Why Composite Bounds Need the Chain Rule
An accumulation function records the signed integral of an integrand as one or more endpoints move. If an endpoint is simply \(x\), the Fundamental Theorem of Calculus connects the derivative of the accumulation to the integrand at that endpoint. But an endpoint can also be a function of \(x\), such as \(x^2\), \(2x+1\), or \(\sin x\). In that case, the endpoint itself changes at a rate that must be included.
The key idea is to evaluate the integrand at a moving endpoint and then multiply by the rate at which that endpoint moves. If both endpoints move, the upper endpoint contributes positively and the lower endpoint contributes negatively.
Here \(t\) is the variable of integration, while \(x\) is the input that controls the limits. Keeping those roles distinct helps prevent errors: evaluate \(f\) at the endpoint value, but differentiate the endpoint as a function of \(x\).
Composite Upper Limits
For a fixed lower endpoint \(r\), consider \(J(x)=\int_r^{g(x)} f(t)\,dt\). The Fundamental Theorem of Calculus, Part 1 says that the accumulation function \(H(y)=\int_r^y f(t)\,dt\) satisfies \(H'(y)=f(y)\) when \(f\) is continuous. Since \(J(x)=H(g(x))\), the chain rule gives \(J'(x)=f(g(x))g'(x)\).
The factor \(g'(x)\) matters because the upper endpoint may move faster or slower than \(x\). The integrand value alone describes the rate of accumulation per unit change in the endpoint, not per unit change in \(x\).
Worked Example: A Squared Upper Limit
Differentiate \(A(x)=\int_1^{x^2}(t^2+1)\,dt\).
The integrand is \(f(t)=t^2+1\), and the upper limit is \(g(x)=x^2\), so \(g'(x)=2x\). The composite upper-limit rule gives
Check by evaluating the integral. An antiderivative of \(t^2+1\) is \(F(t)=t^3/3+t\), so
Differentiating this expression gives \(A'(x)=2x^5+2x\), matching the result from the theorem. At \(x=1\), the original integral is zero because both limits equal 1, and the derivative formula gives \(2(1)^5+2(1)=4\); the explicit expression also has derivative 4 there.
Composite Lower Limits
A moving lower endpoint has a negative contribution. For a fixed upper endpoint \(r\), let \(K(x)=\int_{g(x)}^r f(t)\,dt\). Reversing the bounds gives \(K(x)=-\int_r^{g(x)}f(t)\,dt\). Applying the composite upper-limit rule to the integral on the right, then keeping the negative sign, yields \(K'(x)=-f(g(x))g'(x)\).
This negative sign reflects the orientation of the integral. As the lower endpoint moves to the right while the upper endpoint stays fixed, the interval of accumulation becomes shorter. The lower endpoint's contribution is therefore subtracted.
Worked Example: A Composite Lower Limit
Differentiate \(B(x)=\int_{x^2}^{5}(t+1)\,dt\).
Here \(f(t)=t+1\) and the lower limit is \(a(x)=x^2\), with \(a'(x)=2x\). Therefore,
To check, an antiderivative of \(t+1\) is \(F(t)=t^2/2+t\). Thus
Its derivative is \(-2x^3-2x\), as required. At \(x=1\), the formula gives \(-4\); differentiating the explicit expression at 1 also gives \(-4\).
Two Composite Limits
When both limits depend on \(x\), combine the upper- and lower-limit contributions. The upper one is \(f(b(x))b'(x)\); the lower one is \(-f(a(x))a'(x)\). Neither term can generally be omitted, even if the endpoints happen to have the same value at a particular input.
Write the integrand as \(f(t)\), the lower limit as \(a(x)\), and the upper limit as \(b(x)\).
Find \(a'(x)\) and \(b'(x)\) using ordinary differentiation rules.
Evaluate \(f\) at each endpoint, multiply by that endpoint's derivative, and subtract the lower-endpoint contribution.
Worked Example: Both Limits Depend on \(x\)
Differentiate \(C(x)=\int_{x^2}^{2x}t\,dt\).
The integrand is \(f(t)=t\). The upper limit \(b(x)=2x\) has derivative \(b'(x)=2\), and the lower limit \(a(x)=x^2\) has derivative \(a'(x)=2x\). Hence
Check by direct evaluation. Since an antiderivative of \(t\) is \(t^2/2\),
Differentiating gives \(C'(x)=4x-2x^3\), agreeing with the endpoint formula. For a numerical check at \(x=1\), the integral is \(2-\tfrac12=\tfrac32\), and both derivative calculations give \(4-2=2\).
The formula also works when the lower limit is greater than the upper limit. For example, at \(x=-1\), the bounds are 1 and \(-2\); the integral is still defined with its signed orientation. The derivative formula gives \(-4+2=-2\), the same value obtained from \(4x-2x^3\) at \(-1\).
Why the Formula Works
The two-limit rule follows directly from the Fundamental Theorem of Calculus, Part 1 and the chain rule. This derivation also explains why the lower endpoint has a minus sign.
Let \(c\) be any fixed point in an interval on which \(f\) is continuous, and define \(H(y)=\int_c^y f(t)\,dt\). The Fundamental Theorem gives \(H'(y)=f(y)\).
For any \(x\) where the endpoints lie in the interval, additivity of definite integrals gives \(\int_{a(x)}^{b(x)}f(t)\,dt=\int_c^{b(x)}f(t)\,dt-\int_c^{a(x)}f(t)\,dt=H(b(x))-H(a(x))\).
Applying the chain rule to each term gives \(\frac{d}{dx}H(b(x))=H'(b(x))b'(x)=f(b(x))b'(x)\) and \(\frac{d}{dx}H(a(x))=H'(a(x))a'(x)=f(a(x))a'(x)\).
Since the second composition is subtracted, \(I'(x)=f(b(x))b'(x)-f(a(x))a'(x)\). This proves the stated formula under the continuity and differentiability conditions.
Common Mistakes and Checks
A frequent mistake is to evaluate the integrand at a composite endpoint but forget to multiply by the endpoint's derivative. For example, if \(D(x)=\int_1^{x^2}(t+1)\,dt\), an incorrect calculation is \(D'(x)=x^2+1\). That expression accounts for the integrand at \(x^2\), but not for the speed \(2x\) of the endpoint. The correction is
At \(x=2\), the incorrect answer is \(2^2+1=5\), while the correct derivative is \((4+1)(4)=20\). The difference is not a small algebraic detail: it is the chain rule.
Other useful checks include the following:
- Check the sign. A lower endpoint contributes a minus sign, even if its derivative is positive.
- Check both terms. With two moving endpoints, evaluate the integrand at both endpoints and include both endpoint derivatives.
- Do not assume the upper limit is larger. Definite integrals are signed, so the formula remains valid when the bounds reverse order.
- Check the hypotheses. The integrand must be continuous at the endpoint values in a neighborhood of the input, and the limits must be differentiable there, for this form of the theorem to apply.
For a quick structural check, if the upper and lower limits coincide at \(x=x_0\), the integral itself is zero at that input. Its derivative need not be zero: the endpoints can move at different rates, so the two terms in the derivative formula may not cancel.
Check Your Understanding
For each question, identify the endpoint derivatives as well as the integrand values. When possible, check by evaluating the integral first.
- Differentiate \(P(x)=\int_0^{3x+1}(t^2+2)\,dt\).
- Differentiate \(Q(x)=\int_{x}^{4}(t^2+1)\,dt\).
- Differentiate \(R(x)=\int_{x^2}^{x+2}t^2\,dt\).
- In \(S(x)=\int_{x^3}^{2}f(t)\,dt\), where \(f\) is continuous, identify the lower-limit contribution to \(S'(x)\).
- Explain why \(\int_{x^2}^{2x}f(t)\,dt\) can be differentiated by the endpoint formula even at inputs where \(x^2>2x\), provided the continuity and differentiability conditions hold.
