AP Calculus BC Unit 6 Integration and Accumulation of Change Questions

Practice AP Calculus BC Unit 6 questions on accumulation, Riemann sums, definite integrals, antiderivatives, and improper integrals.

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 6.11 Integrating Using Integration by Parts 6.12 Integrating Using Linear Partial Fractions 6.13 Evaluating Improper Integrals 6.14 Selecting Techniques for Antidifferentiation

[Maximum number: 4]

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

[ 2 ]

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.

[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.

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.

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: 2]

The figure above shows the graph of the piecewise-linear function f. For $-4 \leq x \leq 12$ , the function g is defined by $g(x)=\int_{2}^{x} f(t) d t$ .

For $-4 \leq x \leq 12$ , find all intervals for which $g(x) \leq 0$ .

[Maximum number: 7]

The graph of the differentiable function f, shown for $-6 \leq x \leq 7$ , has a horizontal tangent at x=-2 and is linear for $0 \leq x \leq 7$ . Let R be the region in the second quadrant bounded by the graph of f, the vertical line x=-6, and the x - and y-axes. Region R has area 12.

(a)

The function g is defined by $g(x)=\int_{0}^{x} f(t) d t$ . Find the values of g(-6), g(4), and g(6).

[ 3 ]

}

$g(-6)=\int_{0}^{-6} f(t) d t=-\int_{-6}^{0} f(t) d t=-12$

g(-6)

1 point

$g(4)=\int_{0}^{4} f(t) d t=\frac{1}{2} \cdot 4 \cdot 2=4$

g(4)

1 point

$g(6)=\int_{0}^{6} f(t) d t=\frac{1}{2} \cdot 4 \cdot 2-\frac{1}{2} \cdot 2 \cdot 1=3$

g(6)

1 point

Scoring notes:

- Supporting work is not required for any of these values. However, any supporting work that is

shown must be correct to earn the corresponding point.

- Special case: A response that explicitly presents $g(x)=\int_{-6}^{x} f(t) d t$ does not earn the first point it

would have otherwise earned. The response is eligible for all subsequent points for correct answers,

or for consistent answers with supporting work.

○ Note: $\int_{-6}^{-6} f(t) d t=0, \int_{-6}^{4} f(t) d t=16, \int_{-6}^{6} f(t) d t=15$

- Labeled values may be presented in any order. Unlabeled values are read from left to right and

from top to bottom as g(-6), g(4), and g(6), respectively. A response that presents only 1 or 2

values must label them to earn any points.

(b)

The function h is defined by $h(x)=\int_{-6}^{x} f^{\prime}(t) d t$ . Find the values of $h(6), h^{\prime}(6)$ , and $h^{\prime \prime}(6)$ . Show the work that leads to your answers.

Write your responses to this question only on the designated pages in the separate Free Response booklet. Write your solution to each part in the space provided for that part.

[ 4 ]

The function h is defined by $h(x)=\int_{-6}^{x} f^{\prime}(t) d t$ . Find the values of $h(6), h^{\prime}(6)$ , and $h^{\prime \prime}(6)$ . Show

the work that leads to your answers.

$h(6)=\int_{-6}^{6} f^{\prime}(t) d t=f(6)-f(-6)=-1-0.5=-1.5$

Uses Fundamental
Theorem of Calculus

1 point

h(6) with supporting
work

1 point

$h^{\prime}(x)=f^{\prime}(x)$ , so $h^{\prime}(6)=f^{\prime}(6)=-\frac{1}{2}$ .

$h^{\prime}(6)$

1 point

$h^{\prime \prime}(x)=f^{\prime \prime}(x)$ , so $h^{\prime \prime}(6)=f^{\prime \prime}(6)=0$ .

$h^{\prime \prime}(6)$

1 point

Scoring notes:

- Labeled values may be presented in any order.

- Unlabeled values are read from left to right and from top to bottom as $h(6), h^{\prime}(6)$ , and $h^{\prime \prime}(6)$ ,

respectively. A response that presents only 1 or 2 values must label them in order to earn any

points.

- A response of h(6)=-1.5 does not earn either of the first 2 points. A response of

h(6)=f(6)-f(-6) earns the first point but not yet the second point.

- A response of h(6)=-1-0.5 is the minimum work required to earn both of the first 2 points.

- To earn the third point a response must state either $h^{\prime}(x)=f^{\prime}(x)$ or $h^{\prime}(6)=f^{\prime}(6)$ , and provide

an answer of $-\frac{1}{2}$ .

- The fourth point is earned for a response of $h^{\prime \prime}(6)=0$ , with or without supporting work.

- A response that has one or more linkage errors does not earn the first point it would have otherwise

earned. For example, $h^{\prime}(x)=f^{\prime}(6)=-\frac{1}{2}$ does not earn the third point but is eligible for the fourth

point even in the presence of another linkage error, such as $h^{\prime \prime}(x)=f^{\prime \prime}(6)=0$ .

Total for part (c)

for question 4

9 points

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Covered syllabus points

Question 1

Question (a)

Question (b)

Question 2

Question 3

Question 4

Question (a)

Question (b)