Why a Nonlinear Upper Limit Needs a Chain Factor
An integral with a fixed lower endpoint and a changing upper endpoint measures accumulated change up to that endpoint. If the endpoint itself depends on \(x\), then changing \(x\) changes the integral in two linked ways: it determines where the integrand is evaluated, and it determines how quickly the endpoint moves.
The Fundamental Theorem of Calculus identifies the integrand's value at the endpoint. The chain rule accounts for how quickly the endpoint changes. Together, they give a concise formula for differentiating an integral whose upper limit is a differentiable function, whether that function is linear or nonlinear.
The Formula and Its Conditions
Let \(I\) be an interval containing \(a\) and all values of \(g(x)\) under consideration. Suppose \(f\) is continuous on \(I\), and \(g\) is differentiable on an interval \(U\) with \(g(x)\in I\) for every \(x\in U\). Define the accumulation function \(H\) by integrating \(f\) from \(a\) to a variable endpoint \(u\). The Fundamental Theorem of Calculus gives \(H'(u)=f(u)\). The integral with upper endpoint \(g(x)\) is \(H(g(x))\), so the chain rule gives the result.
The variable \(t\) is a dummy variable of integration: it names the input to the integrand while the integral is being formed. The endpoint \(g(x)\), by contrast, depends on \(x\). The formula substitutes \(g(x)\) for \(t\) in \(f(t)\), then multiplies by the derivative of \(g\).
The continuity condition on \(f\) ensures that the accumulation function has derivative \(f\) at the endpoint, as required by the Fundamental Theorem of Calculus. Differentiability of \(g\) ensures that the chain rule applies. In a problem, it is important to check that the endpoint remains in a region where the integrand is continuous and that the endpoint function is differentiable on the domain being considered.
Why the Formula Works
The argument uses two results: the Fundamental Theorem of Calculus for an accumulation function and the ordinary chain rule for a composition of differentiable functions. Here is the full composition argument.
For a variable endpoint \(u\), set \(H(u)=\int_a^u f(t)\,dt\). Then the integral in question is \(F(x)=H(g(x))\).
Because \(f\) is continuous on the relevant interval, the Fundamental Theorem of Calculus gives \(H'(u)=f(u)\).
Differentiate \(H(g(x))\): \(F'(x)=H'(g(x))g'(x)\). Substituting \(H'(g(x))=f(g(x))\) gives \(F'(x)=f(g(x))g'(x)\).
This explains why the endpoint derivative is not an optional adjustment. When \(x\) changes, the endpoint moves at rate \(g'(x)\). The integral's instantaneous change is the integrand's value at the moving endpoint multiplied by that rate.
Worked Examples
Worked Example: A Polynomial Upper Endpoint
Differentiate \(F(x)=\int_2^{x^2+1}(t^3-2t)\,dt\).
Here \(f(t)=t^3-2t\) is continuous for every real \(t\), and \(g(x)=x^2+1\) is differentiable for every real \(x\). First differentiate the endpoint:
Next evaluate the integrand at the endpoint:
Multiply by the endpoint derivative:
For a check, an antiderivative of \(t^3-2t\) is \(G(t)=t^4/4-t^2\), since \(G'(t)=t^3-2t\). Therefore \(F(x)=G(x^2+1)-G(2)\), and \(G(2)=2^4/4-2^2=4-4=0\). Differentiating this expression gives \(G'(x^2+1)(2x)=((x^2+1)^3-2(x^2+1))(2x)\), which agrees with the formula.
Worked Example: A Trigonometric Upper Endpoint
Differentiate \(A(x)=\int_0^{\sin x} e^{t^2}\,dt\).
The integrand \(f(t)=e^{t^2}\) is continuous for every real \(t\), and \(g(x)=\sin x\) is differentiable for every real \(x\). Its derivative is \(g'(x)=\cos x\). Evaluating the integrand at the endpoint gives \(f(g(x))=e^{(\sin x)^2}\). Hence
The two factors have different roles: \(e^{(\sin x)^2}\) is the integrand evaluated at the endpoint, and \(\cos x\) is the rate at which the endpoint changes. As a check at \(x=0\), the endpoint is \(\sin 0=0\), so the formula gives \(A'(0)=e^0\cos 0=1\cdot1=1\).
Worked Example: A Square-Root Upper Endpoint
For \(x>-2\), differentiate \(B(x)=\int_1^{\sqrt{x+3}}\ln(t)\,dt\).
The domain restriction gives \(x+3>1\), so the endpoint \(\sqrt{x+3}\) is greater than \(1\). The integrand \(\ln(t)\) is continuous on the interval of integration, and the endpoint is differentiable for \(x>-2\). Differentiating that endpoint gives
Evaluate the integrand at the endpoint and multiply by this derivative:
For a check, \(G(t)=t\ln(t)-t\) is an antiderivative of \(\ln(t)\), because \(G'(t)=\ln(t)+1-1=\ln(t)\). Thus the integral equals \(G(\sqrt{x+3})-G(1)\). Since \(G(1)=-1\), this is \(\sqrt{x+3}\ln(\sqrt{x+3})-\sqrt{x+3}+1\). Differentiating gives \(\ln(\sqrt{x+3})/(2\sqrt{x+3})\), which equals \(\ln(x+3)/(4\sqrt{x+3})\), as required.
A Common Mistake: Leaving Out the Endpoint Derivative
A frequent error is to substitute the moving endpoint into the integrand but forget to multiply by the derivative of that endpoint. For example, consider \(C(x)=\int_0^{x^2+1}t^2\,dt\). A wrong calculation is
This has evaluated \(t^2\) at \(x^2+1\), but it has not accounted for how quickly \(x^2+1\) changes. The correct calculation includes its derivative:
To check, an antiderivative of \(t^2\) is \(t^3/3\), so \(C(x)=(x^2+1)^3/3-1/3\). Differentiation gives \((x^2+1)^2(2x)\), confirming the corrected answer. For instance, at \(x=1\), the incorrect expression gives \(4\), while the correct derivative is \(4\cdot2=8\).
Another way to avoid this mistake is to keep the two factors visible until the final line: first write \(f(g(x))\), then write \(g'(x)\), and only then multiply. Do not differentiate the integrand with respect to its input; the formula requires its value at the endpoint, not its derivative there.
How to Apply the Rule Reliably
For an integral of the form \(\int_a^{g(x)} f(t)\,dt\), a short checklist keeps the roles of the expressions clear:
- Identify the integrand \(f(t)\) and the endpoint function \(g(x)\).
- Check that \(f\) is continuous on the relevant interval and \(g\) is differentiable on the domain in question.
- Substitute the endpoint into the integrand to get \(f(g(x))\).
- Differentiate the endpoint to get \(g'(x)\).
- Multiply the two factors and, if helpful, simplify.
The endpoint derivative may be zero at particular values of \(x\). The formula still applies there when its hypotheses hold: if \(g'(x)=0\), then the derivative of the integral is zero at that point. This follows directly from the product \(f(g(x))g'(x)\), not from assuming that the integrand itself vanishes.
This rule is useful whenever an accumulated quantity has a boundary that changes nonlinearly with an input. It separates the value of the rate at the current boundary from the speed at which that boundary moves. Keeping those two contributions distinct also makes it easier to check whether a proposed derivative has the correct factors.
Check Your Understanding
For each question, identify the endpoint function and use the nonlinear upper-limit rule where appropriate.
- What conditions on the integrand and the endpoint function are needed for the formula to apply?
- Differentiate \(\int_0^{x^3+2}(1+t^2)\,dt\).
- For \(\int_1^{\cos x+2}\sqrt{1+t}\,dt\), what are the integrand evaluated at the endpoint and the endpoint derivative?
- Explain why \(\int_0^{x^2+1}t^2\,dt\) does not have derivative \((x^2+1)^2\).
- If \(g'(c)=0\) and the formula's hypotheses hold, what is the derivative of \(\int_a^{g(x)}f(t)\,dt\) at \(x=c\)?
