Areas, Volumes, and Arc Lengths 297
- 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.
- (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.
- 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.
- (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).
- (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
)