From an Accumulated Quantity to Its Derivative
An integral with a fixed starting point records the total accumulated contribution from that point up to a variable endpoint. For example, if a quantity is added at a rate \(f(t)\), then \(\int_a^x f(t)\,dt\) is the accumulation between the fixed point \(a\) and the current point \(x\).
The Fundamental Theorem of Calculus gives a direct way to differentiate this accumulation. When the integrand is continuous, the derivative at \(x\) is simply the integrand evaluated at \(x\). The main task in this tutorial is to use that rule correctly in concrete cases, including examples where the integrand is polynomial, trigonometric, or exponential.
The Fixed-Start Derivative Rule
The relevant part of the Fundamental Theorem of Calculus says that if \(f\) is continuous at \(x\), then the accumulation function \(F(x)=\int_a^x f(t)\,dt\) is differentiable there and its derivative is \(f(x)\). In particular, if \(f\) is continuous throughout an interval, the formula holds at every interior point of that interval.
Notice what the rule does and does not require. The lower limit \(a\) is fixed, and the upper limit is the variable \(x\). The integrand must be continuous at the point where the derivative is taken. No antiderivative needs to be found before applying the rule. If an explicit formula for \(F\) is also useful, an antiderivative can be used to compute the integral and check the result.
Why the Rule Works
The derivative measures the change in \(F\) over a small change in its input. For \(h\ne 0\), additivity of definite integrals gives $$ \frac{F(x+h)-F(x)}{h} = \frac{1}{h}\int_x^{x+h} f(t)\,dt. $$ As \(h\) approaches zero, this quotient is the average value of \(f\) over the short interval from \(x\) to \(x+h\). Continuity makes those nearby values approach \(f(x)\).
Using \(F(u)=\int_a^u f(t)\,dt\), subtract \(F(x)\) from \(F(x+h)\). The shared portion from \(a\) to \(x\) cancels, leaving \(\int_x^{x+h}f(t)\,dt\).
Continuity at \(x\) means that for any \(\varepsilon>0\), all \(t\) sufficiently close to \(x\) satisfy \(|f(t)-f(x)|<\varepsilon\).
For sufficiently small nonzero \(h\), every point between \(x\) and \(x+h\) is close enough to \(x\). Therefore $$ \left|\frac{1}{h}\int_x^{x+h} f(t)\,dt-f(x)\right| = \left|\frac{1}{h}\int_x^{x+h}(f(t)-f(x))\,dt\right| \le \frac{1}{|h|}\int_{\min(x,x+h)}^{\max(x,x+h)}\varepsilon\,dt =\varepsilon. $$ Thus the difference quotient approaches \(f(x)\), which proves \(F'(x)=f(x)\).
This argument also explains the continuity condition: it guarantees that over a sufficiently short interval, the integrand stays close to its value at the endpoint. The rule is especially efficient in computation because it identifies the derivative without evaluating the accumulated total first.
Worked Examples
Worked Example: A Polynomial Integrand
Differentiate $$ F(x)=\int_{-1}^{x}(3t^2-4t+1)\,dt. $$
The integrand \(f(t)=3t^2-4t+1\) is a polynomial, so it is continuous for every real \(t\). The fixed-start derivative rule applies at every real \(x\): $$ F'(x)=3x^2-4x+1. $$ To check this result directly, first find an antiderivative of the integrand: $$ A(t)=t^3-2t^2+t, $$ since $$ A'(t)=3t^2-4t+1. $$ Evaluate at the endpoints: $$ F(x)=A(x)-A(-1) =(x^3-2x^2+x)-\bigl((-1)^3-2(-1)^2+(-1)\bigr). $$ The lower-endpoint value is $$ A(-1)=-1-2-1=-4, $$ so $$ F(x)=x^3-2x^2+x+4. $$ Differentiating this explicit expression gives $$ F'(x)=3x^2-4x+1, $$ in agreement with the theorem. As an additional endpoint check, \(F(-1)=(-1)^3-2(-1)^2+(-1)+4=0\), as it must for an integral whose two limits are both \(-1\).
Worked Example: A Trigonometric Integrand
Let $$ G(x)=\int_0^x(2\sin t-\cos t)\,dt. $$ Find \(G'(x)\), and check it by evaluating the integral explicitly.
The integrand \(2\sin t-\cos t\) is continuous for every real \(t\). The fixed-start derivative rule therefore gives $$ G'(x)=2\sin x-\cos x. $$ For a check, an antiderivative is $$ A(t)=-2\cos t-\sin t, $$ because $$ A'(t)=2\sin t-\cos t. $$ Now evaluate at the upper and lower limits: $$ G(x)=A(x)-A(0) =(-2\cos x-\sin x)-(-2\cos 0-\sin 0). $$ Since \(\cos 0=1\) and \(\sin 0=0\), $$ G(x)=-2\cos x-\sin x+2. $$ Differentiating this expression gives $$ G'(x)=2\sin x-\cos x, $$ which matches the result from the theorem. Also, \(G(0)=-2-0+2=0\), consistent with the integral from \(0\) to \(0\).
Worked Example: An Exponential Integrand
Differentiate and check $$ H(x)=\int_1^x(e^{2t}+3e^t)\,dt. $$
Both exponential terms are continuous for every real \(t\), so the integrand is continuous. Applying the fixed-start derivative rule immediately yields $$ H'(x)=e^{2x}+3e^x. $$ For an explicit check, an antiderivative is $$ A(t)=\frac{1}{2}e^{2t}+3e^t, $$ because $$ A'(t)=\frac{1}{2}(2e^{2t})+3e^t=e^{2t}+3e^t. $$ Therefore $$ H(x)=A(x)-A(1) =\left(\frac{1}{2}e^{2x}+3e^x\right) -\left(\frac{1}{2}e^2+3e\right). $$ The terms involving \(e^2\) and \(e\) are constants with respect to \(x\). Differentiating gives $$ H'(x)=\frac{1}{2}(2e^{2x})+3e^x=e^{2x}+3e^x. $$ The derivative agrees with the theorem. Substitution of \(x=1\) also gives \(H(1)=0\), as required by the integral's equal limits.
A Direct Rule and an Antiderivative Check
The examples illustrate two useful routes to the same answer. If only the derivative is requested, the shortest route is to check continuity and substitute \(x\) into the integrand. If an explicit expression for the accumulation function is requested, evaluate the definite integral using an antiderivative and then differentiate the resulting expression to check it.
| Task | Useful method |
|---|---|
| Find the derivative of \(\int_a^x f(t)\,dt\) | Apply the fixed-start derivative rule: the derivative is \(f(x)\), provided \(f\) is continuous at \(x\). |
| Find an explicit formula for \(\int_a^x f(t)\,dt\) | Find an antiderivative \(A\) and calculate \(A(x)-A(a)\). |
| Check a derivative obtained from the rule | Differentiate the explicit formula, if one has been calculated, and compare with \(f(x)\). |
In the antiderivative method, the subtraction \(A(x)-A(a)\) matters: the lower endpoint contributes a constant to the accumulation function. That constant affects the value of the function, even though its derivative is zero.
Common Mistake: Differentiating the Lower-Endpoint Contribution
Consider $$ P(x)=\int_1^x 3t^2\,dt. $$ An incorrect calculation is to evaluate the integral as \(x^3-1\) and then claim $$ P'(x)=3x^2-1. $$ The error is treating the constant \(-1\) as if its derivative were \(-1\). In fact, the derivative of any constant is zero: $$ \frac{d}{dx}(x^3-1)=3x^2-0=3x^2. $$ This agrees with the fixed-start derivative rule, which gives \(P'(x)=3x^2\) directly. The explicit integral also checks correctly: $$ P(x)=x^3-1^3=x^3-1, \qquad P(1)=1-1=0. $$ The lower limit determines the constant in the accumulation function, but it is not differentiated as a variable.
What to Check Before Applying the Rule
- Confirm that the lower limit is fixed and the upper limit is \(x\).
- Identify the integrand and evaluate it at \(x\), not at the dummy variable \(t\).
- Check that the integrand is continuous at the point where the derivative is requested.
- If finding the entire accumulation function, remember to subtract the antiderivative's value at the fixed lower limit.
These checks separate the immediate derivative calculation from the distinct task of evaluating the integral. For example, \(\int_1^x(e^t+t)\,dt\) has derivative \(e^x+x\) because its integrand is continuous; evaluating the integral first is optional when only the derivative is needed.
Why Fixed-Start Accumulations Matter
Accumulation functions turn a rate into a total accumulated change. The derivative rule says that the instantaneous rate of change of that total recovers the current value of the rate being accumulated. It is useful in problems involving distance accumulated from velocity, total quantity accumulated from a flow rate, or area accumulated under a graph.
The rule is also a practical reminder of how the two operations are related: integration constructs an accumulation, and differentiation recovers the continuous integrand. The fixed starting point sets the baseline value, while the variable endpoint determines the current total. Keeping those roles clear makes the calculation both short and reliable.
Check Your Understanding
For each question, state any continuity condition needed and show the calculation that supports your answer.
- Find the derivative of \(F(x)=\int_2^x(t^3+2t)\,dt\).
- Find the derivative of \(G(x)=\int_0^x(\cos t+4\sin t)\,dt\).
- Find the derivative of \(H(x)=\int_{-1}^x e^t\,dt\), and check it by writing an explicit formula for \(H(x)\).
- For \(Q(x)=\int_3^x 5t^4\,dt\), explain why the fixed lower limit affects \(Q(x)\) but does not contribute a nonzero term to \(Q'(x)\).
- A student writes \(\frac{d}{dx}\int_0^x(t^2+1)\,dt=x^2+1\). Explain why the rule applies and why the dummy variable \(t\) is replaced by \(x\) in the answer.
