Why a Variable Lower Limit Changes the Sign
An integral with a variable endpoint measures accumulation between that endpoint and a fixed point. When the variable is the lower endpoint, increasing it removes a short piece from the interval of accumulation. That change produces a negative sign in the derivative.
For a continuous function \(f\) and a fixed number \(b\), consider the function
The integration variable is \(t\); \(x\) determines the lower endpoint, while \(b\) remains fixed. The derivative is evaluated at \(x\), so the integrand in the result is \(f(x)\), not \(f(t)\).
Where the Negative Sign Comes From
There are two connected ways to understand the sign. First, reversing the bounds of a definite integral changes its sign:
The integral on the right has \(x\) as its upper endpoint. By the Fundamental Theorem of Calculus, differentiating that accumulation with respect to its upper endpoint gives \(f(x)\); the minus sign in front then gives \(-f(x)\). This is a useful shortcut, but a difference-quotient argument shows directly why it works.
Deriving the Lower-Bound Rule
Let \(G(x)=\int_x^b f(t)\,dt\), and suppose \(f\) is continuous at \(x\). For a small nonzero change \(h\), the difference between the new and old accumulations is
The cancellation follows from additivity of definite integrals. Dividing by \(h\), the difference quotient becomes
As \(h\) approaches zero, the average value of \(f\) over the short interval from \(x\) to \(x+h\) approaches \(f(x)\), by continuity. The minus sign remains. More explicitly, for any \(\varepsilon>0\), continuity at \(x\) gives a \(\delta>0\) such that \(|f(t)-f(x)|<\varepsilon\) whenever \(|t-x|<\delta\). If \(0<|h|<\delta\), then
This proves that the difference quotient approaches \(-f(x)\), so \(G'(x)=-f(x)\). The argument also works when \(h<0\): oriented integrals account for the reversed direction, and the same absolute-value bound applies.
Worked Examples
Worked Example: A Polynomial Integrand
Differentiate \(G(x)=\int_x^5(2t+1)\,dt\), and check the result by evaluating the integral first.
The integrand \(2t+1\) is continuous, so the lower-bound rule applies. Directly, an antiderivative with respect to \(t\) is \(t^2+t\), giving
Differentiate this expression:
The lower-bound rule gives \(-(2x+1)=-2x-1\), in agreement. For a numerical check at \(x=2\), the integral equals \(30-4-2=24\), while the derivative there is \(-2(2)-1=-5\). Differentiating the explicit formula gives that same slope.
Worked Example: A Trigonometric Integrand
Let \(H(x)=\int_x^{\pi/2}\sin t\,dt\). Find \(H'(x)\), then check both the function and its derivative by evaluating the integral.
Since \(\sin t\) is continuous, the lower-bound rule gives
For an independent check, an antiderivative of \(\sin t\) is \(-\cos t\). Therefore
Differentiating \(H(x)=\cos x\) gives \(H'(x)=-\sin x\), as required. At \(x=\pi/6\), the function value is \(\cos(\pi/6)=\sqrt{3}/2\), and the derivative is \(-\sin(\pi/6)=-1/2\). Both match the expressions obtained from the evaluated integral.
Worked Example: Checking the Sign with a Linear Integrand
Differentiate \(J(x)=\int_x^0(t-1)\,dt\). Evaluate \(J\) first to make the sign visible.
An antiderivative of \(t-1\) is \(\frac{t^2}{2}-t\). Substitution at the endpoints gives
Thus
The rule gives \(-(x-1)\), since the integrand evaluated at the lower endpoint is \(x-1\). At \(x=3\), direct evaluation gives \(J(3)=3-\frac{9}{2}=-\frac{3}{2}\), and the derivative is \(J'(3)=1-3=-2\). The integral check is also \(\int_3^0(t-1)\,dt=0-(\frac{9}{2}-3)=-\frac{3}{2}\); differentiating the explicit formula confirms the slope \(-2\).
A Common Sign Mistake
A frequent error is to treat a variable lower endpoint as if it were a variable upper endpoint. For example, applying the wrong sign to \(K(x)=\int_x^4 t^2\,dt\) gives the incorrect calculation
The lower-bound rule instead gives \(K'(x)=-x^2\). Evaluating the integral confirms the correction:
The mistake is not a matter of whether the integrand is positive or negative. It is caused by confusing which endpoint moves. When the lower endpoint moves to the right, the interval of accumulation loses the thin slice at its left edge; its contribution is subtracted.
Using the Rule Reliably
For an integral with fixed upper endpoint, write the variable and fixed endpoints clearly, then evaluate the integrand at the variable endpoint and attach the lower-bound sign. The continuity condition matters: continuity of \(f\) at the point \(x\) ensures that the average value over the shrinking interval approaches \(f(x)\). In the ordinary AP Calculus setting, a continuous integrand on the interval of interest provides this condition.
Keep the roles of \(t\) and \(x\) distinct. The letter \(t\) is a placeholder that runs through the integration interval; after the integral is differentiated, the result is a function of \(x\). In particular, one does not keep \(t\) in the derivative.
The sign also makes sense from the integral’s orientation. If \(f\) is positive near \(x\), moving the lower endpoint right removes positive accumulated contribution, so the function decreases locally and its derivative is negative. If \(f\) is negative near \(x\), removing that negative contribution increases the integral, and \(-f(x)\) is positive. The formula handles both cases without requiring a separate rule.
This result is useful whenever a quantity is defined by accumulation from a moving starting point to a fixed destination. The derivative tells how quickly that accumulated quantity changes at the current starting point. Remembering both the endpoint’s role and the integral’s orientation makes the sign a consequence of the setup rather than a rule to guess.
Check Your Understanding
For each question, identify the variable endpoint and use the negative sign only when the variable is the lower endpoint.
- If \(P(x)=\int_x^3 f(t)\,dt\) and \(f\) is continuous, what is \(P'(x)\)?
- Differentiate \(Q(x)=\int_x^2(4t-1)\,dt\). State the continuity condition that permits the rule.
- Evaluate \(R(x)=\int_x^\pi\cos t\,dt\), then differentiate your expression and compare it with the lower-bound rule.
- Explain, using the difference in accumulated integrals, why \(\int_x^b f(t)\,dt\) has a negative sign in its derivative.
- A student says \(\frac{d}{dx}\int_x^5 t^3\,dt=x^3\). Identify the error and give the corrected derivative.
