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The Fundamental Theorem of Calculus · Tutorial 1117 of 1495

The Missing Chain Factor Error

Use the Fundamental Theorem of Calculus together with the chain rule to differentiate integrals whose endpoints depend on the input.

Intermediate 9 min read

What You'll Learn

  • State the conditions for differentiating an integral with a moving endpoint.
  • Apply the chain factor when an upper endpoint is a function of the input.
  • Account for the sign and chain factor when the lower endpoint moves.
  • Diagnose omitted and incorrectly calculated endpoint derivatives.
  • Check a variable-limit derivative by differentiating an evaluated integral.

Why a Chain Factor Can Go Missing

An integral with a moving endpoint changes when the input changes for two related reasons: the endpoint moves, and the integrand has a value at that endpoint. The Fundamental Theorem of Calculus identifies the integrand value. The chain rule accounts for how quickly the endpoint moves.

The common error is to include the first part and omit the second. For an integral whose upper endpoint is \(g(x)\), the derivative is not generally just the integrand evaluated at \(g(x)\). It must also be multiplied by \(g'(x)\).

Key idea: When differentiating an integral with a moving endpoint, multiply the integrand's value at that endpoint by the derivative of the endpoint.

The Moving-Endpoint Rule

Let \(f\) be continuous on an interval containing the endpoint values, and let \(g\) be differentiable on an open interval of input values. If \(a\) is fixed, define

$$ H(x)=\int_a^{g(x)} f(t)\,dt. $$

Then the Fundamental Theorem of Calculus, together with the chain rule, gives the derivative below. The variable \(t\) is the integration variable; the endpoint that depends on \(x\) is \(g(x)\).

Theorem: If \(f\) is continuous on an interval containing \(g(x)\) and the fixed endpoint \(a\), and \(g\) is differentiable, then
$$ \frac{d}{dx}\int_a^{g(x)} f(t)\,dt = f(g(x))g'(x). $$

If the moving endpoint is the lower endpoint instead, the orientation of the integral introduces a minus sign. Under the same continuity and differentiability conditions,

$$ \frac{d}{dx}\int_{g(x)}^a f(t)\,dt = -f(g(x))g'(x). $$

More generally, if both endpoints depend differentiably on \(x\), then for continuous \(f\),

$$ \frac{d}{dx}\int_{p(x)}^{q(x)}f(t)\,dt = f(q(x))q'(x)-f(p(x))p'(x). $$

The minus sign on the lower-endpoint term expresses the fact that increasing a lower bound removes area from the integral. These formulas require continuity of the integrand at the endpoint values involved and differentiability of the moving endpoints. If those assumptions are not satisfied, the formulas cannot automatically be applied.

Why the Chain Factor Appears

For a fixed lower endpoint \(a\), define an accumulation function \(A\) by integrating from \(a\) to a variable endpoint \(u\). By the Fundamental Theorem of Calculus, \(A\) is differentiable where \(f\) is continuous, and its derivative is the integrand:

$$ A(u)=\int_a^u f(t)\,dt, \qquad A'(u)=f(u). $$

The integral with moving endpoint \(g(x)\) is \(A(g(x))\). Differentiating this composition using the chain rule gives the result:

$$ \frac{d}{dx}A(g(x)) = A'(g(x))g'(x) = f(g(x))g'(x). $$

This derivation also shows exactly where the missing factor error comes from. The Fundamental Theorem supplies \(f(g(x))\); the chain rule supplies \(g'(x)\). Both are required unless \(g'(x)=1\).

Worked Examples

Worked Example: A Linear Upper Endpoint

Differentiate

$$ F(x)=\int_2^{3x-1}(t^2+1)\,dt. $$

The integrand is \(f(t)=t^2+1\), and the upper endpoint is \(g(x)=3x-1\). Its derivative is

$$ g'(x)=3. $$

Apply the moving upper-endpoint rule, keeping the endpoint derivative as a separate factor:

$$ F'(x) = \bigl((3x-1)^2+1\bigr)\cdot 3 = 3\bigl((3x-1)^2+1\bigr). $$

To check, an antiderivative of \(t^2+1\) is \(t^3/3+t\). Evaluating the integral gives

$$ F(x) = \frac{(3x-1)^3}{3}+(3x-1) - \left(\frac{2^3}{3}+2\right) = \frac{(3x-1)^3}{3}+(3x-1)-\frac{14}{3}. $$

Differentiating this expression directly yields

$$ F'(x) = \frac{1}{3}\cdot 3(3x-1)^2\cdot 3+3 = 3(3x-1)^2+3 = 3\bigl((3x-1)^2+1\bigr). $$

This agrees with the moving-endpoint rule. For an additional value check, at \(x=1\) the endpoint is \(2\), so the integral is zero, and the derivative formula gives \(F'(1)=3(1+1)=6\).

Worked Example: A Nonlinear Upper Endpoint

Differentiate

$$ G(x)=\int_0^{x^2+1}\cos t\,dt. $$

Here \(f(t)=\cos t\) and \(g(x)=x^2+1\). The endpoint derivative is

$$ g'(x)=2x. $$

The integrand evaluated at the endpoint is \(\cos(x^2+1)\). Multiplying by the endpoint derivative gives

$$ G'(x)=\cos(x^2+1)\cdot 2x = 2x\cos(x^2+1). $$

For a direct check, an antiderivative of \(\cos t\) is \(\sin t\), and \(\sin 0=0\). Thus

$$ G(x)=\sin(x^2+1)-\sin 0 = \sin(x^2+1). $$

Differentiate the evaluated expression:

$$ G'(x)=\cos(x^2+1)\cdot 2x = 2x\cos(x^2+1). $$

The two methods agree. At \(x=1\), for example, the derivative is \(2\cos 2\); the factor \(2\) comes from the rate of change of the endpoint \(x^2+1\) at \(x=1\).

Worked Example: A Nonlinear Lower Endpoint

Differentiate

$$ J(x)=\int_{x^2}^{4}(t+2)\,dt. $$

The moving endpoint \(x^2\) is the lower endpoint, so its contribution has a minus sign. Its derivative is \(2x\), and the integrand evaluated there is \(x^2+2\). Therefore,

$$ J'(x)=-(x^2+2)(2x) = -2x(x^2+2). $$

Check by evaluating the integral. An antiderivative of \(t+2\) is \(t^2/2+2t\), so

$$ J(x) = \left(\frac{4^2}{2}+2\cdot4\right) - \left(\frac{(x^2)^2}{2}+2x^2\right) = 16-\frac{x^4}{2}-2x^2. $$

Differentiating gives

$$ J'(x)=0-\frac{4x^3}{2}-4x = -2x^3-4x = -2x(x^2+2). $$

This matches the formula, including its minus sign. At \(x=1\), the derivative is \(-2(1)(1+2)=-6\), which is also the value of \(-2x^3-4x\) at \(1\).

Diagnosing the Missing Factor

When an answer contains only the integrand evaluated at the endpoint, inspect the endpoint itself. If it is anything other than \(x\) (or another function with derivative \(1\) at the input under consideration), calculate its derivative explicitly.

1
Identify the moving endpoint.
Write it as \(g(x)\), or as \(p(x)\) if it is the lower endpoint.
2
Evaluate the integrand there.
For an upper endpoint, use \(f(g(x))\); for a lower endpoint, use \(f(p(x))\).
3
Multiply by the endpoint derivative.
Use \(g'(x)\) for the upper endpoint and \(-p'(x)\) for the lower endpoint.
4
Check the result.
If practical, evaluate the integral using an antiderivative and differentiate that expression.

Common Mistakes and Their Corrections

For example, consider the claim that differentiating \(\int_1^{x^2}t^3\,dt\) gives \(x^6\). This evaluates the integrand at \(x^2\), but it omits the derivative of that endpoint. Since \(\frac{d}{dx}(x^2)=2x\), the correct calculation is

$$ \frac{d}{dx}\int_1^{x^2}t^3\,dt = (x^2)^3(2x) = 2x^7. $$

Checking the calculation directly makes the omission visible. An antiderivative of \(t^3\) is \(t^4/4\), so

$$ \int_1^{x^2}t^3\,dt = \frac{(x^2)^4}{4}-\frac{1^4}{4} = \frac{x^8}{4}-\frac14, \qquad \frac{d}{dx}\left(\frac{x^8}{4}-\frac14\right) = 2x^7. $$

A second error is to include an endpoint factor but calculate it incorrectly. For an upper endpoint \(x^2+1\), the chain factor is \(2x\), not \(x\) or \(2\). Differentiate the entire endpoint expression before substituting it into the formula.

Finally, do not lose track of which endpoint moves. The upper endpoint contributes positively and the lower endpoint negatively. A moving lower bound is not handled by merely copying the upper-bound formula: its sign must be reversed.

Why the Rule Is Useful

Variable-limit integrals describe accumulated quantities as the boundary of accumulation changes. Differentiating such an integral converts a change in accumulated amount into the value at the moving boundary, adjusted for the boundary's rate of motion. That adjustment is precisely the endpoint derivative.

The rule is therefore more than a convenient formula. It keeps track of two different rates: the rate described by the integrand and the rate at which the boundary moves. Omitting either one changes the answer. A reliable final check is to evaluate the integral when feasible and differentiate the resulting expression independently.

Check Your Understanding

For each question, identify the moving endpoint, its derivative, and any sign required before simplifying.

  1. Differentiate \(\int_0^{4x+1} t^2\,dt\). What factor would be missing if you used only the integrand at the upper endpoint?
  2. Differentiate \(\int_2^{x^3} \sin t\,dt\). State the endpoint derivative and the resulting derivative.
  3. Differentiate \(\int_{1-x}^{5} (t^2+1)\,dt\). Why does the lower endpoint contribution have a minus sign?
  4. A student claims that \(\frac{d}{dx}\int_0^{\cos x}e^t\,dt=e^{\cos x}\). Identify the omitted factor and write the corrected result.
  5. For \(\int_{x^2}^{x+3} t\,dt\), write the derivative using the two-endpoint formula. Which two endpoint derivatives appear?

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