MAT1322, Final exam Practice sheet:
Integration
Area, Volumes, Applications of Integrals
Compute the area of the following regions:
1. Bounded region delimited by
y
=
e
x
, y
= 1
, x
= 2.
2. Bounded region delimited by
y
= sin(
x
)
, y
= cos(
x
)
, x
= 0
, y
= 0
, x
=
π
2
.
3. Bounded region delimited by
y
= 3
x

6
, y
= 9
/x, x
= 4.
Solution :
1. This region is from
x
= 0 to
x
= 2. Throughout this interval, the curve
y
=
e
x
lies above
y
= 1. Thus, the area
A
is
A
=
2
0
(
e
x

1)
dx
=
e
x

x
2
0
=
e
2

2

(
e
0

0) =
e
2

3
2. This region consists of two parts: one from
x
= 0 to
x
=
π
4
in which
y
= cos(
x
) lies above
y
= sin(
x
), and the other from
x
=
π
4
to
x
=
π
2
in which
y
= sin(
x
) lies above
y
= cos(
x
).
Thus, the total area of these two regions is
A
=
π/
4
0
(cos(
x
)

sin(
x
))
dx
+
π/
2
π/
4
(sin(
x
)

cos(
x
))
dx
=
sin(
x
) + cos(
x
)
π/
4
0
+

cos(
x
)

sin(
x
)
π/
2
π/
4
= (sin(
π/
4) + cos(
π/
4))

(sin(0) + cos(0)) + (

cos(
π/
2)

sin(
π/
2))

(

cos(
π/
4)

sin(
π/
4))
=
√
2
2
+
√
2
2

(0 + 1) + (0

1)


√
2
2

√
2
2
= 2
√
2

2
3. This region spans the interval from
x
= 3 (the positive point of intersection of
y
= 3
x

6
and
y
= 9
/x
) to
x
= 4, and througout this interval, the line
y
= 3
x

6 lies above the curve
y
= 9
/x
. Thus, the area
A
is
A
=
4
3
(
3
x

6

9
x
)
dx
=
3
x
2
2

6
x
+
9
x
2
4
3
=
(
3
·
16
2

6(4) +
9
16
)

(
3
·
9
2

6(3) +
9
9
)
=
65
16
Compute the volume of the following solids:
1.
y
=
e
x
, y
= 1
, x
= 2 rotated around
y
=

2.
2.
y
= sin(
x
)
, y
= cos(
x
)
, x
= 0
, y
= 0
, x
=
√
2
2
about
y
=

1.
1
3.
y
= 3
x

6
, y
= 9
/x, y
= 0
, x
= 4 about
x
=

2.
Solution :
1.
133
.
0630975
2.
4
.
126940766
3. We use the method of cylindrical shells.
The functions
y
1
(
x
) = 3
x

6 and
y
2
(
x
) = 9
/x
intersect at

1 and 3, however we only need the positive intersect. The function
y
1
crosses
the
x
axis at
x
= 2 (draw a sketch). The volume has to be compute in two parts: using
y
1
from
x
= 2 to
x
= 3 and
y
2
from
x
= 3 to
x
= 4. We rotate around the axis
x
=

2, hence
the radius for the method of cylindrical shells becomes
r
(
x
) = (
x

(

2)) =
x
+ 2. Volume 1:
V
1
= 2
π
3
2
(
x
+ 2)(3
x

6)
dx
= 2
π
3
2
3
x
2

12
dx
= 14
π
= 43
.
98229716
.
Volume 2:
V
2
= 2
π
4
3
(
x
+ 2)
·
9
x
dx
= 89
.
08474367
.
So the total volume is
V
=
V
1
+
V
2
= 133
.
0670408.
Applications of integrals:
1. A cable that weighs 2
kg/m
is used to lift 800
kg
of coal up to a mine shaft which is 500
m
deep. Find the work done.
Solution :
The total work is the sum of the work from lifting the coal and the cable. We
denote by
x
the depth from the top of the mine.
The mass of a tiny segment of the cable of length Δ
x
is 2Δ
x
kg. Hence its weight is 2
g
Δ
x
N, where
g
is the gravitational constant.
Also the work required to lift the segment of cable of length Δ
x
to the height
x
is
Δ
W
= (2
g
Δ
x
)
×
x
J
.
Finally, the work required to lift the cable is
W
cable
=
500
0
2
gxdx
= 2
g
×
1
2
x
2
500
0
=
g
×
500
2
2
,
45
×
10
6
J
by using
g
= 9
.
81 m/s
2
.
The work to lift the coal is just
W
coal
= 800
×
500 = 400000 J
(since the force of the bag of coal does not change with the depth).
The total work is hence
W
=
W
cable
+
W
coal
2
.
85
×
10
6
J
.
2