Math 10260 Exam 1  Solutions  Spring 2006
1. Let us set 1980 to stand for
t
= 0
.
The total amount of oil consumed during 1980 to 2000 =
The area under the graph
≈
[
r
(5) +
r
(15)] Δ
t
= (6 + 8)
·
10 = 140
.
2. Using integration by substitution
u
= ln
x
. Then
du
=
1
x
dx
,
2
1
ln
x
x
dx
=
ln 2
0
udu
.
3. Total change of profit =
30
20
[

0
.
5
q
+ 100]
dq
=

0
.
5
2
[
q
2
+ 100
q
]
30
20
=

0
.
5
2
(30)
2
+ 100(30)


0
.
5
2
(20)
2
+ 100(20)
= 875
.
4. CS = [ the area under the demand curve from 0 to
q
e
] –
q
e
·
p
e
= the area of the triangle =
1
2
·
10
·
2 = 10
.
PS =
q
e
·
p
e
– [ the area under the supply curve from 0 to
q
e
] = 3
·
10  18 = 12.
5. Since
g
(
x
)
≥
f
(
x
) on the interval [

2
,
1], then the area of the region between the curves of
f
(
x
) and
g
(
x
) =
1

2
[
g
(
x
)

f
(
x
)]
dx
=
1

2
[2

x

x
2
]
dx
6. Selling price = present value (PV) =
8
0
84000
e

0
.
06
t
dt
=
84000

0
.
06
e

0
.
06
t

8
0
=
84000

0
.
06
[
e

0
.
06
·
8

1] = $533,703.
7. The rate of change of
M
=
dM
dt
= rate due to interest – rate due to payment
= 0
.
06
M

9
,
600
.
Being able to pay off the loan in 20 years means that;
M
(20) = 0
.
8.
dp
dt
= 0
.
02
p

0
.
004
p
2
= 0
.
02
p
(1

0
.
004
0
.
02
p
) = 0
.
02
p
(1

0
.
2
p
) = 0
.
02
p
(1

1
5
p
)
.
Therefore
K
= 5
millions.