Two Different Questions About Integrals
An integral problem may ask for a number, or it may define a function and ask for its derivative. These tasks are related by the Fundamental Theorem of Calculus, but they call for different parts of the theorem. Choosing the right one is the first step in a reliable solution.
For example, evaluating \(\int_a^b f(t)\,dt\) asks for the signed accumulation over a fixed interval. By contrast, differentiating \(A(x)=\int_a^x f(t)\,dt\) asks how that accumulation changes as the endpoint moves. If both endpoints depend on \(x\), each endpoint contributes to the derivative, with the lower endpoint entering with a minus sign.
Choose the Task Before Calculating
Use endpoint evaluation when the problem asks for the value of a definite integral with fixed bounds and an antiderivative is available. Use the differentiation form of the theorem when the problem asks for the derivative of an integral-defined function. These are not interchangeable instructions: one produces a number, while the other produces a function.
When both bounds move, let \(u\) and \(v\) be differentiable, with their values in an interval on which \(f\) is continuous. The corresponding formula is
The upper endpoint contributes its integrand value times its rate of change. The lower endpoint is subtracted because increasing the lower bound removes a slice of accumulated area.
Why the Moving-Bounds Formula Works
The formula follows by combining endpoint evaluation with the chain rule. This gives a short justification and makes the signs easier to remember. Fix any number \(c\) in the interval of continuity, and define \(F(y)=\int_c^y f(t)\,dt\). By the Fundamental Theorem of Calculus, \(F'(y)=f(y)\).
Additivity of definite integrals gives \(\int_{u(x)}^{v(x)}f(t)\,dt=F(v(x))-F(u(x))\). Differentiating each composed term by the chain rule yields the result:
Continuity of \(f\) on an interval containing the endpoint values ensures that the Fundamental Theorem applies there. Differentiability of the bounds is needed to use the chain rule. Before applying the formula, check these conditions rather than assuming that every integral-defined function is differentiable everywhere.
Worked Example: Differentiate an Integral with Two Moving Bounds
Worked Example: Differentiate an Integral with Two Moving Bounds
Let \(H(x)=\int_{x^2}^{2x+1}(t^2+1)\,dt\). Find \(H'(x)\), then check the result using an antiderivative.
The integrand \(f(t)=t^2+1\) is continuous for every real \(t\), and both bounds are differentiable. Here the upper bound is \(v(x)=2x+1\), so \(v'(x)=2\); the lower bound is \(u(x)=x^2\), so \(u'(x)=2x\). Therefore
To check, an antiderivative of \(t^2+1\) is \(F(t)=\frac{t^3}{3}+t\). Thus \(H(x)=F(2x+1)-F(x^2)\), and differentiating this expression gives
which agrees with the moving-bounds calculation.
Worked Example: Evaluate a Definite Integral and Track Units
Worked Example: Evaluate a Definite Integral and Track Units
Suppose a flow rate is \(R(t)=(2\text{ L}/\text{h}^2)t+1\text{ L}/\text{h}\), where \(t\) is measured in hours. Find the amount that flows from \(t=1\text{ h}\) to \(t=3\text{ h}\).
This asks for the value of an integral over fixed bounds, so use endpoint evaluation. An antiderivative is \(F(t)=(1\text{ L}/\text{h}^2)t^2+(1\text{ L}/\text{h})t\), because differentiating it gives \(R(t)\). At the upper endpoint,
At the lower endpoint,
Subtract lower from upper:
The units check: a rate measured in liters per hour, integrated over hours, gives an amount in liters. The answer is \(10\text{ L}\).
Signs, Reversed Bounds, and Direction of Motion
Keep the minus sign on the lower-endpoint term even when the lower endpoint is decreasing. Its derivative may itself be negative, so the two minus signs can produce a positive contribution. Similarly, reversing fixed bounds changes the sign of the integral: \(\int_b^a f(t)\,dt=-\int_a^b f(t)\,dt\). These sign rules reflect signed accumulation, not a change in the integrand.
Worked Example: Check the Signs with Opposing Endpoint Motion
Let \(J(x)=\int_x^{2x}(t^2+1)\,dt\). Find \(J'(x)\) at \(x=1\), and verify it by evaluating the integral as a function of \(x\).
Both bounds are differentiable and \(t^2+1\) is continuous. The upper bound has derivative \(2\), and the lower bound has derivative \(1\). Therefore
For a direct check, use \(F(t)=\frac{t^3}{3}+t\):
Both calculations give \(8\). In particular, the lower-endpoint contribution is subtracted: at \(x=1\), it is \(f(1)\cdot 1=2\), not an additional \(2\).
Worked Example: A Composite Upper Endpoint
Worked Example: A Composite Upper Endpoint
Let \(A(x)=\int_0^{\sin x}(t^2+2)\,dt\). Find \(A'(x)\), then evaluate it at \(x=\frac{\pi}{6}\).
The integrand is continuous and the upper bound \(g(x)=\sin x\) is differentiable. The variable-bound formula includes the derivative of that bound:
Since \(\sin(\frac{\pi}{6})=\frac12\) and \(\cos(\frac{\pi}{6})=\frac{\sqrt3}{2}\),
For a check, \(F(t)=\frac{t^3}{3}+2t\) is an antiderivative, so \(A(x)=\frac{\sin^3 x}{3}+2\sin x\). Differentiating gives \(\sin^2 x\cos x+2\cos x=(\sin^2 x+2)\cos x\), the same derivative.
Continuity Is a Condition, Not a Decoration
If \(f\) is continuous throughout an interval containing the relevant endpoint values, the stated FTC formulas apply. If \(f\) is not continuous, do not assume that the formulas automatically apply at every point. The integral may still be defined, but differentiability or the derivative formula at a discontinuity needs separate checking.
For instance, define \(f(t)=-1\) for \(t<0\) and \(f(t)=1\) for \(t\geq 0\), and let \(B(x)=\int_{-1}^x f(t)\,dt\). For \(x<0\) near zero, \(B(x)=-(x+1)\); for \(x>0\) near zero, \(B(x)=-1+x\). The left derivative at zero is \(-1\), while the right derivative is \(1\). Thus \(B'(0)\) does not exist. This example shows why a continuity check matters; it does not say that every discontinuous integrand makes its accumulation function nondifferentiable everywhere.
A Common Mistake: Adding the Lower-Bound Term
A frequent error is to differentiate both endpoint contributions with a plus sign. For \(J(x)=\int_x^{2x}(t^2+1)\,dt\), the incorrect calculation at \(x=1\) is
This treats an increasing lower bound as if it added area. The corrected calculation subtracts that contribution:
The direct check \(J'(x)=7x^2+1\) gives \(J'(1)=8\), confirming the correction. When uncertain, rewrite the integral as \(F(v(x))-F(u(x))\) and differentiate; this makes the lower-bound minus sign explicit.
A Reliable Workshop Routine
For a mixed Fundamental Theorem problem, identify what the question asks for before choosing a formula. Then check the hypotheses and the bounds. A useful sequence is:
If the problem asks for a definite integral’s value, evaluate endpoints. If it asks for a derivative, differentiate the accumulation function.
Confirm continuity of the integrand on the relevant interval and differentiability of any moving bounds.
For moving bounds, include the derivative of each bound and subtract the lower-bound contribution.
Differentiate an antiderivative expression when practical, verify the sign using bound direction, and confirm units in applied problems.
The Fundamental Theorem connects accumulation and differentiation, but the question determines which direction of that connection to use. Careful choices about the task, continuity, endpoint motion, signs, and units turn a mixed problem into a sequence of manageable checks.
Check Your Understanding
For each question, identify the relevant formula or issue before calculating.
- For \(K(x)=\int_1^{x^2}(t+3)\,dt\), what formula gives \(K'(x)\), and what factor comes from differentiating the upper bound?
- If \(M(x)=\int_{x}^{x+2} f(t)\,dt\) and \(f\) is continuous, write \(M'(x)\) in terms of \(f\).
- What is the relationship between \(\int_b^a f(t)\,dt\) and \(\int_a^b f(t)\,dt\)?
- Why must a moving lower endpoint contribute with a minus sign?
- If a continuous rate has units of liters per hour, what units does its definite integral over time have?
