Orientation Determines the Sign
An integral with a moving endpoint records how accumulated quantity changes as that endpoint moves. The sign of the change depends not only on the value of the integrand, but also on whether the endpoint moves in the direction that adds to the accumulation or in the direction that removes from it.
The key idea is orientation. Reversing the order of the limits reverses the sign of the integral:
This convention makes an integral a signed accumulation, not simply a nonnegative area. In particular, an integral whose lower limit moves has a negative derivative contribution. The endpoint’s motion matters too: if the endpoint itself moves in the negative direction, its negative velocity can make the total contribution positive.
Moving a Lower Endpoint
Let \(f\) be continuous and let \(c\) be fixed. The Fundamental Theorem of Calculus, together with the chain rule, gives
provided \(g\) is differentiable and its values lie in the interval where \(f\) is continuous. To handle a moving lower endpoint, reverse the limits first:
Differentiating both sides shows where the minus sign comes from:
The integrand is evaluated at the moving endpoint. The factor \(g'(x)\) accounts for how quickly that endpoint moves, and the negative sign reflects its position as a lower limit.
Worked Example: A Moving Lower Limit
Differentiate \(G(x)=\int_x^3(u^2+1)\,du\), and check the result by evaluating the integral explicitly.
Here \(f(u)=u^2+1\) and the lower endpoint is \(g(x)=x\), so \(g'(x)=1\). The lower-limit formula gives
For a direct check, an antiderivative of \(u^2+1\) is \(u^3/3+u\). Therefore,
Differentiating this expression gives \(G'(x)=-x^2-1\), as required. At \(x=1\), for example, \(G(1)=12-\frac13-1=\frac{32}{3}\), while \(G'(1)=-(1+1)=-2\). As the lower endpoint moves right, it removes accumulated quantity from the interval, so the derivative is negative.
Endpoint Motion and the Chain Factor
The sign of a lower-limit contribution and the direction of endpoint motion are separate factors. If a lower endpoint moves left as \(x\) increases, then its velocity is negative. The lower-limit minus sign and the negative velocity multiply to give a positive contribution.
For instance, if the moving upper endpoint is \(g(x)\), its contribution is \(f(g(x))g'(x)\). If \(g'(x)<0\), that contribution is negative when \(f(g(x))>0\). The endpoint is an upper limit, but it is moving left, so the interval loses accumulation at that end. Conversely, a lower endpoint moving left can increase the accumulation because it extends the interval in the direction that adds it.
Worked Example: A Decreasing Upper Endpoint
Let \(A(x)=\int_1^{4-2x}(u+1)\,du\). Find \(A'(x)\), and verify it by differentiating an explicit expression for \(A(x)\).
The upper endpoint is \(g(x)=4-2x\), so \(g'(x)=-2\). The upper-limit formula gives
For an independent check, an antiderivative of \(u+1\) is \(u^2/2+u\). Substitution of the endpoints gives
Expanding the square, \((4-2x)^2=16-16x+4x^2\). Hence
Differentiating gives \(A'(x)=4x-10\), matching the formula. At \(x=1\), the endpoint is \(2\), so \(A(1)=2(1)^2-10(1)+21/2=5/2\), and \(A'(1)=4-10=-6\). The upper endpoint moves left at speed \(2\); since the integrand at \(u=2\) is \(3\), the accumulation decreases at rate \(3\cdot2=6\).
Two Moving Endpoints
If both endpoints move, their contributions have opposite orientation signs. Let \(a\) and \(b\) be differentiable, and let \(f\) be continuous over the interval containing their values. Then
The upper endpoint contributes \(f(b(x))b'(x)\). The lower endpoint contributes \(-f(a(x))a'(x)\). This formula combines the moving-upper-limit rule with the moving-lower-limit rule; it does not require that \(a(x)<b(x)\) for every \(x\). The integral is oriented, so the same sign convention applies if the endpoint order is reversed.
One way to derive the formula is to insert a fixed point \(c\) and split the accumulation:
Differentiate the first term using the moving-upper-limit rule, and differentiate the second term in the same way before applying the minus sign already in front of it. The result is
This derivation explains why the lower endpoint does not receive the same sign as the upper endpoint: its contribution comes from subtracting the accumulation from \(c\) to \(a(x)\).
Worked Example: Contributions from Both Endpoints
Let \(H(x)=\int_{x^2}^{3x}2u\,du\). Find \(H'(x)\), and check the answer both from the formula and from an explicit expression.
The integrand is \(f(u)=2u\), the upper endpoint is \(b(x)=3x\) with \(b'(x)=3\), and the lower endpoint is \(a(x)=x^2\) with \(a'(x)=2x\). The endpoint formula gives
To verify this, an antiderivative of \(2u\) is \(u^2\), so
Differentiating gives \(H'(x)=18x-4x^3\), confirming the result. At \(x=1\), the endpoints are \(1\) and \(3\), and \(H(1)=9-1=8\). The upper endpoint contributes \(f(3)\cdot3=6\cdot3=18\); the lower endpoint contributes \(-f(1)\cdot2=-2\cdot2=-4\). Their total is \(18-4=14\), which agrees with \(H'(1)=18-4=14\).
The endpoint-motion interpretation makes the subtraction concrete: near \(x=1\), the upper endpoint moves right and adds accumulation, while the lower endpoint also moves right and removes accumulation. The net rate is the upper contribution minus the lower contribution.
A Reliable Sign Check
Before simplifying a variable-limit derivative, identify each endpoint’s role and velocity. This prevents the common mistake of treating both limits as if they were upper limits.
Write down which function is the lower limit and which is the upper limit.
Differentiate each endpoint function with respect to the variable.
Use a plus sign for the upper endpoint and a minus sign for the lower endpoint.
Ask whether the endpoint motion adds to or removes from the accumulated interval.
Common Sign Error
A frequent error is to write the moving-lower-limit rule without its minus sign. For example, an incorrect calculation for \(J(x)=\int_x^2(u+1)\,du\) is
This treats the lower endpoint as though it were an upper endpoint. The correction is to reverse the limits, or use the lower-limit formula directly:
The explicit calculation confirms the correction. Since an antiderivative is \(u^2/2+u\),
and differentiating gives \(J'(x)=-x-1=-(x+1)\), not \(x+1\). For \(x=1\), the incorrect sign predicts \(2\), whereas the correct derivative is \(-2\). Since increasing \(x\) moves the lower endpoint right and shortens the interval, a positive rate is inconsistent with the direction of accumulation.
Why the Signs Matter
Variable-limit integrals often represent a net amount accumulated over an interval. A moving endpoint changes which portion of the integrand is included. The sign records whether that change adds to or removes from the total. This is essential when the integrand is a rate, such as a rate of flow or production: an endpoint’s speed alone is not enough to determine the accumulation rate; its role as an upper or lower limit matters as well.
For reliable work, keep the factors distinct: evaluate the integrand at the endpoint, multiply by the endpoint’s derivative, and apply the orientation sign. If a bound is a complicated function, its derivative may itself be negative. Do not replace that derivative by its magnitude: its sign is part of the formula.
Check Your Understanding
For each question, keep the endpoint’s position and direction separate when deciding the sign.
- For continuous \(f\), what is the derivative of \(\int_{g(x)}^c f(u)\,du\) when \(g\) is differentiable?
- If an upper endpoint has negative velocity and the integrand there is positive, what sign does its contribution have?
- Differentiate \(\int_{x}^{x+2}3u\,du\) using the two-endpoint formula.
- In the derivative of \(\int_{a(x)}^{b(x)}f(u)\,du\), why is the lower-endpoint term subtracted?
- How can differentiating an explicit antiderivative expression help detect a sign error?
