Interpreting the Behavior | AP Calc AB Unit 6 Study Guide

The Core Idea: Interpreting the Behavior of Accumulation Functions Involving Area

An accumulation function measures the net accumulation of a quantity whose rate of change is described by another function. Specifically, a function defined as an integral, $g (x) = \int_{a}^{x} f (t) d t$ , represents the accumulated net signed area under the curve of $f (t)$ starting from a fixed point $a$ and ending at a variable point $x$ . The fundamental concept of this topic is the profound connection between the integrand, $f (t)$ , and the accumulation function, $g (x)$ .

The behavior of $g (x)$ —where it increases, decreases, its concavity, and where it has extrema or points of inflection—is entirely determined by the properties of $f (t)$ . By analyzing the function $f (t)$ (e.g., whether it is positive, negative, increasing, or decreasing), we can describe the characteristics of $g (x)$ without ever needing to find an explicit antiderivative for $f (t)$ . This relationship is a direct application of the Fundamental Theorem of Calculus.

Key Formulas/Rules/Theorems

The primary theorem governing accumulation functions is the Second Fundamental Theorem of Calculus.

1. The Accumulation Function Definition

If $f$ is a continuous function on an interval containing $a$ , then the function $g (x)$ defined by

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

is called an accumulation function. It represents the net signed area under the curve of $f$ from $t = a$ to $t = x$ .

2. The Second Fundamental Theorem of Calculus

This theorem establishes the derivative of an accumulation function. If $g (x) = \int_{a}^{x} f (t) d t$ , then the derivative of $g (x)$ is the integrand function evaluated at $x$ .

$\frac{d}{d x} [\int_{a}^{x} f (t) d t] = f (x)$

This means $g^{'} (x) = f (x)$ . In other words, the function $f (x)$ is the rate of change of the accumulation function $g (x)$ .

3. The Definite Integral as Net Change

The definite integral of a rate of change gives the net change in the original quantity. If $F^{'} (x)$ is the rate of change of a quantity $F (x)$ , then the net change in $F (x)$ from $x = a$ to $x = b$ is:

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

This is the First Fundamental Theorem of Calculus and is directly related to understanding accumulation.

Understanding the Relationship Between $f (x)$ and $g (x)$

The core of this topic is translating the properties of the function $f (x)$ into properties of the accumulation function $g (x) = \int_{a}^{x} f (t) d t$ . Since $g^{'} (x) = f (x)$ and $g^{''} (x) = f^{'} (x)$ , we can establish a clear dictionary of relationships.

If the function $f (x)$ ...	Then the accumulation function $g (x)$ ...	Justification
is positive ( $f (x) > 0$ )	is increasing	$g^{'} (x) = f (x) > 0$
is negative ( $f (x) < 0$ )	is decreasing	$g^{'} (x) = f (x) < 0$
changes from negative to positive	has a local minimum	$g^{'} (x)$ changes from $-$ to $+$
changes from positive to negative	has a local maximum	$g^{'} (x)$ changes from $+$ to $-$
is increasing	is concave up	$g^{''} (x) = f^{'} (x) > 0$
is decreasing	is concave down	$g^{''} (x) = f^{'} (x) < 0$
has a local extremum (a "turn")	has a point of inflection	$g^{''} (x) = f^{'} (x)$ changes sign

Core Concepts & Rules

An accumulation function, $g (x) = \int_{a}^{x} f (t) d t$ , calculates the net signed area under the curve of $f$ from a constant $a$ to a variable $x$ .
The derivative of an accumulation function is simply the integrand: $\frac{d}{d x} g (x) = f (x)$ . This is the most critical connection.
The value of $f (x)$ at any point gives the slope (the rate of change) of $g (x)$ at that same point.
Where $f (x)$ is above the x-axis ( $f (x) > 0$ ), $g (x)$ is increasing.
Where $f (x)$ is below the x-axis ( $f (x) < 0$ ), $g (x)$ is decreasing.
Critical points for $g (x)$ occur where $g^{'} (x) = f (x) = 0$ .
The slope of $f (x)$ (i.e., $f^{'} (x)$ ) determines the concavity of $g (x)$ .
Where $f (x)$ is increasing, $g (x)$ is concave up.
Where $f (x)$ is decreasing, $g (x)$ is concave down.
Points of inflection for $g (x)$ occur where $g^{''} (x) = f^{'} (x)$ changes sign, which corresponds to local extrema on the graph of $f (x)$ .

Step-by-Step Example 1: Analytical Application

Let the function $g (x)$ be defined by $g (x) = \int_{2}^{x} (t^{2} - 3 t - 4) d t$ . Determine the intervals on which $g (x)$ is decreasing and the intervals on which $g (x)$ is concave down.

Step 1: Find the derivative of $g (x)$ to analyze its behavior.

Using the Second Fundamental Theorem of Calculus, $g^{'} (x) = f (x)$ .

$g^{'} (x) = x^{2} - 3 x - 4$

Step 2: Find where $g (x)$ is decreasing.

The function $g (x)$ is decreasing where its derivative, $g^{'} (x)$ , is negative. We find the critical points by setting $g^{'} (x) = 0$ .

$x^{2} - 3 x - 4 = 0$

$(x - 4) (x + 1) = 0$

The critical points are $x = 4$ and $x = - 1$ . Now, we create a sign chart for $g^{'} (x)$ .

Interval	Test Value	$g^{'} (x) = (x - 4) (x + 1)$	Behavior of $g (x)$
$(- \infty, - 1)$	$x = - 2$	$(-) (-) = +$	Increasing
$(- 1, 4)$	$x = 0$	$(-) (+) = -$	Decreasing

| $(4, \infty)$ | $x = 5$ | $(+)(+)=+` | Increasing | Based on the sign chart, $g(x)$ is decreasing on the interval $(- 1, 4)$ because $g^{'} (x) < 0$ on this interval.

Step 3: Find the second derivative of $g (x)$ to analyze its concavity.

We differentiate $g^{'} (x)$ to find $g^{''} (x)$ .

$g^{''} (x) = \frac{d}{d x} (x^{2} - 3 x - 4) = 2 x - 3$

Step 4: Find where $g (x)$ is concave down.

The function $g (x)$ is concave down where its second derivative, $g^{''} (x)$ , is negative. We find potential inflection points by setting $g^{''} (x) = 0$ .

$2 x - 3 = 0 ⟹ x = \frac{3}{2}$

Now, we create a sign chart for $g^{''} (x)$ .

Interval	Test Value	$g^{''} (x) = 2 x - 3$	Concavity of $g (x)$
$(- \infty, 3/2)$	$x = 0$	$-$	Concave Down

| $(3/2, \infty)$ | $x = 2$ | $+` | Concave Up | Based on the sign chart, $g(x)$ is concave down on the interval $(- \infty, 3/2)$ because $g^{''} (x) < 0$ on this interval.

Step-by-Step Example 2: Exam-Style Application

The graph of a continuous function $f$ , consisting of three line segments and a semicircle, is shown below for the interval $- 2 \leq t \leq 6$ . Let $g$ be the function defined by $g (x) = \int_{0}^{x} f (t) d t$ .

(Imagine a graph of $f (t)$ that goes through $(- 2, 0)$ , $(0, 2)$ , $(2, 0)$ , has a semicircle below the x-axis from $(2, 0)$ to $(6, 0)$ with center (4,0) and radius 2.)

a) Find the values of $g (2)$ and $g (6)$ .

$g (2) = \int_{0}^{2} f (t) d t$ . This is the area of the triangle above the axis from $t = 0$ to $t = 2$ .
Area = $\frac{1}{2} \times base \times height = \frac{1}{2} \times 2 \times 2 = 2$ . So, $g (2) = 2$ .
$g (6) = \int_{0}^{6} f (t) d t = \int_{0}^{2} f (t) d t + \int_{2}^{6} f (t) d t$ . The second integral is the area of the semicircle below the axis.
Area of semicircle = $\frac{1}{2} π r^{2} = \frac{1}{2} π (2)^{2} = 2 π$ . Since it's below the axis, the signed area is $- 2 π$ .
$g (6) = g (2) + (- 2 π) = 2 - 2 π$ .

b) On what intervals is $g (x)$ increasing?

$g (x)$ is increasing where $g^{'} (x) = f (x) > 0$ .
Looking at the graph, $f (x)$ is positive (above the x-axis) on the interval $(- 2, 2)$ .

c) Find the x-coordinate of all points of inflection of $g (x)$ . Justify your answer.

$g (x)$ has a point of inflection where $g^{''} (x) = f^{'} (x)$ changes sign. This occurs where the slope of $f (x)$ changes sign, which corresponds to local extrema on the graph of $f$ .
The slope of $f (x)$ changes from positive to negative at $x = 0$ .
The slope of $f (x)$ changes from negative to positive at $x = 4$ .
Therefore, $g (x)$ has points of inflection at $x = 0$ and $x = 4$ .

Using Your Calculator

For an accumulation function $g (x) = \int_{a}^{x} f (t) d t$ where $f (t)$ is given as a complex equation (e.g., $f (t) = e^{- t^{2}} cos (t)$ ), a calculator is essential for analysis.

Let $g (x) = \int_{1}^{x} sin (t^{2}) d t$ .

**1. Evaluating the function at a point, e.g., $g (3) ‘ : * * * U se t h e n u m er i c a l in t e g r a t i o n f e a t u re ‘ f n I n t ‘. * O n t h e h o m escree n, e n t er : ‘ f n I n t (s in (T^{2}), T, 1, 3) ‘. * T hi sc a l c u l a t es$ \int_1^3 \sin(t^2) dt$.

2. Evaluating the derivative at a point, e.g., $g^{'} (2)$ :

Remember that $g^{'} (x) = f (x) = sin (x^{2})$ . This does not require a calculator's derivative command.
Simply evaluate $f (2)$ : $sin (2^{2}) = sin (4)$ .

3. Evaluating the second derivative at a point, e.g., $g^{''} (2)$ :

Remember that $g^{''} (x) = f^{'} (x)$ . We need to find the derivative of $f (x) = sin (x^{2})$ at $x=2`. * Use the numerical derivative feature `nDeriv`. * On the home screen, enter: `nDeriv(sin(X^2), X, 2)`. **4. Graphing the accumulation function $g(x)$ :**
In your Y= editor, set Y1 = fnInt(sin(T^2), T, 1, X).
Graphing Y1 will show the graph of $g (x)$ . This can be slow.
To verify the relationships, you can set `Y2 = sin(X^2) $(the graph of $f(x)`) and `Y3 = nDeriv(sin(X^2), X, X)` (the graph of $f'(x)`). * You can then visually confirm: where `Y2` is positive, `Y1` is increasing. Where `Y3` is positive, `Y1` is concave up. ## AP Exam Quick Hit ### Common Question Types - **Graphical Analysis:** Given the graph of a function $f$ , you will be asked to analyze $g (x) = \int_{a}^{x} f (t) d t$ . This includes finding values of $g$ (by calculating areas), finding where $g$ is increasing/decreasing (where $f$ is positive/negative), and finding points of inflection for $g$ (where $f$ has local extrema).

Finding Absolute Extrema: You will be asked to find the absolute maximum or minimum value of $g (x) = \int_{a}^{x} f (t) d t$ on a closed interval. This requires the Candidates Test: find critical points where $f (x) = 0$ , then evaluate $g (x)$ at these critical points and at the endpoints of the interval. The largest value is the absolute maximum.

Common Mistakes

Confusing $f$ and $g$ : A very common error is to identify the maximum value on the graph of $f$ and claim it is the maximum value of $g$ . Remember, the y-values of $f$ represent the slope of $g$ . The maximum value of $g$ occurs at a point where $f$ changes from positive to negative, and its value is an accumulated area.
Forgetting the Initial Condition: When finding a value like $g (4) = \int_{0}^{4} f (t) d t$ , students often forget to add the value of the integral from the starting point to an intermediate point. For example, $g (4) = g (2) + \int_{2}^{4} f (t) d t$ .
Mixing up Increasing vs. Concave Up: Students mistakenly think $g$ is concave up where $f$ is positive. This is incorrect. $g$ is increasing where $f$ is positive. $g$ is concave up where $f$ is increasing (i.e., $f^{'} > 0$ ).
Ignoring the Starting Point $a$ : The value of $g (x)$ depends on the starting point $a$ of the integral $\int_{a}^{x} f (t) d t$ . A different $a$ will shift the graph of $g (x)$ vertically, changing its absolute extrema values (but not their locations). Also, remember that $g (a) = \int_{a}^{a} f (t) d t = 0$ .

Interpreting the Behavior of Accumulation Functions Involving Area - AP Calculus AB Study Guide