AP Calculus AB Unit 6: Integration

Explore AP Calculus Unit 6 questions on approximating, representing, and evaluating definite integrals and connecting integration with derivatives.

Covered syllabus points

6.1 Exploring Accumulations of Change 6.2 Approximating Areas with Riemann Sums 6.3 Riemann Sums, Summation Notation, and Definite Integral Notation 6.4 The Fundamental Theorem of Calculus and Accumulation Functions 6.5 Interpreting the Behavior of Accumulation Functions Involving Area 6.6 Applying Properties of Definite Integrals 6.7 The Fundamental Theorem of Calculus and Definite Integrals 6.8 Finding Antiderivatives and Indefinite Integrals: Basic Rules and Notation 6.9 Integrating Using Substitution 6.10 Integrating Functions Using Long Division and Completing the Square 6.14 Selecting Techniques for Antidifferentiation

[Maximum number: 1]

The graph of the function f shown above consists of two line segments and a semicircle. Let g be defined by $g(x)=\int_{0}^{x} f(t) d t$ . What is the value of g(5) ?

$-1.5+2 \pi$

$2 \pi$

$1.5+2 \pi$

$4.5+2 \pi$

[Maximum number: 2]

A student starts reading a book at time t=0 minutes and continues reading for the next 10

minutes. The rate at which the student reads is modeled by the differentiable function R, where

R(t) is measured in words per minute. Selected values of R(t) are given in the table shown.

Table for Question Unit 6: Integration and Accumulation of Change question 2 — AP Calculus AB

Use a trapezoidal sum with the three subintervals indicated by the data in the table to

approximate the value of $\int_{0}^{10} R(t) d t$ . Show the work that leads to your answer.

C Use a trapezoidal sum with the three subintervals indicated by the data in the table to approximate the

value of $\int_{0}^{10} R(t) d t$ . Show the work that leads to your answer.

$\begin{aligned} & \int_{0}^{10} R(t) d t & \approx \frac{R(0)+R(2)}{2}(2-0)+\frac{R(2)+R(8)}{2}(8-2) & +\frac{R(8)+R(10)}{2}(10-8) \end{aligned}$	Form of trapezoidal sum	Point 5 (P5)
$\begin{aligned} & =\frac{90+100}{2}(2-0)+\frac{100+150}{2}(8-2)+\frac{150+162}{2}(10-8) & =\frac{190}{2}(2)+\frac{250}{2}(6)+\frac{312}{2}(2)=190+750+312=1252 \end{aligned}$	Answer with supporting work	Point 6 (P6)
Scoring Notes for Part C

- Read "=" as " $\approx$ " for P5.

- The form of a trapezoidal sum includes three terms, each of which includes a product of two factors,

where one of the factors incorporates the $\frac{1}{2}$ as part of the product. To earn P5, at least five of the six

factors must be correct. If any of the six factors is incorrect, the response does not earn P6. Consider

the following examples:

○ $\frac{90+100}{2}(2-0)+\frac{100+150}{2}(8-2)+\frac{150+162}{2}(10-8)$ earns P5 and is sufficient to earn P6.

○ $\frac{190}{2}(2)+\frac{250}{2}(6)+\frac{312}{2}(2)$ earns P5 and is sufficient to earn P6.

$\circ \quad \frac{1}{2}((R(0)+R(2))(2)+(R(2)+R(8))(6)+(R(8)+R(10))(2))$ earns P5 and is eligible for P6.

○ $\frac{90+100}{2}(2)+\frac{100+150}{2}(2)+\frac{150+162}{2}(2)$ earns P5 but is not eligible for P6.

(Note that the factor of 2 in the second term of this expression is incorrect.)

- Special case: A response of (90+100)+(100+150) 3+(150+162) earns both P5 and P6.

- To be eligible for P6, a response must have earned P5.

Special case: A response of $95 \cdot 2+125 \cdot 6+156 \cdot 2$ earns P6 but does not earn P5.

- A response of $\frac{90+100}{2}(2-0)+\frac{100+150}{2}(8-2)+\frac{150+162}{2}(10-8)$ or equivalent banks P6

(i.e., subsequent errors in simplification will not be considered in scoring for P6).

- A response of $\frac{(90 \cdot 2+100 \cdot 6+150 \cdot 2)+(100 \cdot 2+150 \cdot 6+162 \cdot 2)}{2}$ or equivalent earns both P5

and P6. (Note that the average of the left Riemann sum and right Riemann sum is equivalent to the

trapezoidal sum.)

- A completely correct left Riemann sum (e.g., $90 \cdot 2+100 \cdot 6+150 \cdot 2=1080$ ) or a completely

correct right Riemann sum (e.g., $100 \cdot 2+150 \cdot 6+162 \cdot 2=1424$ ) earns P5 but does not earn P6.

[Maximum number: 1]

For a certain continuous function f, the right Riemann sum approximation of $\int_{0}^{2} f(x) d x$ with n subintervals of equal length is $\frac{2(n+1)(3 n+2)}{n^{2}}$ for all n. What is the value of $\int_{0}^{2} f(x) d x$ ?

$Graph of $f$$

Graph of $f$

[Maximum number: 6]

The continuous function f is defined on the closed interval $-6 \leq x \leq 12$ . The graph of f,

consisting of two semicircles and one line segment, is shown in the figure.

$Graph of $f$$

Graph of $f$

Let g be the function defined by $g(x)=\int_{6}^{x} f(t) d t$ .

(a)

Find $g^{\prime}(8)$ . Give a reason for your answer.

[ 2 ]


$g^{\prime}(x)=f(x)$	Considers $g^{\prime}(x)=f(x)$	Point 1 (P1)
$g^{\prime}(8)=f(8)=1$	Answer	Point 2 (P2)
Scoring Notes for Part A
- A response that does not earn P1 can earn P2 with an implied application of the Fundamental Theorem of Calculus (e.g., $g^{\prime}(8)=1$ or f(8)=1 ).

(b)

Find all values of x in the open interval -6<x<12 at which the graph of g has a point of

inflection. Give a reason for your answer.

[ 2 ]

B Find all values of x in the open interval -6<x<12 at which the graph of g has a point of inflection.

Give a reason for your answer.

The graph of g has a point of inflection where $g^{\prime \prime}=f^{\prime}$ changes sign, which is where $g^{\prime}=f$ changes from decreasing to increasing or vice versa.	Answer	Point 3 (P3)
The graph of g has points of inflection at x=-3 and x=6 because f changes from decreasing to increasing there.
The graph of g also has a point of inflection at x=3 because f changes from increasing to decreasing there.

Scoring Notes for Part B

- P3 is earned only for an answer of x=-3, x=3, and x=6. If any other/additional values of x in

-6<x<12 are declared to be points of inflection, the response does not earn either P3 or P4.

Consideration of x=-6 or of x=12 does not impact scoring.

- To earn P4, a response must tie the reason to the given graph of f.

○ A response of " g has a point of inflection at x=-3, x=3, and x=6 because f changes from

increasing to decreasing or decreasing to increasing there" earns both P3 and P4.

○ A response of " g has a point of inflection at x=-3, x=3, and x=6 because the slope of f

changes sign there" earns both P3 and P4.

○ A response of " g has a point of inflection at x=-3, x=3, and x=6 because f attains relative

extrema there" earns both P3 and P4.

○ A response of " g has a point of inflection at x=-3, x=3, and x=6 because g changes

concavity there" earns P3 but not P4.

○ A response of " g has a point of inflection at x=-3, x=3, and x=6 because $g^{\prime \prime}=f^{\prime}$

changes sign there" earns P3 but not P4.

○ A response that relies upon an ambiguous term such as "the function" or "the graph" does not

earn P4.

- Special case: A response with two of the three correct x-values with correct reasoning and no

other/additional values of x declared to be points of inflection earns P4 but not P3.

(c)

Find g(12) and g(0). Label your answers.

[ 2 ]

C Find g(12) and g(0). Label your answers.

$g(12)=\int_{6}^{12} f(t) d t=\frac{1}{2} \cdot 6 \cdot 3=9$	g(12)	Point 5 (P5)
$g(0)=\int_{6}^{0} f(x) d x=-\int_{0}^{6} f(x) d x=-\frac{\pi}{2} 3^{2}=-\frac{9 \pi}{2}$	g(0)	Point 6 (P6)

Scoring Notes for Part C

- Unlabeled values do not earn either P5 or P6.

- P5 is earned for a response of g(12)=9, with or without supporting work.

- P6 is earned for a response of $g(0)=-\frac{9 \pi}{2}$ , with or without supporting work.

Note: Incorrect communication between the label " g(0) " and the answer will be treated as scratch

work and will not impact scoring. For example, $g(0)=\int_{0}^{6} f(x) d x=-\frac{9 \pi}{2}$ earns P6.

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Covered syllabus points

Unit 6: Integration and Accumulation of Change question 1

Unit 6: Integration and Accumulation of Change question 2

Unit 6: Integration and Accumulation of Change question 3

Unit 6: Integration and Accumulation of Change question 4

Question (a)

Question (b)

Question (c)