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AP Calculus AB study notes

6.4.3 Differentiating Functions Defined by Integrals (FTC Part 1)

AP Syllabus focus:
‘Apply the Fundamental Theorem of Calculus, Part 1, to state that the derivative of an accumulation function equals the integrand evaluated at x.’

The Fundamental Theorem of Calculus, Part 1, connects differentiation and integration by showing how accumulation functions behave and how their derivatives reveal underlying instantaneous rates.

Understanding Accumulation Functions Defined by Integrals

An accumulation function represents the total accumulated change of a quantity from a fixed starting point up to a variable upper limit. In this subsubtopic, the focus is on how to differentiate such functions when they are expressed as definite integrals with a variable bound. This idea forms the core of the Fundamental Theorem of Calculus, Part 1 (FTC Part 1), which provides a powerful link between rate functions and the accumulation of change over an interval.

When a function is defined using a definite integral whose upper limit is a variable, the function’s derivative can often be found without integrating at all. Instead, FTC Part 1 states that the derivative of the accumulation function equals the original integrand evaluated at the variable upper limit, provided the integrand is continuous on the interval.

FTC Part 1 and Its Meaning

FTC Part 1 tells us how accumulation behaves instantaneously. Suppose an accumulation function increases over an interval by summing contributions from a rate function. Then its derivative measures the instantaneous rate of accumulation — precisely the value of the rate at that point. This relationship is foundational in AP Calculus AB because it allows students to transition between integrals as totals and derivatives as rates.

Before using FTC Part 1, students must understand how integrals define new functions and how the structure of the integral affects differentiation.

Key Structure of an Accumulation Function

A typical accumulation function has the following form:

  • A constant lower limit, representing the initial point of accumulation.

  • A variable upper limit, usually written as xx.

  • A continuous integrand, representing the rate of change being accumulated.

g(x)=axf(t),dt g(x) = \int_{a}^{x} f(t), dt
g(x) g(x) = Accumulation function (units depend on ff and tt)
a a = Fixed lower limit of integration
x x = Variable upper limit
f(t) f(t) = Continuous integrand representing a rate of change

Because the upper limit is variable, changes in xx alter the accumulated total, making gg behave like any differentiable function whose derivative depends on the rate function ff

The shaded region represents the accumulated area under a continuous function from x=ax = a to a variable endpoint xx. As xx moves, the value of the accumulation function F(x)=axf(t),dtF(x) = \int_a^x f(t), dt changes with the size of this region. This image directly supports interpreting accumulation functions as “area from a fixed start to a moving boundary.” Source.

A student must be aware that the integrand uses tt as the variable of integration, which is independent of xx. This separation allows xx to serve as a boundary while the integral itself accumulates values of f(t)f(t).

Applying the Fundamental Theorem of Calculus, Part 1

FTC Part 1 directly describes the derivative of an accumulation function.

g(x)=f(x) g'(x) = f(x)
g(x) g'(x) = Derivative of the accumulation function
f(x) f(x) = Value of the integrand evaluated at the upper limit

This result means that a definite integral does more than compute total change; it also constructs a function whose derivative retrieves the original rate. For AP students, this provides a crucial bridge between the two major branches of calculus.

When applying this relationship, it is essential that the integrand be continuous on the interval of interest. Continuity guarantees that the accumulation process is well-behaved and differentiable at every point in the domain.

How Differentiation Works in This Context

Differentiating a function given as an integral does not involve traditional differentiation rules until the structure of the integral changes. In its simplest form — constant lower limit, variable upper limit, no transformations applied to the limit — the differentiation becomes immediate through FTC Part 1.

Students should observe that the upper limit of integration determines which value of the integrand becomes the derivative. Conceptually, the small increase in accumulated area as xx grows corresponds exactly to the height of the graph of ff at xx. Thus, a thin vertical slice added at xx represents the instantaneous rate, reinforcing the geometric intuition behind FTC Part 1.

The blue and red shaded regions show the accumulated area up to xx and the extra area added from xx to x+hx+h. The expressions A(x+h)A(x)A(x+h)-A(x) and hf(x)h \cdot f(x) compare the true additional area with the area of a thin rectangle whose height is the function value at xx. Although the formulas and the fraction A(x+h)A(x)h\dfrac{A(x+h)-A(x)}{h} go slightly beyond the AP statement, they concretely illustrate how the derivative of the area function approaches f(x)f(x) as hh becomes small. Source.

Variations and Extensions within AP Calculus AB

While full generalizations are explored in higher mathematics, AP Calculus AB focuses on understanding the fundamental structure:

  • The lower limit must be constant.

  • The variable upper limit must be the differentiation variable.

  • The integrand must be continuous.

Under these conditions, students apply FTC Part 1 directly, even when the integrand is algebraic, trigonometric, exponential, or a combination of these.

Occasionally, the upper limit may involve a more complex expression. Although FTC Part 1 still applies, such cases typically require an additional chain rule step. These instances remain aligned with the same essential idea: the derivative of an accumulated quantity reflects the rate at the boundary of accumulation.

Interpreting FTC Part 1 in Context

In applied problems, the accumulation function often represents a total amount — distance, volume, mass, or another physical quantity — derived from a rate function. FTC Part 1 connects this total to the instantaneous rate. When applying the theorem, students learn to:

  • Identify the integrand as the underlying rate of change.

  • Recognize the accumulation function as representing a total amount or net change.

  • Understand that the derivative of this total retrieves the instantaneous rate at the upper limit.

These ideas allow students to interpret accumulation functions fluently using graphical, numerical, or verbal descriptions of the rate function.

Summary of Student Responsibilities for This Subsubtopic

Students working with FTC Part 1 should be able to:

  • Explain why the derivative of an accumulation function equals the integrand evaluated at the upper limit.

  • Recognize functions defined by integrals and determine when FTC Part 1 applies.

  • Use proper interpretation of variables in the integrand versus the variable limit.

  • Apply differentiation rules when the structure of the upper limit requires additional steps.

These skills form an essential part of understanding the relationship between total change and instantaneous rate within AP Calculus AB.

FAQ

FTC Part 1 does not directly apply when the lower limit is the variable and the upper limit is constant. The sign of the expression changes because reversing limits reverses the sign of the integral.

If H(x) = ∫ from x to a of f(t) dt, then H′(x) = –f(x).

This negative sign arises because increasing x reduces the accumulated area rather than increasing it.

FTC Part 1 requires continuity of the integrand at the point where the derivative is being taken. A single removable discontinuity away from that point does not prevent the theorem from applying.

If the integrand is discontinuous at the same point where the derivative is evaluated, the derivative of the accumulation function may fail to exist.

If the integrand is negative throughout the interval, the accumulation function decreases as the upper limit increases.

Key consequences:
• The accumulation function is strictly decreasing.
• Its derivative is always negative.
• The graph of the accumulation function will slope downward, though its curvature depends on how the integrand varies.

FTC Part 1 bypasses integration entirely because it describes a structural relationship between a rate function and the function defined by accumulating that rate.

Instead of computing total area, it identifies the instantaneous rate of change directly as the integrand evaluated at the upper bound.

This makes FTC Part 1 one of the most powerful shortcuts in calculus.

The symbol used in the integral (t, u, or any other letter) is only a dummy variable; it has no effect on the application of FTC Part 1.

What matters is:
• The integrand’s value at the upper limit.
• The upper limit matching the differentiation variable.
• The continuity of the integrand.

The variable of integration simply keeps the expression organised and avoids confusion when functions contain multiple variables.

Practice Questions

Question 1 (1–3 marks)
A function G is defined by G(x) = ∫ from 2 to x of h(t) dt, where h is continuous.
(a) State G′(x).
(b) If h(4) = –3, determine the value of G′(4).

Question 1
(a) 1 mark: Correct application of FTC Part 1: G′(x) = h(x).
(b) 1 mark: Substitution at x = 4 giving G′(4) = h(4) = –3.

Question 2 (4–6 marks)
Let f be a continuous function and define F(x) = ∫ from 1 to x of (3f(t) – 2) dt.
(a) Find F′(x).
(b) Given that f(2) = 5, compute F′(2).
(c) Suppose instead the function is written as H(x) = ∫ from 1 to g(x) of (3f(t) – 2) dt, where g is differentiable. Using the chain rule and the Fundamental Theorem of Calculus Part 1, express H′(x) in terms of f, g, and g′.

Question 2
(a) 1 mark: Correct application of FTC Part 1: F′(x) = 3f(x) – 2.
(b) 1 mark: Substitution of x = 2 to obtain F′(2) = 3f(2) – 2.
1 mark: Correct numerical result: F′(2) = 13.
(c) 1 mark: Recognition that the derivative must be evaluated at the upper limit: 3f(g(x)) – 2.
1 mark: Correct multiplication by g′(x) from the chain rule.
1 mark: Fully correct final expression: H′(x) = (3f(g(x)) – 2) g′(x).

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