Math 1100-6
December 15, 2008
Page 1
Math 1100: Assignment #7
Due: Friday Dec 12, 2008 @ 10:30am
No late assignments will be accepted.
Please complete the following questions for the assignment. Please be mindful of “good
presentation” criteria outlined in the syllabus as you’re preparing to hand it in.
§13.2
17.
Z
4√
4x + 9 dx =
0
=
=
=
¯
3 ¯4
2
2
(4x + 9) ¯
12
0
³
´
3
3
1
(25) 2 − 9 2
6
1
(125 − 27)
6
49
3
22.
Z
2
e
4x
dx =
0
=
1 4x ¯¯2
e ¯
4
0
¢
1¡ 8
e −1
4
26.
Z
0
1
3x3
dx =
4x4 + 9
=
=
¯1
3
¯
ln(4x4 + 9)¯
16
0
3
(ln(13) − ln(9))
16
µ ¶
3
13
· ln
16
9
38. Note that y = x2 + 3x + 2 = (x + 2)(x + 1), so y < 0 for −2 ≤ x ≤ −1, so y > 0 in the
domain of integration (i.e. we don’t need to worry about correcting for regions where the
function is negative).
¶¯
1 3 3 2
¯3
A=
(x + 3x + 2) dx =
x + x + 2x ¯
3
2
−1
−1
27
1 3
= 9+
+6+ − +2
2
3 2
88
=
3
Z
3
µ
2
Math 1100-6
December 15, 2008
44.
Z
Z
1
3
0
x dx =
−1
Z
3
1
x dx +
−1
x3 dx
0
1 1
= − + =0
4 4
46.
Z
Z
2
2
2
6(2x − x ) dx = 6
1
(2x − x2 ) dx
1
2
= 6 =4
3
49. a) The depreciation is the area of the triangle + the area of the square, i.e.
1
· 10 · 30000 + 10 · 30000
2
= $450, 000
D =
b) Using the definite integral, we have
Z
10
0
µ
¶
1 2 ¯¯10
3000(20 − t) dt = 3000 20t − t ¯
2
0
= 3000 (200 − 50 − 0 + 0)
= $450, 000
62. We want to calculate:
Z
10
0
¢ ¯¯10
1 ¡ 3
20t − t4 ¯
16875
0
1
104
=
16875
= 59.25%
60t2 − 4t3
dt =
16875
§13.3
6.
Z
1
−1
¶¯
1 4 1 2
¯1
x + x + 2x ¯
4
2
−1
1 1
1 1
=
+ +2− − +2
4 2
4 2
= 4
¡ 3
¢
x + x + 2 dx =
µ
Page 2
Math 1100-6
December 15, 2008
Page 3
13. Note that f (x) > g(x) in the domain of integration, therefore, the area is
µ
¶¯
Z 2
2 3
¯2
2
A=
(2x + 2) dx =
x + 2x ¯
3
0
0
28
=
3
16. We need to find the point where the graphs intersect. To this end, we set:
x2 − 2x + 1 = x2 − 5x + 4
→
3x − 3 = 0
→
x=1
Therefore, the area is:
Z 2
(3x − 3) dx =
¯2
3 2
¯
x − 3x¯
2
1
3
= 6−6− +3
2
7
=
2
1
23. We need to find when f (x) = g(x) to find the domain of integration. To this end, we set
→
3
= 4−x
x
0 = x2 − 4x + 3 = (x − 3)(x − 1)
Therefore, the area is:
¶
µ
¶¯
Z 3µ
3
1 2
¯3
4−x−
dx =
4x − x − 3 ln(x) ¯
x
2
1
0
1
9
= 12 − − 3 ln(3) − 4 + − 0
2
2
= 4 − 3 ln(3)
√
25. Note that y = x + 3 = 2 when x = 1, so the area is:
¶¯
µ
Z 1
√
¡
¢
3
2
¯1
2 − x + 3 dx =
2x − (x + 3) 2 ¯
3
−3
−3
16
= 2−
+6+0
3
8
=
3
28.
1
2−0
Z
0
2
µ
¶
1 3 ¯¯2
2
(2x − x ) dx =
x − x ¯
3
0
8
= 4−
3
4
=
3
2
Math 1100-6
December 15, 2008
Page 4
38. The equation for the inventory in the first 3 months is y = −400x + 1300. Therefore, the
average inventory for the first 3 months is:
1
3−0
Z
¢ ¯¯3
1¡
−200x2 + 1300x ¯
3
0
= 700
3
(−400x + 1300) dx =
0
§13.4
20. At the equilibrium quantity, the price is p = $4. Therefore, the consumer surplus is:
Z
0
12
¯12
100
¯
dx − 4 · 12 = 50 ln(2x + 1)¯ − 48
1 + 2x
0
= 250 ln(5) − 48
= $354.36
26. The profit function is:
P (x) = px − C(x)
1
= − x3 − 2x2 + 30x − 500 − 2x2 − 10x
3
1 3
= − x − 4x2 + 20x − 500
3
We now want to maximize the profit. To this end, set
dP
= 0 = x2 + 8x − 20
dx
√
−8 ± 64 + 40
→
x =
2
= −4 ± 6
= 2
The price at this level is:
1
74
p = − 22 − 2 · 2 + 30 =
3
3
The consumer surplus at x = 2 is:
Z
2
1
74
(− x2 − 2x + 30) dx −
·2
3
3
µ0
¶¯
1
¯2 148
=
− x3 − x2 + 30x ¯ −
9
3
0
52
=
9
CS =
Math 1100-6
December 15, 2008
Page 5
28. First we need to find the equilibrium quantity by setting:
p = 36 = 0.1x2 + 3x + 20
→
0 = x2 + 30x − 160
√
x = −15 + 385
So, the producers surplus is:
√
Z −15+ 385
√
¡
¢
P S = (−15 + 385)36 −
0.1x2 + 3x + 20 dx
√
µ0
¶¯
√
0.1 3 3 2
¯−15+ 385
= (−15 + 385)36 −
x + x + 20x ¯
3
2
0
= $38.62
32. We are given that the supply and demand functions are:
ps = 4x + 4
pd = 49 − x2
First we need to know the equilibrium price and quantity, so we set:
ps = 4x + 4 = 49 − x2 = pd
0 = x2 + 4x − 44
√
−4 ± 16 + 4 · 44
x =
√2
→
x = −2 + 4 3 = 4.93
√
ps = pd = 4(−1 + 4 3) = 23.71
→
→
Therefore, the producer surplus is:
Z 4.93
P S = 4.93 · 23.71 − 4
(x + 1) dx
0
= $48.57
36. We are given that the demand and supply functions are:
ps = 160 + 4x + x2
pd = 280 − 4x − x2
To find the equilibrium, we need to find when ps − pd = 0
0 = x2 + 4x − 60 = (x + 10)(x − 6)
This means that
ps = pd = 160 + 24 + 36 = 220
So, we find the producer surplus is:
Z 6
P S = 220 · 6 −
(160 + 4x + x2 ) dx
0
= 216
→
x=6
Math 1100-6
December 15, 2008
§13.7
4.
Z
∞
5
¯b
1
1
−4 ¯
dx
=
−
lim
(x
−
1)
¯
(x − 1)5
4 b→∞
5
£
¤
1
= − lim (b − 1)−4 − 4−4
4 b→∞
1
=
1024
10.
Z
∞
1
2
1 x2 ¯¯b
e ¯
b→∞ 2
1
³
´
1
2
= lim
eb − e
b→∞ 2
= ∞
x ex dx =
lim
18.
Z
∞
−∞
¯b
1
¯
¯
a→−∞ b→∞ 2(x2 + 1) a
¶
µ
1
1
+ lim
= lim − 2
2
a→−∞ 2(a + 1)
b→∞
2(b + 1)
= 0+0=0
x
dx =
(x2 + 1)2
lim
lim −
§14.2
1.
z = x4 − 5x2 + 6x − 5y + 7
∂z
= 4x3 − 10x + 6
∂x
∂z
= −5
∂y
5.
f (x, y) = (x3 + 2y 2 )3
∂f
= 3(x3 + 2y 2 )2 (3x2 )
∂x
∂f
= 3(x3 + 2y 2 )2 (4y)
∂y
Page 6
Math 1100-6
December 15, 2008
Page 7
15.
f (x, y) = ln(xy + 1)
∂f
y
=
∂x
xy + 1
∂f
x
=
∂y
xy + 1
19.
z = 5x3 − 4xy
∂z
= 15x2 − 4y
∂x
→
(x, y, z) = (1, 2, −3)
∂z
(1, 2, −3) = 15 − 8 = 7
∂x
20.
z = 5x3 − 4xy
∂z
= −4x
∂y
→
(x, y, z) = (1, 2, −3)
∂z
(1, 2, −3) = −4
∂y
37.
f (x, y) =
∂f
∂x
∂2f
∂x2
→
→
∂ 2 f ¯¯
¯
∂x2 (x,y)=(−1,4)
∂f
∂y
∂2f
∂y 2
∂ 2 f ¯¯
¯
∂y 2 (x,y)=(−1,4)
=
=
=
=
=
=
2x
x2 + y 2
2(x2 + y 2 ) − 2x(2x)
x2 − y 2
=
−2
(x2 + y 2 )2
(x2 + y 2 )2
·
¸
(2x)(x2 + y 2 )2 − (x2 − y 2 )(4x)(x2 + y 2 )
4x(3y 2 − x2 )
−2
=
−
(x2 + y 2 )4
(x2 + y 2 )3
4(11)
88
=−
(1 + 4)3
4913
4xy
− 2
(x + y 2 )2
£ 2
¤
−4x
−4x
(x3 − 3y 2 )
(x + y 2 )2 − y(4y)(x2 + y 2 ) = 2
2
2
4
(x + y )
(x + y 2 )3
88
4913