Unit IV Exam
Reviewer
Evaluate
Z
cot
x
(
cot
x
−
tan
x
)
dx
Z
cot
x
(
cot
x
−
tan
x
)
dx
=
Z
(
cot
2
x
−
1
)
dx
=
Z
(
csc
2
x
−
1
−
1
)
dx
=
Z
(
csc
2
x
−
2
)
dx
Evaluate
Z
x
5
(
x
3
−
1
)
8
dx
Let
u
=
x
3
−
1
. This implies that
(1)
du
=
3
x
2
dx
or
1
3
du
=
x
2
dx
and
(2)
x
3
=
u
+
1
.
Z
x
5
(
x
3
−
1
)
8
dx
=
Z
x
3
(
x
3
−
1
)
8
x
2
dx
=
Z
(
u
+
1
)
u
8
1
3
du
=
1
3
Z
(
u
9
+
u
8
)
du
=
1
3
u
10
10
+
u
9
9
+
C
=
1
3
(
x
3
−
1
)
10
10
+
(
x
3
−
1
)
9
9
Evaluate
Z
1
0
(
|
2
−
3
x
|
+
1
)
dx
We first take note that
|2
−
3
x
|
=
2
−
3
x x
≤
2
3
−2
+
3
x x
>
2
3
Z
1
0
(
|2
−
3
x
|
+
1
)
dx
=
Z
23
0
(
2
−
3
x
+
1
)
dx
+
Z
1
2 3
(
−2
+
3
x
+
1
)
dx
=
Z
23
0
(
3
−
3
x
)
dx
+
Z
1
2 3
(
−1
+
3
x
)
dx
=
3
x
−
3
x
2
2
2 30
+
−
x
+
3
x
2
2
1
2 3=
2
−
2
3
−
0
+
1
2
−
0
=
11
Evaluate
Z
π2/
16π2
/
36sec
2
√
x
√
x
dx
Let
u
=
√
x. This implies
du
=
1
2
√
x
dx. Also,
x
=
π
2
36
−→
u
=
π
6
and
x
=
π
2
16
−→
u
=
π
4
Z
π2/
16π2
/
36sec
2
√
x
√
x
dx
=
Z
π/
4π
/
62 sec
2
u du
=
2 tan
u
π
/
4π
/
6=
2 tan
π
4
−
2 tan
π
6
=
2
−
2
√
Find
b
such that the average value of
h
(
x
) =
3
x
2
on
[
0,
b
]
is equal to 1.
Since
h
(
x
) =
3
x
2
is continuous in
R
, then
h
is also continuous on
[
0,
b
]
.
By the MVT for integrals,
have
=
Z
b
0
3
x
2
dx
b
−
0
=
x
3
b
0
b
=
b
3
b
=
b
2
Given
G
(
x
) =
Z
tan
x
4 πx
cos
(t
−
1
)
t
dt
. Find:
(a)
G(
π
/4
)
(b)
G
0
(
π
/4
)
(a)
G
(
π
/4
) =
Z
1
1
cos
(
t
−
1
)
t
dt
=
0
(b)
G
(
x
) =
Z
1
4 π
x
cos
(
t
−
1
)
t
dt
+
Z
tan
x
1
cos
(
t
−
1
)
t
dt
G
(
x
) =
−
Z
4π
x
1
cos
(
t
−
1
)
t
dt
+
Z
tan
x
1
cos
(
t
−
1
)
t
dt
G
0
(
x
) =
−
cos
(
4x
π
−
1
)
4x
π
4
π
+
cos
(
tan
x
−
1
)
tan
x
sec
2
x
Elsa threw her crown upwards with an initial velocity of 24
ft
/
s
, from a
height of 40
ft
. From the same height at the same time, she released her
cape and just let it go. Assume that no other forces except acceleration due
gravity of -32
ft
/
s
2
affects the crown and cape.
1
What is the maximum height of the crown?
2
Which item is moving faster at time
t
=
1
. Explain your answer.
acrown
(
t
)
=
−32
vcrown
(
t
)
=
−32
t
+
C
1
vcrown
(
0
)
=
C
1
=
24
vcrown
(
t
)
=
−32
t
+
24
scrown
(
t
)
=
−16
t
2
+
24
t
+
C
2
scrown
(
0
)
=
C
2
=
40
scrown
(
t
)
=
−16
t
2
+
24
t
+
40
Elsa threw her crown upwards with an initial velocity of 24
ft
/
s
, from a
height of 40
ft
. From the same height at the same time, she released her
cape and just let it go. Assume that no other forces except acceleration due
gravity of -32
ft
/
s
2
affects the crown and cape.
1
What is the maximum height of the crown?
2
Which item is moving faster at time
t
=
1
. Explain your answer.
acape
(
t
)
=
−32
vcape
(
t
)
=
−32
t
+
C
1
vcape
(
0
)
=
C
1
=
0
vcape
(
t
)
=
−32
t
scape
(
t
)
=
−16
t
2
+
C
2
scape
(
0
)
=
C
2
=
40
scape
(
t
)
=
−
16
t
2
+
40
x
y
R
(
−3, 0
)
(
−1, 0
)
(
1, 4
)
y
=
2
x
3
+
2
y
=
x
+
3
Let
R
be the region bounded by the
graphs
y
=
2
x
3
+
2
,
y
=
x
+
3
and
the
x-axis. Set-up the integral
1.
the definite integral for the area
of
R
2.
the definite integral for the
perimeter of
R
3.
the definite integral for the
volume of the solid generated
when
R
is revolved about
x
y
R
(
−3, 0
)
(
−1, 0
)
(
1, 4
)
y
=
2
x
3
+
2
y
=
x
+
3
Area of
R
: The best method uses
horizontal strips. So our integral is
in terms of
y.
A
R
=
Z
4
0
3
r
y
−
2
2
−
(
y
−
3
)
!
dy
If we use vertical strips,
A
R
=
Z
−
1
−
3
((
x
+
3
)
−
0
)
dx
+
Z
1
−
1
x
y
R
(
−3, 0
)
(
−1, 0
)
(
1, 4
)
y
=
2
x
3
+
2
y
=
x
+
3
Perimeter of
R
: We set-up with
respect to
x.
L
R
= (
−1
−
(
−3
))
+
Z
1
−
3
q
1
+ (
1
)
2
dx
+
Z
1
−
1
q
x
y
y
=
0
R
(
−3, 0
)
(
−1, 0
)
(
1, 4
)
y
=
2
x
3
+
2
y
=
x
+
3
Volume of the SOR when
R
is
revolved about
y
=
0
using
cylindrical shells
V
=
2
π
r h w
V
R
=
Z
4
0
2π
y
3
r
y
−
2
2
−
(
y
−
3
)
!
x
y
x
=
1
R
(
−3, 0
)
(
−1, 0
)
(
1, 4
)
y
=
2
x
3
+
2
y
=
x
+
3
Volume of the SOR when
R
is
revolved about
x
=
1
using washers
V
=
π
(
r
2
2
−
r
2
1
)
h
V
R
=
Z
4
0
π
1
−
(
y
−
3
)
2
−
1
−
q
3y
−
2
2
2
#
More items for review
I.
Evaluate the following integrals.
1.
Z
x
5
(
x
3
+
1
)
4
dx
2.
Z
sin
x
(
1
+
sin
x
cos
x
)
dx
3.
Z
x
csc
2
(
tan
x
2
)
cos
2
(
x
2
)
dx
4.
Z
0
π
/
3sec
x
tan
x
√
32
−
sec
x dx
II.
Let
R
be the region bounded by the graphs of
x
=
1
−
y
2
,
x
+
y
+
1
=
0
, and the
x-axis. Set-up the definite integral(s) for
1
the area of
R
.
2the perimeter of
R
.