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6.5 Interpreting the Behavior of Accumulation Functions Involving Area

1 min readjune 18, 2024

6.5 Interpreting the Behavior of Accumulation Functions Involving Area

Welcome back to AP Calculus! 🌶️

In this guide, we'll analyze the behavior of accumulation functions using the Fundamental Theorem of Calculus. We'll focus on graphical, numerical, analytical, and verbal representations to gain a comprehensive understanding of integrally-defined functions.

👩‍🏫The Fundamental Theorem Of Calculus

The Fundamental Theorem of Calculus links accumulation functions, which are adapted from infinite Riemann sums, to antiderivatives, which “undo” a derivative. This theory states that if

F(n)=anf(t)dtF(n)=\int_{a}^{n}f(t)dt

Then,

F(n)=f(n)F’(n) = f(n)

The first equation above states that F(n)F(n) is the accumulation function of f(t)f(t) which equals the area under the curve of f(t)f(t) between the boundaries made by the lines x=ax =a and x=nx = n. Here, a is some constant number while n is a variable. Visually, this looks like the following:

Untitled

Image Courtesy of Julianna Fontanilla

The second equation says that the derivative of F(n)F(n) is f(n)f(n). This also means that the antiderivative of f(n)f(n) is F(n)F(n), and that F(n)=f(n)F'(n) = f(n).

To put these 2 pieces of information together, the area under a function is equal to the value of its antiderivative (calculated with the same bounds, of course).

📒 Using the Fundamental Theorem Of Calculus

Just like previous functions, these integrally defined functions can be defined by their various characteristics listed below. Some AP multiple-choice or free-response questions will ask you to analyze these factors given a graph or a function with an integral.

Quick Refresher Chart

If F(x) is…then F’(x)and F’'(x)
Increasing+———
Decreasing-———
Relative Maximum0 and changes from + to -is -
Relative Minimum0 and changes from - to +is +
Concave Up———is +
Concave Down———is -
Inflection Point———Changes Sign

It’s important to note that finding increasing/decreasing ranges uses the first derivative while finding concave up/down ranges uses the second derivative. In addition, a relative minimum or maximum cannot be an inflection point.

When working with analysis integral problems where you must find one of the above items in the table remember that you are working backwards. Instead of finding or drawing the derivative of the given numerical graph or function, use the information as the already-found derivative of your target integral. For example, if you are told that

G(x)=ax6n2dnG(x) =\int_a^x 6n^2dn

Realize that 6x26x^2 is equal to the first derivative and find critical points or the second derivative accordingly.


🤔 Practicing with Accumulation Functions

Almost all AP questions on this topic will fall into 1 of 3 types: graph, table, or equation. These are all equally likely to occur. We will be going over a graph problem!

📈 Graph Questions

Graph questions almost always tell you that a function G(x)=axf(t)dtG(x) =\int_a^x f(t)dt and give you a graph or image of the function f(t)f(t).

The most important thing to remember with this type of problem is that the graph given is not your target function. Instead, the graph is the derivative of your target function. Let’s try an official College Board FRQ from the 2022 exam! (All credits to College Board.)

Untitled

Courtesy of College Board

Let ff be a differentiable function with f(4)=3f(4)=3. On the interval 0x70 \leq x \leq 7, the graph of ff', the derivative of ff, consists of a semicircle and two line segments, as shown in the figure above.

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

(b) Find the x-coordinates of all points of inflection on the graph of ff for 0<x<70<x<7. Justify your answer.

(c) Let gg be the function defined by g(x)=f(x)xg(x)=f(x)-x. On what intervals, if any, is gg decreasing for 0x70 \leq x \leq 7? Show the analysis that leads to your answer.

(d) For the function gg defined in part (c), find the absolute minimum value on the interval 0x70 \leq x \leq 7. Justify your answer.

AP Calc 2022 #3a

Part A of this question asks us to find f(0)f(0) and f(5)f(5). Since we are told that the graph above is the graph of f(x)f'(x), this means that the original equation of…

f(x)=axf(x)dxf(x) = \int_a^x f'(x) dx

Using the Fundamental Theorem of Calculus, we know that this equation is equal to the accumulation function, and is represented on the f(x)f'(x) graph by the area between the curve and the x-axis. This area is the change in f(x)f(x) over the given bounds.

Using mathematical operations, we can set up 2 equations to find f(0)f(0) and f(5)f(5).

The first equation is

f(0)+04f(x)dx=f(4)f(0) + \int_0^4 f'(x) dx = f(4)

which can be simplified to

f(0)=f(4)04f(x)dxf(0) = f(4)- \int_0^4 f'(x) dx

We know that f(4)=3f(4) = 3 from the problem and that 04f(x)dx\int_0^4 f'(x) dx is the area under f(x)f'(x) between x = 0 and x = 4, given the definition of the accumulation function. Our problem tells us that f(x)f'(x) is composed of a semicircle and 2 line segments, and we can see that the area between 0 and 4 is the semicircle area. The area of this semicircle is

π2r2=π222=2π\frac \pi 2*r^2 =\frac \pi2* 2^2 = 2\pi

Remember that since this area is below the x-axis, it is considered negative!

Plugging this into the our equation above gives

f(0)=3(2π)=3+2πf(0) = 3- (-2\pi)=3+2\pi

In the same way, you can set up your equation for f(5)f(5).

f(4)+45f(x)dx=f(5)f(4) + \int_4^5f'(x)dx = f(5)

Again, we know that f(4)=3f(4) = 3. The area of f(x)f'(x) from x=4x=4 and x=5x=5 lies on the first straight line which has a slope of 2064=1\frac {2-0}{6-4} = 1, using the known points of (4,0) and (6,2). Using this information in a slope equation OR by looking at the graph, we know that when x=5x =5 then y=1y =1. This means that the triangle area between x=4x=4 and x=5x=5 has a height of 1 and a width of 1, with a total area of 12\frac 12. Putting this into our earlier equation, we get

f(5)=3+12=3.5f(5) = 3 + \frac 12=3.5

Remember to write down your integrals AND label your final answers as f(0) and f(5) to earn all points on the free-response portion!

AP Calc 2022 #3b

This part asks for the x-coordinates of any points of inflection of the function f(x)f(x). Since we are given the graph of f(x)f’(x), we need to look at the derivative of this graph to find points of inflection (2nd derivative) of the original function f(x)f(x).

Remember that points of inflection occur when the second derivative of a function changes signs. On the graph given to us, this occurs where the slope of the derivative changes signs. This occurs at x = 2, where the slope changes from negative to positive, and x = 6, where the slope changes from positive to negative. Since the slope of a function at a point is the 1st derivative, the slope of the first derivative at a point is the second derivative!

AP Calc 2022 #3c

Our new equation is defined by

g(x)=f(x)xg(x) =f(x) -x

so the new derivative is

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

A function is decreasing when its derivative is negative, so g(x) is decreasing when

0>f(x)10>f'(x)-1

or

1>f(x)1>f'(x)

According to the graph, this occurs between 0<x<50<x<5.

AP Calc 2022 #3d

This part asks for the absolute minimum function of g(x)g(x). A local minimum occurs when g(x)=0g(x) =0and g(x)g(x) changes from negative to positive. Our candidates are the two endpoints of x=0x=0 and x=7x=7 as well as the local minimum x=5x = 5. Just as a reminder, g(x)=f(x)1g'(x) = f'(x) -1 so g(x)=0g'(x) =0 when f(x)=1f'(x) =1 which only occurs within the function when x=5x = 5.

Testing all 3 candidates for the absolute minimum gives…

xf(x)g(x) = f(x) -x
03 + 2pi3 + 2pi
53.53.5 - 5 = -1.5
77.56.5 -7 =-0.5

f(0)f(0) and f(5)f(5) were both imported from Part A.

f(7)f(7) was found by adding f(5)+57f(x)dx=3.5+3=6.5f(5) + \int_5^7f'(x)dx =3.5 + 3 = 6.5 through geometry.

Untitled

Courtesy of College Board and Julianna Fontanilla

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


📝 Practice Tips

Here are some tips and tricks when approaching these question types:

  1. 💭 Understand the Relationship: If F(x)=axf(t)dtF(x)=\int_{a}^{x}f(t)dt, this means that F(x)=f(x)F’(x) = f(x) and F(x)=f(x)F’'(x) = f'(x).
  2. 📈 Use Graphical Information: Identify notable points on a graph, such as zeroes, minimums, and maximums. Always relate these back to the original integral equation.
  3. 🧠 Evaluate Extrema: Evaluate function values at critical points (zeroes) to determine relative and absolute extrema of the original integral. Don’t forget to also evaluate endpoints!
  4. Clarify the Question: Make sure you are looking for the zeroes of the correct derivative! It gets very easy to confuse F(x), F’(x), and F’’(x). Write Tip 1 somewhere on your paper.

This topic may be difficult, but the most important thing to remember is that area under a curve = the antiderivative. Practice using both concepts with different integral questions to get comfortable with the idea. As always, the best way to improve your understanding is to try lots of practice problems and familiarize yourself with the setup of common AP questions covering this topic. You’ve got this! 👍🏼

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6.5 Interpreting the Behavior of Accumulation Functions Involving Area

1 min readjune 18, 2024

6.5 Interpreting the Behavior of Accumulation Functions Involving Area

Welcome back to AP Calculus! 🌶️

In this guide, we'll analyze the behavior of accumulation functions using the Fundamental Theorem of Calculus. We'll focus on graphical, numerical, analytical, and verbal representations to gain a comprehensive understanding of integrally-defined functions.

👩‍🏫The Fundamental Theorem Of Calculus

The Fundamental Theorem of Calculus links accumulation functions, which are adapted from infinite Riemann sums, to antiderivatives, which “undo” a derivative. This theory states that if

F(n)=anf(t)dtF(n)=\int_{a}^{n}f(t)dt

Then,

F(n)=f(n)F’(n) = f(n)

The first equation above states that F(n)F(n) is the accumulation function of f(t)f(t) which equals the area under the curve of f(t)f(t) between the boundaries made by the lines x=ax =a and x=nx = n. Here, a is some constant number while n is a variable. Visually, this looks like the following:

Untitled

Image Courtesy of Julianna Fontanilla

The second equation says that the derivative of F(n)F(n) is f(n)f(n). This also means that the antiderivative of f(n)f(n) is F(n)F(n), and that F(n)=f(n)F'(n) = f(n).

To put these 2 pieces of information together, the area under a function is equal to the value of its antiderivative (calculated with the same bounds, of course).

📒 Using the Fundamental Theorem Of Calculus

Just like previous functions, these integrally defined functions can be defined by their various characteristics listed below. Some AP multiple-choice or free-response questions will ask you to analyze these factors given a graph or a function with an integral.

Quick Refresher Chart

If F(x) is…then F’(x)and F’'(x)
Increasing+———
Decreasing-———
Relative Maximum0 and changes from + to -is -
Relative Minimum0 and changes from - to +is +
Concave Up———is +
Concave Down———is -
Inflection Point———Changes Sign

It’s important to note that finding increasing/decreasing ranges uses the first derivative while finding concave up/down ranges uses the second derivative. In addition, a relative minimum or maximum cannot be an inflection point.

When working with analysis integral problems where you must find one of the above items in the table remember that you are working backwards. Instead of finding or drawing the derivative of the given numerical graph or function, use the information as the already-found derivative of your target integral. For example, if you are told that

G(x)=ax6n2dnG(x) =\int_a^x 6n^2dn

Realize that 6x26x^2 is equal to the first derivative and find critical points or the second derivative accordingly.


🤔 Practicing with Accumulation Functions

Almost all AP questions on this topic will fall into 1 of 3 types: graph, table, or equation. These are all equally likely to occur. We will be going over a graph problem!

📈 Graph Questions

Graph questions almost always tell you that a function G(x)=axf(t)dtG(x) =\int_a^x f(t)dt and give you a graph or image of the function f(t)f(t).

The most important thing to remember with this type of problem is that the graph given is not your target function. Instead, the graph is the derivative of your target function. Let’s try an official College Board FRQ from the 2022 exam! (All credits to College Board.)

Untitled

Courtesy of College Board

Let ff be a differentiable function with f(4)=3f(4)=3. On the interval 0x70 \leq x \leq 7, the graph of ff', the derivative of ff, consists of a semicircle and two line segments, as shown in the figure above.

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

(b) Find the x-coordinates of all points of inflection on the graph of ff for 0<x<70<x<7. Justify your answer.

(c) Let gg be the function defined by g(x)=f(x)xg(x)=f(x)-x. On what intervals, if any, is gg decreasing for 0x70 \leq x \leq 7? Show the analysis that leads to your answer.

(d) For the function gg defined in part (c), find the absolute minimum value on the interval 0x70 \leq x \leq 7. Justify your answer.

AP Calc 2022 #3a

Part A of this question asks us to find f(0)f(0) and f(5)f(5). Since we are told that the graph above is the graph of f(x)f'(x), this means that the original equation of…

f(x)=axf(x)dxf(x) = \int_a^x f'(x) dx

Using the Fundamental Theorem of Calculus, we know that this equation is equal to the accumulation function, and is represented on the f(x)f'(x) graph by the area between the curve and the x-axis. This area is the change in f(x)f(x) over the given bounds.

Using mathematical operations, we can set up 2 equations to find f(0)f(0) and f(5)f(5).

The first equation is

f(0)+04f(x)dx=f(4)f(0) + \int_0^4 f'(x) dx = f(4)

which can be simplified to

f(0)=f(4)04f(x)dxf(0) = f(4)- \int_0^4 f'(x) dx

We know that f(4)=3f(4) = 3 from the problem and that 04f(x)dx\int_0^4 f'(x) dx is the area under f(x)f'(x) between x = 0 and x = 4, given the definition of the accumulation function. Our problem tells us that f(x)f'(x) is composed of a semicircle and 2 line segments, and we can see that the area between 0 and 4 is the semicircle area. The area of this semicircle is

π2r2=π222=2π\frac \pi 2*r^2 =\frac \pi2* 2^2 = 2\pi

Remember that since this area is below the x-axis, it is considered negative!

Plugging this into the our equation above gives

f(0)=3(2π)=3+2πf(0) = 3- (-2\pi)=3+2\pi

In the same way, you can set up your equation for f(5)f(5).

f(4)+45f(x)dx=f(5)f(4) + \int_4^5f'(x)dx = f(5)

Again, we know that f(4)=3f(4) = 3. The area of f(x)f'(x) from x=4x=4 and x=5x=5 lies on the first straight line which has a slope of 2064=1\frac {2-0}{6-4} = 1, using the known points of (4,0) and (6,2). Using this information in a slope equation OR by looking at the graph, we know that when x=5x =5 then y=1y =1. This means that the triangle area between x=4x=4 and x=5x=5 has a height of 1 and a width of 1, with a total area of 12\frac 12. Putting this into our earlier equation, we get

f(5)=3+12=3.5f(5) = 3 + \frac 12=3.5

Remember to write down your integrals AND label your final answers as f(0) and f(5) to earn all points on the free-response portion!

AP Calc 2022 #3b

This part asks for the x-coordinates of any points of inflection of the function f(x)f(x). Since we are given the graph of f(x)f’(x), we need to look at the derivative of this graph to find points of inflection (2nd derivative) of the original function f(x)f(x).

Remember that points of inflection occur when the second derivative of a function changes signs. On the graph given to us, this occurs where the slope of the derivative changes signs. This occurs at x = 2, where the slope changes from negative to positive, and x = 6, where the slope changes from positive to negative. Since the slope of a function at a point is the 1st derivative, the slope of the first derivative at a point is the second derivative!

AP Calc 2022 #3c

Our new equation is defined by

g(x)=f(x)xg(x) =f(x) -x

so the new derivative is

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

A function is decreasing when its derivative is negative, so g(x) is decreasing when

0>f(x)10>f'(x)-1

or

1>f(x)1>f'(x)

According to the graph, this occurs between 0<x<50<x<5.

AP Calc 2022 #3d

This part asks for the absolute minimum function of g(x)g(x). A local minimum occurs when g(x)=0g(x) =0and g(x)g(x) changes from negative to positive. Our candidates are the two endpoints of x=0x=0 and x=7x=7 as well as the local minimum x=5x = 5. Just as a reminder, g(x)=f(x)1g'(x) = f'(x) -1 so g(x)=0g'(x) =0 when f(x)=1f'(x) =1 which only occurs within the function when x=5x = 5.

Testing all 3 candidates for the absolute minimum gives…

xf(x)g(x) = f(x) -x
03 + 2pi3 + 2pi
53.53.5 - 5 = -1.5
77.56.5 -7 =-0.5

f(0)f(0) and f(5)f(5) were both imported from Part A.

f(7)f(7) was found by adding f(5)+57f(x)dx=3.5+3=6.5f(5) + \int_5^7f'(x)dx =3.5 + 3 = 6.5 through geometry.

Untitled

Courtesy of College Board and Julianna Fontanilla

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


📝 Practice Tips

Here are some tips and tricks when approaching these question types:

  1. 💭 Understand the Relationship: If F(x)=axf(t)dtF(x)=\int_{a}^{x}f(t)dt, this means that F(x)=f(x)F’(x) = f(x) and F(x)=f(x)F’'(x) = f'(x).
  2. 📈 Use Graphical Information: Identify notable points on a graph, such as zeroes, minimums, and maximums. Always relate these back to the original integral equation.
  3. 🧠 Evaluate Extrema: Evaluate function values at critical points (zeroes) to determine relative and absolute extrema of the original integral. Don’t forget to also evaluate endpoints!
  4. Clarify the Question: Make sure you are looking for the zeroes of the correct derivative! It gets very easy to confuse F(x), F’(x), and F’’(x). Write Tip 1 somewhere on your paper.

This topic may be difficult, but the most important thing to remember is that area under a curve = the antiderivative. Practice using both concepts with different integral questions to get comfortable with the idea. As always, the best way to improve your understanding is to try lots of practice problems and familiarize yourself with the setup of common AP questions covering this topic. You’ve got this! 👍🏼