AP Calculus BC Unit 6: Integration and Accumulation

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 ]

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.

[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 BC

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: 2]

Figure for Question Unit 6: Integration and Accumulation of Change question 3 — AP Calculus BC

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 $g(x) \leq 0$ .

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Covered syllabus points

Unit 6: Integration and Accumulation of Change question 1

Question (a)

Question (b)

Unit 6: Integration and Accumulation of Change question 2

Unit 6: Integration and Accumulation of Change question 3