AP Syllabus focus:
‘Use the Fundamental Theorem of Calculus, Part 2, to evaluate ∫ₐᵇ f(x) dx as F(b) − F(a), where F is any antiderivative of f.’
Evaluating definite integrals becomes significantly more efficient using the Fundamental Theorem of Calculus, Part 2, which links antiderivatives to accumulated change and provides a powerful computational shortcut.
Understanding the Role of the Fundamental Theorem of Calculus, Part 2
The Fundamental Theorem of Calculus, Part 2 (FTC Part 2) establishes a direct connection between definite integrals and antiderivatives, allowing integrals to be computed without relying on geometric reasoning or numerical approximations. This connection expresses that if a function is continuous on a closed interval, its accumulated net change over that interval can be captured entirely by evaluating an antiderivative at the interval’s endpoints. Because of this, FTC Part 2 becomes one of the most essential tools in introductory calculus, transforming integration from an area-finding problem into a problem of determining antiderivatives.
When applying this theorem, it is important to identify an antiderivative—a function whose derivative equals the integrand. Once an antiderivative is known, the theorem guarantees that the definite integral is simply the difference between its values at the upper and lower bounds.
= A continuous function on the interval
= Any antiderivative of , meaning
= Lower and upper limits of integration, representing endpoints of the interval
A central idea in this relationship is that definite integrals represent net accumulation, meaning intervals where the function is positive contribute positive area and intervals where the function is negative contribute negative area. This interpretation remains consistent regardless of the method of evaluation, but FTC Part 2 gives a more algebraic pathway to obtaining this accumulated value.
The curve appears above the x-axis with the region between the curve and the interval shaded, illustrating accumulated area. This visual represents the definite integral of a nonnegative function over an interval. The precise curve and bounds are not essential and serve only to provide context for the geometric meaning of integration. Source.
Requirements and Conditions for Using FTC Part 2
Continuity and Antiderivatives
For FTC Part 2 to apply, the integrand must be continuous, or at least have only removable or mild discontinuities that still allow an antiderivative to exist. When the function meets this requirement, an antiderivative will be guaranteed, and therefore the definite integral can be evaluated through the difference of antiderivative values.
Antiderivative: A function such that for a given function .
This result is especially important because it means any antiderivative will work. Since antiderivatives can differ by a constant, the difference remains the same regardless of which specific antiderivative is chosen.
A key conceptual point is that definite integrals do not depend on the constant of integration. This contrasts with indefinite integrals, where the constant is essential because the expression represents a family of functions rather than a single numerical value.
Structure and Purpose of Endpoint Evaluation
Why Endpoint Substitution Works
FTC Part 2 provides an elegant solution to the accumulation problem because the antiderivative captures all instantaneous rate-of-change information of the integrand. Substituting the upper endpoint, , computes the total accumulated amount up to the end of the interval, while substituting the lower endpoint, , captures the accumulated amount up to the start. Subtracting these values isolates the change occurring only between and , which aligns perfectly with the concept of definite integrals as measurements of net accumulation.
Step-by-Step Logical Process
Use the following structure to correctly apply FTC Part 2:
Identify the integrand and confirm continuity on the interval.
Find any antiderivative of the integrand.
Evaluate the antiderivative at the upper limit.
Evaluate the antiderivative at the lower limit.
Subtract: upper value minus lower value.
These steps ensure consistent and accurate evaluation while reinforcing the broader conceptual message that integration is the reverse process of differentiation.
Relationship to the Concept of Accumulated Change
The capacity of FTC Part 2 to compute accumulated change reflects one of the defining ideas of calculus: accumulated effects arise from summing infinitely many contributions, but antiderivatives make that infinite summation unnecessary. By applying an antiderivative, the process of accumulating rates is condensed into a single subtraction.
This reinforces the idea that definite integrals provide net change, not just area. For instance, if the integrand switches sign, the integral reflects this through net accumulation rather than total unsigned area.
The graph shows regions where lies above or below the x-axis, shaded with “+” and “−” to indicate positive and negative contributions. This demonstrates that the definite integral represents net accumulation rather than total unsigned area. Extra stylistic elements such as color coding appear but do not introduce concepts beyond the AP syllabus. Source.
Importance for AP Calculus AB
FTC Part 2 is one of the most frequently applied theorems in AP Calculus AB. Students must be comfortable:
Recognizing when a definite integral requires FTC Part 2
Identifying and using antiderivatives correctly
Applying the upper-minus-lower evaluation framework
Understanding why the constant of integration is unnecessary
Connecting the result to the idea of net accumulation
Mastery of this subsubtopic supports later topics involving substitution, integration techniques, and contextual interpretation of integrals, all of which depend heavily on the principles introduced by FTC Part 2.
FAQ
If an elementary antiderivative is not available, the Fundamental Theorem of Calculus Part 2 cannot be applied directly in symbolic form.
In such cases, students typically:
• Use numerical methods such as trapezoidal or midpoint approximations.
• Rely on technology-assisted integration when permitted.
Even though FTC Part 2 provides the theoretical foundation for definite integrals, not all integrals can be evaluated exactly, so approximation becomes essential.
All antiderivatives differ by a constant, but when computing F(b) minus F(a), that constant subtracts out.
For example, if two antiderivatives differ by +C, then both F(b) and F(a) increase by C, making the difference identical.
As a result, the constant of integration has no effect on the value of a definite integral, which is why it is omitted.
Net change corresponds to the signed result of a definite integral, reflecting increases and decreases together.
To identify net change:
• Look for quantities that can both increase and decrease, such as velocity, rate of flow, or rate of profit.
• Determine whether the question references displacement, net accumulation, or overall change.
Total change usually requires integrating the absolute value instead, which is not the default interpretation in FTC Part 2.
Continuity ensures that an antiderivative exists on the entire interval, which is essential for the upper–lower evaluation F(b) − F(a).
If the integrand has a removable or mild discontinuity, the integral may still exist, but FTC Part 2 technically relies on the existence of a differentiable antiderivative.
For exam purposes, students should check for continuity unless the problem explicitly states otherwise.
FTC Part 2 works because integration undoes differentiation in the context of accumulation.
The process involves:
• Identifying a function whose derivative matches the integrand.
• Using that function to capture all accumulated rate information across the interval.
This reversal principle creates the shortcut that replaces summing infinitely many tiny contributions with the single computation F(b) − F(a).
Practice Questions
Question 1 (1–3 marks)
The function f is continuous on the interval [2, 7]. An antiderivative of f is F.
Given that F(7) = 12.4 and F(2) = 5.1, evaluate the definite integral from 2 to 7 of f(x) dx.
Question 1
• Correct application of FTC Part 2: integral equals F(7) − F(2). (1 mark)
• Correct numerical result: 12.4 − 5.1 = 7.3. (1 mark)
Question 2 (4–6 marks)
Let g be a continuous function on [0, 5], and let G be an antiderivative of g. Some values of G are shown below:
G(0) = −3
G(2) = 4
G(5) = 10
(a) Use the Fundamental Theorem of Calculus, Part 2, to compute the definite integral from 0 to 5 of g(x) dx.
(b) The graph of g is known to be negative on [0, 1] and positive on (1, 5]. Explain how this sign information is reflected in your answer to part (a).
Question 2
(a)
• States or uses the relation integral from 0 to 5 of g(x) dx = G(5) − G(0). (1 mark)
• Substitutes values correctly: 10 − (−3). (1 mark)
• Correct evaluation: 13. (1 mark)
(b)
• States that negative values of g on [0, 1] contribute negative accumulation and positive values on (1, 5] contribute positive accumulation. (1 mark)
• Explains that the positive contribution outweighs the negative, resulting in a positive net value of 13. (1 mark)
