The Fundamental Theorem | AP Calc BC Unit 6 Study Guide

The Core Idea: The Fundamental Theorem of Calculus and Accumulation Functions

The Fundamental Theorem of Calculus (FTC) establishes the profound, inverse relationship between differentiation and integration. It provides the essential bridge that connects the two major branches of calculus. The theorem is presented in two parts. The first part introduces the concept of an accumulation function, a function defined as a definite integral, $F (x) = \int_{a}^{x} f (t) d t$ . It reveals that the rate of change of this accumulating area is precisely the original function $f (x)$ . In essence, taking the derivative of an integral "undoes" the integration.

The second part of the theorem leverages this inverse relationship to provide a practical method for calculating the exact value of a definite integral. Instead of relying on the cumbersome limit of Riemann sums, the FTC allows us to evaluate $\int_{a}^{b} f (x) d x$ by simply finding any antiderivative of the integrand $f (x)$ , which we call $F (x)$ , and calculating the change in that antiderivative over the interval, $F (b) - F (a)$ . This transforms the problem of finding area into a more straightforward algebraic process.

Key Formulas/Rules/Theorems

The Fundamental Theorem of Calculus (FTC)

The theorem is composed of two distinct but related parts.

Part 1: The Derivative of an Accumulation Function

If a function $f$ is continuous on the interval $[a, b]$ , then the function $F$ defined by

$F (x) = \int_{a}^{x} f (t) d t$

is an antiderivative of $f$ . This means that the derivative of $F (x)$ is $f (x)$ :

$F^{'} (x) = \frac{d}{d x} [\int_{a}^{x} f (t) d t] = f (x)$

for all $x$ in the open interval $(a, b)$ .

Part 2: The Evaluation Theorem

If a function $f$ is continuous on the interval $[a, b]$ and $F$ is any antiderivative of $f$ (meaning $F^{'} (x) = f (x)$ ), then the definite integral of $f$ from $a$ to $b$ is given by:

$\int_{a}^{b} f (x) d x = F (b) - F (a)$

This is often written using the notation $[F (x)]_{a}^{b}$ or $F (x)_{a}^{b}$ to represent $F (b) - F (a)$ .

The FTC and the Chain Rule

When the upper limit of integration is a differentiable function of $x$ , say $u (x)$ , the Chain Rule must be applied when differentiating the integral:

$\frac{d}{d x} [\int_{a}^{u (x)} f (t) d t] = f (u (x)) \cdot u^{'} (x)$

Understanding Accumulation Functions

An accumulation function measures the net accumulation of a quantity whose rate of change is given by some function. The standard form is $g (x) = \int_{a}^{x} f (t) d t$ , where $f (t)$ represents the rate at which a quantity is changing. The function $g (x)$ then gives the total net change or accumulation of that quantity from a fixed starting point $t = a$ to a variable endpoint $t = x$ .

A key insight is that an accumulation function is a particular antiderivative of the integrand.

The function $g (x) = \int_{a}^{x} f (t) d t$ is the specific antiderivative of $f$ that has a value of zero at $x = a$ . We can see this by evaluating $g (a) = \int_{a}^{a} f (t) d t = 0$ .
To represent an antiderivative that satisfies a different initial condition, say $h (a) = C$ , we can write the function as $h (x) = C + \int_{a}^{x} f (t) d t$ . When we evaluate this at $x = a$ , we get $h (a) = C + \int_{a}^{a} f (t) d t = C + 0 = C$ , satisfying the desired condition. This form is crucial for solving initial value problems where a rate and an initial value are known.

Core Concepts & Rules

FTC Part 1: The derivative of a definite integral with a constant lower limit and an upper limit of $x$ is simply the integrand with $x$ substituted for the variable of integration.
FTC Part 2: To evaluate a definite integral, find any antiderivative of the integrand and compute the difference between its value at the upper limit and its value at the lower limit.
Inverse Relationship: Differentiation and integration are inverse operations. Taking the derivative of an accumulation function returns the original function.
Accumulation Function: A function defined as $g (x) = \int_{a}^{x} f (t) d t$ represents the net accumulated area under the curve of $f$ from $a$ to $x$ .
Initial Condition: The accumulation function $g (x) = \int_{a}^{x} f (t) d t$ is a specific antiderivative of $f$ that always satisfies the condition $g (a) = 0$ .
General Solution: Any particular antiderivative $h (x)$ of a function $f (x)$ that passes through the point $(a, C)$ can be expressed as $h (x) = C + \int_{a}^{x} f (t) d t$ .
Chain Rule Application: If the upper limit of integration is a function $u (x)$ , the derivative of the integral is the integrand evaluated at $u (x)$ , multiplied by the derivative of $u (x)$ .

Step-by-Step Example 1: Evaluating a Definite Integral using FTC Part 2

Problem: Evaluate the definite integral $\int_{1}^{3} (3 x^{2} - 4 x) d x$ .

Step 1: Identify the integrand and find an antiderivative.

The integrand is $f (x) = 3 x^{2} - 4 x$ . We need to find a function $F (x)$ such that $F^{'} (x) = f (x)$ . Using the power rule for antiderivatives:

$F (x) = \frac{3 x ^{2 + 1}}{2 + 1} - \frac{4 x ^{1 + 1}}{1 + 1} = \frac{3 x ^{3}}{3} - \frac{4 x ^{2}}{2} = x^{3} - 2 x^{2}$

(Note: We can ignore the constant of integration +C because it will cancel out in the next step.)

Step 2: Apply the FTC Part 2 formula: $\int_{a}^{b} f (x) d x = F (b) - F (a)$ .

Here, $a = 1$ and $b = 3$ . We will evaluate $F (3) - F (1)$ .

Step 3: Calculate $F (3)$ and $F (1)$ .

$F (3) = (3)^{3} - 2 (3)^{2} = 27 - 2 (9) = 27 - 18 = 9$

$F (1) = (1)^{3} - 2 (1)^{2} = 1 - 2 (1) = 1 - 2 = - 1$

Step 4: Compute the final difference.

$F (3) - F (1) = 9 - (- 1) = 10$

Solution: $\int_{1}^{3} (3 x^{2} - 4 x) d x = 10$ .

Step-by-Step Example 2: Exam-Style Application with a Graph

Problem: The graph of a continuous function $f$ is shown below, consisting of two line segments and a semicircle. Let the function $g$ be defined as $g (x) = \int_{- 2}^{x} f (t) d t$ .

The Fundamental Theorem of Calculus and Accumulation Functions - AP Calculus BC Study Guide