Why a Variable Lower Limit Changes the Sign
An integral with a variable lower limit measures accumulated area from the moving point \(x\) to a fixed point \(b\). As \(x\) moves to the right, the interval from \(x\) to \(b\) loses the portion just passed. This direction of change is why a minus sign appears in the derivative.
The sign can be established precisely by reversing the limits. The Fundamental Theorem of Calculus gives the derivative of an accumulation function with a fixed lower limit and a variable upper limit. Reversing the limits turns the integral in this tutorial into the negative of that familiar form.
Reverse the Limits First
The definite integral changes sign when its limits are reversed. In particular, for any integrable function \(f\) and suitable points \(x\) and \(b\),
This identity applies whether \(x\) is less than or greater than \(b\). The symbol \(t\) is a dummy variable of integration: it records the values through which the integral accumulates, while \(x\) determines the moving endpoint.
Define the fixed-start accumulation function \(G(x)=\int_b^x f(t)\,dt\). When \(f\) is continuous, the Fundamental Theorem of Calculus Part I says that \(G'(x)=f(x)\) at interior points of the interval. The integral with variable lower limit is \(F(x)=\int_x^b f(t)\,dt=-G(x)\). Differentiating this negative gives \(F'(x)=-G'(x)=-f(x)\).
In \(\int_x^b f(t)\,dt\), \(b\) stays fixed and \(x\) moves.
Write \(\int_x^b f(t)\,dt=-\int_b^x f(t)\,dt\).
The derivative of \(\int_b^x f(t)\,dt\) is \(f(x)\), provided \(f\) is continuous at the relevant points.
The original integral is the negative of that accumulation function, so its derivative is \(-f(x)\).
Why the Formula Works: A Difference-Quotient Derivation
The reversal argument gives a quick route from the Fundamental Theorem. A difference-quotient argument also shows directly how the minus sign arises. Let \(F(x)=\int_x^b f(t)\,dt\), where \(f\) is continuous on an interval containing \(x\) and \(b\). For nonzero \(h\) sufficiently close to zero, additivity of the definite integral gives
Divide by \(h\). Since \(h\ne 0\), this yields
To see the limit, compare the average of \(f\) over the short interval with the value \(f(x)\). Continuity at \(x\) means that \(f(t)\) becomes as close as desired to \(f(x)\) whenever \(t\) is sufficiently close to \(x\). More explicitly, for \(h\) small enough that every point between \(x\) and \(x+h\) is close to \(x\),
The right-hand side tends to zero as \(h\) tends to zero, by continuity of \(f\) at \(x\). Therefore the average tends to \(f(x)\). The difference quotient for \(F\) has a minus sign in front of that average, so its limit is \(-f(x)\). Thus \(F'(x)=-f(x)\).
Worked Examples
Worked Example: A Polynomial Integrand
Let \(F(x)=\int_x^3(2t+1)\,dt\). The integrand \(2t+1\) is continuous, so the formula applies. Reverse the limits, then differentiate:
Check by evaluating the integral explicitly. An antiderivative of \(2t+1\) is \(t^2+t\), since \(\frac{d}{dt}(t^2+t)=2t+1\). Hence
Differentiating this expression gives \(F'(x)=-2x-1\), agreeing with the formula. For an additional value check, \(F(1)=12-1-1=10\), and the original integral gives \(\int_1^3(2t+1)\,dt=[t^2+t]_1^3=12-2=10\).
Worked Example: A Trigonometric Integrand
Consider \(H(x)=\int_x^{\pi/2}\cos t\,dt\). The cosine function is continuous, so reversing the limits gives
Check by evaluating the integral: an antiderivative of \(\cos t\) is \(\sin t\), because \(\frac{d}{dt}\sin t=\cos t\). Thus
Its derivative is \(-\cos x\), as required. At \(x=0\), the integral is \(\int_0^{\pi/2}\cos t\,dt=1-0=1\), matching \(H(0)=1-\sin 0=1\).
Worked Example: The Integrand Can Be Negative
Let \(K(x)=\int_x^1(t^2-4)\,dt\). The integrand is continuous, even though it is negative for \(t\) between \(0\) and \(1\). Reversing the limits gives
In particular, the minus sign applies to the whole integrand value: \(-(x^2-4)=-x^2+4\). To check by direct evaluation, an antiderivative of \(t^2-4\) is \(\frac{t^3}{3}-4t\), since its derivative is \(t^2-4\). Therefore
Differentiate the result: \(K'(x)=-x^2+4=4-x^2\), which agrees with the formula. At \(x=0\), direct evaluation gives \(\int_0^1(t^2-4)\,dt=\frac{1}{3}-4=-\frac{11}{3}\); the explicit expression gives \(K(0)=-\frac{11}{3}\) as well.
A Common Sign Error
A frequent mistake is to treat the variable lower limit as though it were a variable upper limit. For example, the incorrect calculation for \(F(x)=\int_x^3(2t+1)\,dt\) is
This loses the effect of the lower endpoint. The Fundamental Theorem gives \(+f(x)\) for an accumulation integral with a fixed lower limit and variable upper limit, but \(\int_x^3 f(t)\,dt\) is the negative of that form. The correction is
The explicit evaluation confirms that the incorrect sign cannot be right: \(F(x)=12-x^2-x\), whose derivative is \(-2x-1\), not \(2x+1\). A reliable check is to reverse the limits before differentiating, rather than trying to memorize a sign in isolation.
What the Sign Says About Accumulation
The derivative records how the integral changes when its lower endpoint moves. If \(f(x)>0\), then \(F'(x)=-f(x)<0\): moving \(x\) slightly to the right removes a small positive contribution from the interval, so the integral decreases. If \(f(x)<0\), then \(F'(x)>0\): moving the lower limit right removes a small negative contribution, so the integral increases. These conclusions follow directly from the formula and do not require the whole integral to be positive or negative.
The formula concerns the value of the integrand at the moving endpoint, not an average over the entire interval. The integral determines the accumulated quantity \(F(x)\); its derivative responds locally, at \(x\). Continuity ensures that the contribution from a very short interval near \(x\), divided by its length, approaches \(f(x)\). The lower-limit orientation contributes the negative sign.
Check Your Understanding
For each question, keep track of which endpoint is fixed and where the minus sign enters.
- If \(F(x)=\int_x^5 f(t)\,dt\) and \(f\) is continuous, what is \(F'(x)\)?
- Rewrite \(\int_x^b f(t)\,dt\) with the limits reversed. What sign appears?
- Find the derivative of \(G(x)=\int_x^2(3t^2-1)\,dt\).
- A learner claims that \(\frac{d}{dx}\int_x^b f(t)\,dt=f(x)\). What step in the reversal argument corrects this claim?
- Why does continuity of \(f\) matter when differentiating the variable-lower-limit integral?
