Why the Endpoint Matters
An integral with a moving endpoint describes accumulated signed area up to a location that changes with \(x\). Its derivative is governed by the value of the integrand at that endpoint, together with the rate at which the endpoint moves.
When the integrand contains nested expressions, it is easy to confuse two different operations: evaluating the integrand at a moving endpoint, and differentiating the integrand as a function of its own input. The variable-limit formula requires the first operation. The chain rule supplies a factor for the moving endpoint.
The Variable-Endpoint Formula
Let \(f\) be continuous on an interval containing the values of two differentiable functions \(a(x)\) and \(b(x)\). Define
The Fundamental Theorem of Calculus, together with the chain rule, gives the derivative of this accumulated quantity.
The upper endpoint contributes the integrand evaluated there, multiplied by the upper endpoint's derivative. The lower endpoint contributes the same kind of product with a minus sign. If one endpoint is constant, its derivative is zero, so its contribution vanishes.
This formula does not require \(f\) itself to be differentiable. Continuity is enough for the Fundamental Theorem of Calculus to give a derivative for an accumulation function. In particular, a nested integrand such as \(f(t)=\sin(t^3+2)\) can be evaluated at a moving endpoint without differentiating its nested expression.
Why the Formula Works
A short derivation shows both the endpoint terms and the minus sign. Choose a constant \(c\) in the interval under consideration, and define \(H(y)=\int_c^y f(t)\,dt\). By the Fundamental Theorem of Calculus, \(H'(y)=f(y)\). Reversing the limits changes the sign of an integral, so
Since \(H\) is differentiable and the endpoint functions are differentiable, the chain rule applies to each composition:
The proof explains why the integrand is evaluated at the endpoints: differentiating \(H(b(x))\) uses \(H'(b(x))=f(b(x))\). It also explains why we do not differentiate \(f\) in this formula: the derivative comes from \(H\), whose derivative is \(f\).
Composed Integrands: Evaluate First
Suppose the integrand has the form \(f(t)=q(r(t))\), where \(q\) and \(r\) are functions and the composition is continuous on the required interval. The upper-endpoint contribution is
Here \(r(b(x))\) is obtained by substituting \(b(x)\) for \(t\) in the integrand. The factor \(b'(x)\) accounts for the endpoint's movement. There is no additional factor from differentiating \(r\) or \(q\), because this calculation evaluates \(f\); it does not differentiate \(f(b(x))\).
Worked Example: A Nested Integrand with a Nonlinear Upper Endpoint
Differentiate
The integrand is \(f(t)=\sin(t^3+2)\). The lower endpoint is \(a(x)=1\), so \(a'(x)=0\). The upper endpoint is \(b(x)=x^2\), so \(b'(x)=2x\).
Evaluate the integrand at the upper endpoint by substituting \(t=x^2\):
Apply the variable-endpoint formula:
As a check at \(x=1\), the formula gives \(F'(1)=2\sin 3\). The upper endpoint is moving at rate \(2\), and the integrand's value there is \(\sin 3\), so their product is \(2\sin 3\), as the formula states.
Two Moving Endpoints
When both endpoints move, each contributes its own endpoint value and chain factor. The lower endpoint's contribution is subtracted, even when its derivative is negative; the formula already accounts for that sign.
Worked Example: A Nested Integrand Between Two Moving Limits
Differentiate
Here \(f(t)=\sqrt{1+t^4}\), \(a(x)=x\), and \(b(x)=x^2+1\). Their derivatives are \(a'(x)=1\) and \(b'(x)=2x\). Evaluate the integrand at each endpoint:
Substitute these values and the endpoint derivatives into the formula:
For a numerical check at \(x=1\), the upper endpoint is \(2\), the lower endpoint is \(1\), and the endpoint rates are \(2\) and \(1\). Thus
This confirms that the upper contribution is weighted by \(2\), while the lower contribution is subtracted with weight \(1\).
Nested Endpoint Functions
The endpoint itself may also be a composition, such as \(\sin x\) or \(x^2+1\). The integrand is still evaluated at the endpoint, and the derivative of that endpoint supplies its chain factor. These are separate roles in the formula.
Worked Example: Trigonometric Endpoints and a Nested Integrand
Let
Set \(f(t)=e^{\cos(t^2+1)}\), \(a(x)=\sin x\), and \(b(x)=x+2\). Then \(a'(x)=\cos x\) and \(b'(x)=1\). Substitution into the integrand gives
Therefore,
At \(x=0\), the upper endpoint is \(2\), the lower endpoint is \(0\), and both endpoints have derivative \(1\). Substitution into the result gives
The nested expressions \(t^2+1\) and \(\cos(t^2+1)\) are evaluated at each endpoint. They are not differentiated when computing these endpoint contributions.
A Common Mistake: Differentiating the Wrong Quantity
Consider again \(f(t)=\sin(t^3+2)\) with the endpoint \(t=x^2\). Differentiating the integrand with respect to its input gives
Evaluating this derivative at \(t=x^2\) gives \(3x^4\cos(x^6+2)\). That is the value of \(f'(t)\) at \(t=x^2\), not the derivative of the integral. If we instead differentiate the endpoint-evaluated integrand \(f(x^2)=\sin(x^6+2)\), the chain rule gives
That is the derivative of \(f(x^2)\), not the derivative of \(\int_1^{x^2}f(t)\,dt\). The integral's derivative is \(2x f(x^2)=2x\sin(x^6+2)\). For example, at \(x=1\), these three expressions are \(3\cos 3\), \(6\cos 3\), and \(2\sin 3\), respectively. They are different quantities.
A Reliable Calculation Order
Writing down the endpoint roles before substituting can prevent sign and chain-factor errors. For an integral with two moving limits, use this sequence:
Write \(f(t)\), \(a(x)\), and \(b(x)\) separately.
Find \(a'(x)\) and \(b'(x)\), including any chain factors.
Substitute \(t=b(x)\) and \(t=a(x)\) into \(f(t)\); do not differentiate \(f\).
Use \(f(b(x))b'(x)-f(a(x))a'(x)\). The upper term is added and the lower term is subtracted.
When the Formula Applies
The stated formula applies when the integrand is a continuous function of the integration variable on an interval containing the endpoint values, and the endpoints are differentiable at the \(x\)-values in question. A nested expression is acceptable when it defines such a continuous integrand on the needed interval. Differentiability of the integrand is not required.
The central practical point is to keep the roles separate. The integrand determines the endpoint values; the endpoint functions determine the chain factors. With those roles identified, even a complicated nested integrand can be handled by the same formula.
Check Your Understanding
For each question, identify the endpoint values and endpoint derivatives before applying the formula.
- Differentiate \(\displaystyle \int_0^{x^3}\cos(t^2+1)\,dt\).
- Differentiate \(\displaystyle \int_{\cos x}^{2x}\sqrt{1+t^2}\,dt\).
- For \(f(t)=\ln(1+t^4)\), explain the difference between \(f'(b(x))\) and the endpoint contribution \(f(b(x))b'(x)\).
- State the continuity and differentiability conditions used in the variable-endpoint formula.
- In one sentence, explain why differentiating the integrand is not part of evaluating its contribution to the derivative of an accumulation integral.
