When the Input Is Below the Basepoint
An accumulation function is often written as \(A(x)=\int_a^x f(t)\,dt\), where \(a\) is a fixed basepoint and \(x\) is the input. If \(x\) is greater than \(a\), the integral runs from left to right along the number line. If \(x\) is less than \(a\), its endpoints appear in the opposite order.
The endpoint order matters. An integral from \(a\) to \(x\), with \(x<a\), is not the same as the integral from \(x\) to \(a\). Reversing the endpoints reverses the sign. This convention lets one formula for an accumulation function apply on either side of its basepoint.
This tutorial focuses on the orientation of the interval. The sign change comes from the direction in which the interval is traversed, not from changing the function or its values.
Definition of an Oriented Integral
For \(u<v\), the definite integral \(\int_u^v f(t)\,dt\) is the usual integral over an interval with endpoints in increasing order. To allow either order of endpoints, define the integral with reversed endpoints by
This is called orientation: the ordered pair of endpoints specifies the direction of integration. In particular, the interval from \(v\) to \(u\) has the opposite orientation to the interval from \(u\) to \(v\).
The dummy variable \(t\) simply indicates the variable of integration. The input \(x\) determines the endpoint, while \(a\) remains fixed. For an input below \(a\), first identify the integral from the smaller endpoint to the larger one, then apply the minus sign.
Why Reversing the Endpoints Changes the Sign
The sign reversal can be justified directly from the Riemann-sum definition of an integral. Take \(u<v\), and partition the interval \([u,v]\) into points \(u=t_0<t_1<\cdots<t_n=v\). A Riemann sum for the integral from \(u\) to \(v\) has terms \(f(\xi_i)(t_i-t_{i-1})\), where \(\xi_i\) is a sample point in \([t_{i-1},t_i]\).
Now traverse the same subintervals in reverse, from \(v\) to \(u\). Each subinterval has the same function value at the corresponding sample point, but its directed width changes from \(t_i-t_{i-1}\) to \(t_{i-1}-t_i\). Since \(t_{i-1}-t_i=-(t_i-t_{i-1})\), every term changes sign. Therefore the reversed sum is the negative of the forward sum. Taking limits of the sums gives the same relation for the integrals.
This argument explains why orientation affects the integral even when the graph of \(f\) has not changed. The reversal changes the sign of every directed width in the sum, so the entire integral changes sign.
A Reliable Method for Inputs Below the Basepoint
When evaluating \(A(x)=\int_a^x f(t)\,dt\) with \(x<a\), use the following steps:
Confirm that \(x<a\), so the accumulation runs from the larger endpoint \(a\) to the smaller endpoint \(x\).
Write \(A(x)=-\int_x^a f(t)\,dt\). Keep the same integrand and put the smaller endpoint first inside the integral.
Use the appropriate integral information, such as a constant-function rectangle or a signed-area calculation.
Negate the value of the increasing-order integral. Reversing the endpoints a second time should return the original value.
The minus sign belongs to the reversed interval. It is not a modification to \(f\), and it should not be omitted just because the integral from the smaller endpoint to the larger one has already been evaluated.
Worked Examples
Worked Example: A Constant Positive Integrand
Let \(A(x)=\int_2^x 3\,dt\). Find \(A(-1)\).
The input is below the basepoint because \(-1<2\). Reverse the endpoints and include a minus sign:
On \([-1,2]\), the constant integrand has height \(3\), and the interval has width \(2-(-1)=3\). Thus the increasing-order integral is the rectangle's signed area:
Therefore,
Check: Reversing back gives \(\int_{-1}^2 3\,dt=9=-A(-1)\), as required by the orientation rule.
Worked Example: A Positive Linear Integrand
Let \(A(x)=\int_2^x (t+3)\,dt\). Find \(A(-1)\) using the area of a trapezoid.
Since \(-1<2\), reverse the endpoints:
On the interval \([-1,2]\), the endpoint values of the integrand are
The graph is a straight line, so the region between the graph and the horizontal axis is a trapezoid with parallel sides \(2\) and \(5\), and width \(2-(-1)=3\). Its area is
Apply the orientation sign:
Check: The forward integral from \(-1\) to \(2\) is \(21/2\), so the reversed integral from \(2\) to \(-1\) must be \(-21/2\). This agrees with the result.
Worked Example: A Negative Constant Integrand
Let \(A(x)=\int_4^x (-2)\,dt\). Find \(A(1)\).
Here \(1<4\), so reverse the endpoints:
The increasing-order integral is a rectangle with height \(-2\) and width \(4-1=3\). Its signed integral is
Negating this value gives
Check: Reversing the endpoints of \(\int_4^1(-2)\,dt=6\) gives \(\int_1^4(-2)\,dt=-6\). The two values are opposites.
The Interval Direction and the Integrand Sign
There are two separate sources of sign in an oriented integral. The integrand can be positive or negative, and the interval can be traversed in either direction. Keep those facts distinct: the integral over a reversed interval is the negative of the integral over the corresponding interval in increasing order, whatever the sign of the integrand.
| Integral situation | What determines the sign? |
|---|---|
| Positive constant on an interval in increasing order | The integral is positive. |
| The same positive constant on the reversed interval | Reversing endpoints negates the integral. |
| Negative constant on an interval in increasing order | The integral is negative. |
| The same negative constant on the reversed interval | Reversing endpoints negates that negative value. |
The last case explains how a reversed integral can be positive even when the integrand is negative. The integral in increasing order is negative, and reversing the endpoints multiplies that value by \(-1\). The orientation rule itself is unchanged.
Common Mistake: Reversing the Endpoints but Keeping the Sign
Suppose \(A(x)=\int_4^x 3\,dt\), and we want \(A(1)\). A common incorrect calculation is
This calculation changes the endpoints but forgets to change the sign. The correct calculation is
A quick check is to reverse the endpoints once more: \(\int_1^4 3\,dt=9\), the opposite of \(\int_4^1 3\,dt=-9\). If both directions give \(9\), the orientation sign has been lost.
Why the Orientation Convention Is Useful
Using oriented endpoints means the accumulation function \(A(x)=\int_a^x f(t)\,dt\) has one consistent definition for inputs on either side of \(a\). There is no need to define a separate formula just because the input crosses the basepoint. Instead, the order of the endpoints carries the direction, and the sign rule handles that direction precisely.
When \(x<a\), the most dependable habit is to write the reversed-endpoint equation before evaluating anything. That makes the direction explicit, keeps the integrand intact, and prevents the sign from being decided by visual intuition alone. The same rule applies whether the increasing-order integral is positive, negative, or zero.
Check Your Understanding
Use endpoint orientation in each question. When useful, first rewrite the integral with endpoints in increasing order.
- If \(A(x)=\int_3^x 2\,dt\), write \(A(-2)\) as the negative of an integral in increasing order, then evaluate it.
- For \(B(x)=\int_5^x (-4)\,dt\), find \(B(2)\). Explain why its value is positive.
- State the relationship between \(\int_p^q f(t)\,dt\) and \(\int_q^p f(t)\,dt\) when \(p<q\).
- A learner claims that reversing endpoints changes the sign of the integrand. Explain what actually changes.
- If the increasing-order integral from \(x\) to \(a\) is \(-7\), where \(x<a\), what is \(\int_a^x f(t)\,dt\)?
