6.5 Interpreting the Behavior of Accumulation Functions Involving Area

An accumulation function $g(x)=\int_a^x f(t)\,dt$ stores the net signed area under $f$ from $a$ to $x$. Its behavior comes from the graph of $f$: $g$ increases where $f$ is positive, decreases where $f$ is negative, and changes concavity where $f$ changes from increasing to decreasing or back. For AP Calculus, read the integrand graph before describing the accumulation function.

Why This Matters for the AP Calculus Exam

This topic builds the skill of reading a function defined by an integral without ever finding a formula for it. On both multiple-choice and free-response questions, you are often handed the graph of $f$ (frequently labeled as a derivative) and asked about the function it defines. You need to connect the sign of $f$ to whether $g$ is increasing or decreasing, the zeros of $f$ to the extrema of $g$, and the slope of $f$ to the concavity of $g$. Free-response questions in AP Calculus regularly use a graph made of semicircles and line segments so you can find exact areas with geometry, then interpret what those areas mean.

Key Takeaways

• $g(x) = \int_a^x f(t)\,dt$ measures net signed area under $f$, so area below the x-axis counts as negative.
• By the Fundamental Theorem of Calculus, $g'(x) = f(x)$ and $g''(x) = f'(x)$.
• $g$ increases where $f$ is positive and decreases where $f$ is negative; extrema of $g$ happen where $f$ changes sign.
• $g$ is concave up where $f$ is increasing and concave down where $f$ is decreasing; inflection points of $g$ occur where $f$ changes from increasing to decreasing or back.
• $g(a) = 0$ always, since the area from $a$ to $a$ is zero.
• When you only have a graph of $f$, use geometry (triangles, rectangles, semicircles) to compute exact accumulated area.

The Fundamental Theorem of Calculus, Briefly

This is a quick summary. For the full setup, see the 6.4 guide.

The Fundamental Theorem of Calculus connects an accumulation function to an antiderivative. If

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

then

$$g'(x) = f(x)$$

The first equation says $g(x)$ is the accumulated area under $f(t)$ between the fixed lower bound $x = a$ and the variable upper bound $x$. Here $a$ is a constant and $x$ varies.

The accumulation function with boundaries $a$ and $x$ and the area under the curve $f(t)$ shaded and labeled as the value of $g$

The second equation says the derivative of $g(x)$ is $f(x)$. Equivalently, $g$ is an antiderivative of $f$. Putting these together: the accumulated area under $f$ equals the value of $g$ at that upper bound.

Reading the Behavior of g from f

An integrally defined function has the same kinds of features as any other function: intervals of increase and decrease, extrema, concavity, and inflection points. Many AP questions hand you a graph or formula for $f$ and ask you to describe $g$.

Quick Reference Chart

Increasing and decreasing come from the first derivative; concavity comes from the second derivative. A relative minimum or maximum cannot also be an inflection point.

The key move with accumulation problems is that you are working backward. Instead of differentiating the given graph or formula, treat it as the already-found derivative of your target function. For example, if

$$G(x) =\int_a^x 6n^2\,dn$$

then $6x^2$ is $G'(x)$, so use it to find critical points or differentiate it again for concavity.

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How to Use This on the AP Calculus Exam

Almost all questions on this topic come as a graph, a table, or an equation, and all three show up regularly. The graph type is the most common, so that is the focus here.

Free Response

Graph questions usually define $G(x) =\int_a^x f(t)\,dt$ and give you the graph of $f(t)$. The graph you see is not your target function. It is the derivative of your target function. Here is an official College Board free-response question from the 2022 exam.

Question 3 from the 2022 AP Calculus AB FRQ Exam. Courtesy of College Board

Let $f$ be a differentiable function with $f(4)=3$. On the interval $0 \leq x \leq 7$, the graph of $f'$, the derivative of $f$, consists of a semicircle and two line segments, as shown in the figure above.

(a) Find $f(0)$ and $f(5)$.

(b) Find the x-coordinates of all points of inflection on the graph of $f$ for $0<x<7$. Justify your answer.

(c) Let $g$ be the function defined by $g(x)=f(x)-x$. On what intervals, if any, is $g$ decreasing for $0 \leq x \leq 7$? Show the analysis that leads to your answer.

(d) For the function $g$ defined in part (c), find the absolute minimum value on the interval $0 \leq x \leq 7$. Justify your answer.

Part (a)

Part (a) asks for $f(0)$ and $f(5)$. Since the graph is $f'(x)$, the original function satisfies

$$f(x) = f(\text{start}) + \int_{\text{start}}^{x} f'(t)\,dt$$

By the Fundamental Theorem of Calculus, the integral of $f'$ equals the net signed area between the curve and the x-axis, which is the change in $f$ over those bounds. Set up two equations.

The first is

$$f(0) + \int_0^4 f'(x)\,dx = f(4)$$

which rearranges to

$$f(0) = f(4)- \int_0^4 f'(x)\,dx$$

Here, $f(4) = 3$, and $\int_0^4 f'(x)\,dx$ is the area under $f'(x)$ from $x = 0$ to $x = 4$, which is the semicircle region. Its area is

$$\frac{\pi}{2}\,r^2 =\frac{\pi}{2}\cdot 2^2 = 2\pi$$

Since this region is below the x-axis, the net signed area is negative. So

$$f(0) = 3- (-2\pi)=3+2\pi$$

For $f(5)$, set up

$$f(4) + \int_4^5 f'(x)\,dx = f(5)$$

The region from $x=4$ to $x=5$ lies under the first line segment, which has slope $\frac{2-0}{6-4} = 1$ using the points $(4,0)$ and $(6,2)$. At $x = 5$, the height is $1$, so the triangle has base $1$ and height $1$, giving area $\frac{1}{2}$. Then

$$f(5) = 3 + \frac{1}{2}=3.5$$

Write down your integral setups and clearly label your final answers as $f(0)$ and $f(5)$.

Part (b)

This asks for the x-coordinates of points of inflection of $f$. Since you are given $f'$, the inflection points of $f$ occur where the slope of $f'$ changes sign (that is, where $f''$ changes sign). On the graph, this happens at $x = 2$, where the slope of $f'$ changes from negative to positive, and at $x = 6$, where it changes from positive to negative.

Part (c)

The new function is

$$g(x) =f(x) -x$$

so

$$g'(x) = f'(x) -1$$

A function decreases when its derivative is negative, so $g$ is decreasing when

$$f'(x)-1 < 0$$

or

$$f'(x) < 1$$

From the graph, this holds on $0<x<5$.

Part (d)

This asks for the absolute minimum of $g(x)$. Since $g'(x) = f'(x) - 1$, a critical point happens where $f'(x) = 1$, which occurs at $x = 5$. The candidates are the endpoints $x=0$ and $x=7$ plus the critical point $x = 5$.

The values $f(0)$ and $f(5)$ come from part (a). The value $f(7)$ comes from $f(5) + \int_5^7 f'(x)\,dx = 3.5 + 3 = 6.5$, using geometry to find the area.

Integral between $x =5$ and $x =7$ broken into a triangle and a rectangle. Courtesy of College Board and Julianna Fontanilla

So the absolute minimum of $g(x)$ is $-1.5$, which occurs at $x=5$.

Problem Solving Tips

1. 💭 Anchor the relationship. If $g(x)=\int_{a}^{x}f(t)\,dt$, then $g'(x) = f(x)$ and $g''(x) = f'(x)$. Write this somewhere on your paper.
2. 📈 Use the graph. Mark zeros, maxima, and minima of $f$, then relate each back to a feature of $g$.
3. 🧠 Check every candidate. Evaluate $g$ at critical points where $f$ changes sign, and don't forget the endpoints.
4. ✅ Track which function you need. It is easy to mix up $g$, $g'$, and $g''$. Confirm you are analyzing the right derivative before you answer.

Common Misconceptions

• The given graph is the function itself. When a problem gives you the graph of $f$ and defines $g(x) = \int_a^x f(t)\,dt$, the graph is $g'$, not $g$. Read features of $g$ off of $f$.
• Area is always positive. Accumulation is net signed area. Regions below the x-axis subtract from the total, so a semicircle below the axis contributes a negative value.
• Extrema of g come from zeros of f', not f. The extrema of $g$ happen where $f$ (which equals $g'$) changes sign, not where $f$ has a maximum or minimum.
• Every corner is an inflection point. A corner or sharp change in $f$ only signals an inflection point of $g$ if the slope of $f$ actually changes sign there.
• Forgetting endpoints in an absolute extremum search. On a closed interval, compare $g$ at the endpoints as well as at the interior critical points.
• Mixing up g, g', and g''. Increasing and decreasing of $g$ come from the sign of $f$; concavity of $g$ comes from whether $f$ is rising or falling.

Related AP Calculus Guides

• Unit 6 Overview: Integration and Accumulation of Change
• 6.11 Integrating Using Integration by Parts
• 6.1 Integration and Accumulation of Change
• 6.12 Integrating Using Linear Partial Fractions
• 6.3 Riemann Sums, Summation Notation, and Definite Integral Notation
• 6.4 The Fundamental Theorem of Calculus and Accumulation Functions