5 Steps to a 5 AP Calculus BC 2019

Areas, Volumes, and Arc Lengths 297

29. The velocity function of a particle moving
    along thex-axis isv(t)=tcos(t^2 +1)
    fort≥0.

(a)If att=0, the particle is at the origin,

find the position of the particle att=2.

(b)Is the particle moving to the right or

left att=2?

(c)Find the acceleration of the particle

att=2 and determine if the velocity of

the particle is increasing or decreasing.

Explain why.

30. (Calculator) givenf(x)=xexand
    g(x)=cosx, find:

(a) the area of the region in the first

quadrant bounded by the graphs

f(x),g(x), andx=0.

(b) The volume obtained by revolving the

region in part (a) about thex-axis.

31. Find the slope of the tangent line to the
    curve defined byr=5 cos 2θat the point
    whereθ=
    3 π
    2

.

32.

∫

2

x^2 − 4 x

dx

33.

∫∞

e

dx

x

12.9 Solutions to Practice Problems

Part A The use of a calculator is not

allowed.

1. (a) F(0)=

∫ 0

0

f(t)dt= 0

F(3)=

∫ 3

0

f(t)dt

=

1

2

( 3 + 2 )( 4 )= 10

F(5)=

∫ 5

0

f(t)dt

=

∫ 3

0

f(t)dt+

∫ 5

3

f(t)dt

= 10 +(−4)= 6

(b) Since

∫ 5

3

f(t)dt≤0,Fis

decreasing on the interval [3, 5].

(c) Att=3,Fhas a maximum value.

(d) F′(x)= f(x),F′(x) is increasing on

(4, 5), which impliesF≤(x)>0.

ThusFis concave upward on (4, 5).

2. (See Figure 12.9-1.)

y

x

x = –1 x = 2

y = x^3

–1 0 2

Figure 12.9-1

A=

∣∣

∣∣

∫ 0

− 1

x^3 dx

∣∣

∣∣+

∫ 2

0

x^3 dx

=

∣

∣∣

∣∣

[

x^4

4

] 0

− 1

∣

∣∣

∣∣+

[

x^4

4

] 2

0

=

∣∣

∣∣

∣

0 −

(− 1 )^4

4

∣∣

∣∣

∣

+

(

24

4

− 0

)

=

1

4

+ 4 =

17

4