In particular, if \(f\) is positive throughout that interval, the accumulation is negative; if \(f\) is negative throughout, the accumulation is positive.
| Position of \(x\) | Useful expression | Interpretation |
|---|---|---|
| \(x<b\) | \(A(x)=\int_x^b f(t)\,dt\) | Signed area from \(x\) to \(b\) |
| \(x=b\) | \(A(b)=0\) | The interval has zero length |
| \(x>b\) | \(A(x)=-\int_b^x f(t)\,dt\) | Negative of the signed area from \(b\) to \(x\) |
The sign of the accumulation is not determined by endpoint order alone: the sign of \(f\) also matters. A negative orientation reverses the sign of the signed area, whatever that signed area may be.
Why Reversing Bounds Changes the Sign
The orientation rule follows from additivity of definite integrals. For integrable \(f\), integrals over adjacent intervals add, and the integral over an interval of zero length is zero. Applying additivity to the points \(x\) and \(b\) gives
Moving the second term to the other side proves the reversal identity:
This also explains why the sign is necessary. The two integrals traverse the same interval in opposite directions, so their contributions cancel when added.
Worked Examples
Worked Example: A Variable Endpoint and a Positive Triangular Area
Let \(f(t)=4-t\), fix \(b=4\), and find \(A(1)=\int_1^4(4-t)\,dt\). On the interval from \(1\) to \(4\), the graph is a line segment from height \(3\) to height \(0\), so the region is a triangle above the axis.
Its base is \(4-1=3\), and its height is \(4-1=3\). Therefore its signed area is
The graph lies above the axis on this interval, so the signed area is positive. Hence \(A(1)=\frac{9}{2}\).
Check by reversing the endpoints. The signed area from \(4\) to \(1\) is \(-\frac{9}{2}\), so
This agrees with the original accumulation.
Worked Example: The Variable Endpoint Lies Above the Fixed Endpoint
Let \(f(t)=-3\), fix \(b=2\), and evaluate \(A(5)=\int_5^2 -3\,dt\). Since \(5>2\), reverse the bounds and include a minus sign:
From \(2\) to \(5\), the graph is a rectangle of width \(5-2=3\) and height \(-3\). Its signed area is
Therefore,
Check directly using the constant integrand and the oriented endpoint difference:
The direct calculation confirms the reversed-bound result. The negative rate and the reversed orientation each contribute a negative sign, giving a positive accumulation.
Worked Example: Combining Positive and Negative Signed Areas
Let \(f(t)=t-3\), fix \(b=4\), and evaluate \(A(0)=\int_0^4(t-3)\,dt\) from signed areas. The graph crosses the horizontal axis at \(t=3\). From \(0\) to \(3\), it forms a triangle below the axis with base \(3\) and height \(3\), so its signed area is
From \(3\) to \(4\), the graph forms a triangle above the axis with base \(1\) and height \(1\), so its signed area is
Add the two signed areas:
Check by reversing the bounds. The signed area from \(4\) to \(0\) is the negative of the signed area from \(0\) to \(4\), namely \(4\). Therefore
which matches the value found by combining the two regions.
A Reliable Evaluation Procedure
When evaluating \(A(x)=\int_x^b f(t)\,dt\) from a graph or signed-area information, keep the fixed endpoint and the variable endpoint in view. The following sequence separates the geometric sign from the orientation sign.
Record the variable endpoint \(x\) and fixed endpoint \(b\). Determine which one is smaller.
For the interval between the endpoints, assign positive area above the horizontal axis and negative area below it. Combine regions using their signs.
If \(x<b\), that signed area is already \(\int_x^b f(t)\,dt\). If \(x>b\), negate the signed area computed from \(b\) to \(x\).
Compare with the endpoint order and the sign of the function on the interval. Confirm that any reversed bounds contributed exactly one additional minus sign.
Common Mistake: Reversing Bounds Without Reversing the Sign
Suppose \(f(t)=2\), \(b=4\), and \(x=1\). The correct value is the area of a rectangle with width \(4-1=3\) and height \(2\):
A common incorrect step is to reverse the bounds but keep the same value:
In fact, the reversed integral is
The correction is to include the minus sign:
Reversing bounds does not preserve an integral's value. It preserves the interval but reverses its orientation, so the sign must change.
Why This Formulation Matters
Writing a variable-lower-endpoint accumulation as a negative integral from the fixed endpoint to the variable endpoint makes the role of orientation explicit. It is especially useful when the available information describes signed area outward from the fixed endpoint, or when the variable endpoint lies to the right of that fixed endpoint.
The essential distinction is between geometric area and signed, oriented accumulation. Geometric area is nonnegative. A definite integral can be negative because the graph lies below the axis, because the interval is traversed in reverse, or because both effects occur. Keeping those effects separate makes the sign of the answer much easier to verify.
Check Your Understanding
Use endpoint orientation and signed area to answer the questions. Include the sign in each answer.
- If \(A(x)=\int_x^b f(t)\,dt\), write an equivalent expression with the endpoints reversed.
- For \(f(t)=2\), \(b=1\), and \(x=4\), evaluate \(A(4)\).
- If \(x<b\) and \(f\) is below the horizontal axis throughout \([x,b]\), what is the sign of \(A(x)\)?
- A signed area from \(b\) to \(x\) is \(-7\), where \(x>b\). Find \(\int_x^b f(t)\,dt\).
- Explain why reversing the bounds without adding a minus sign gives an incorrect result.
