Math 1100-6
November 7, 2008
Page 1
Math 1100: Assignment #6 SOLUTIONS
Due: Wednesday Nov 19, 2008 @ 10:30am
No late assignments will be accepted. Note that this assignment is due on
Wednesday, not Friday.
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.
§12.2
3.
Z
(x2 + 3)4
+C
4
(x2 + 3)3 (2x) dx =
4.
Z
(3x3 + 1)4 (9x2 ) dx =
(3x3 + 1)5
+C
5
(8x4 + 5)3 (32x3 ) dx =
(8x4 + 5)4
+C
4
6.
Z
8.
Z
(4x2 − 3x)4 (8x − 3) dx =
(4x2 − 3x)5
+C
5
12.
Z
3(5 − x)−2
+C
2
3(5 − x)−3 dx =
20.
Z
p
3
3x 5 − x2 dx = −(5 − x2 ) 2 + C
28.
Z
5x3
dx =
(x4 − 8)3
Z
5x3 (x4 − 8)−3 dx = −
8(x4
5
+C
− 8)2
32.
Z
x2 + 1
√
dx =
x3 + 3x + 10
2
3
Z
1
3x3 + 3
√
dx
2 x3 + 3x + 10
Math 1100-6
November 7, 2008
Page 2
du
= 3x3 + 3, therefore, doing u-substitution, we get:
dx
Z
Z
2
1
3x3 + 3
3
1 1
√
√ du
dx =
3
2 x3 + 3x + 10
2
2 u
3√
=
u+C
2
p
3
=
x3 + 3x + 10 + C
2
Let u = x3 + 3x + 10 and note that
44.
40000
(10 + x)2
Z
40000
R = M R dx = 60000x +
10 + x
M R = 60000 −
→
(Note that there is no “+C“ in the above equation because we require R(0) = 0, which
means that C = 0).
48. We are given:
dN
dt
=
1
√
2 t+1
&
N (3) = 2
Solving for N (t) gives:
N (t) =
√
t+1+C
We require N (3) = 2, so
√
3+1+C = 2
→
C=0
√
√
Therefore, N (t) = t + 1, so N (8) = 9 = 3, so 3 units are created after 8 hours.
N (3) =
§12.3
2.
Z
4 e4x dx =
e4x + C
6.
Z
1600 e0.4x dx = 4000 e0.4x + C
12.
Z
4
e1−2x
Z
dx =
4 e2x−1 dx = 2 e2x−1 + C
Math 1100-6
November 7, 2008
Page 3
14.
Z
x3
dx =
e4x4
Z
4
x3 e−4x dx
= −
1 −4x4
+C
e
16
16.
¶
Z µ
Z ³
´
x
5
2
3x2
x e − x dx =
x e3x − 5 e− 3 dx
e3
x
1 3x2
=
e + 15 e− 3 + C
6
22.
Z
x2
dx =
x3 − 9
=
Z
3x2
1
dx
3
x3 − 9
¢
1 ¡ 3
ln x − 9 + C
3
26.
Z
¡ 4
¢
4x3 + 2x
2
dx
=
ln
x
+
x
+C
x4 + x2
30.
Z
¶
Z µ
2x + 3
2
x +
dx
2x + 1
¶
Z µ
2
=
x2 + 1 +
dx
2x + 1
1 3
=
x + x + ln(2x + 1) + C
3
2x3 + x2 + 2x + 3
dx =
2x + 1
32.
Z
x4 − 2x2 + x
dx =
x2 − 2
=
Z µ
x2 +
x
2
x −2
¶
dx
1 3 1
x + ln(x2 − 2) + C
3
2
44.
C 0 (t) =
→
C(t) =
Set C(0) = 0
0 =
∴ Const. =
C(t) =
´
c ³ −bt
b e − a e−at
b−a³
´
c
e−at − e−bt + Const
b−a
c
(1 − 1) + Const
b−a
0
´
c ³ −at
e
− e−bt
b−a
Math 1100-6
November 7, 2008
Page 4
52. a)
S 0 (t) = −147.78 e−0.2t
S(t) = 738.9 e−0.2t + C
→
Set S(0) = 7389 = 738.9 + C
C = 6650.1
→
S(t) = 738.9 e−0.2t + 6650.1
b)
S(10) = 738.9 e−2 + 6650.1
= 6750.10
§12.4
1. We want to solve for C(x) given:
M C = 2x + 100
Z
→
C(0) = 200
M C dx = x2 + 100x + c
C(t) =
Set C(0) = 200 = c
C(t) = x2 + 100x + 200
→
4. We want to solve for C(x) given:
M C = 3x + 50
C(20) = 2000
3 2
C(t) = M C dx =
x + 50x + c
2
Set C(20) = 200 = 600 + 1000 + c
Z
→
→
→
c = −1400
3 2
C(t) =
x + 50x − 1400
2
7. We are given:
M C = 3x + 20
C(80) = 11400
M R = 44 − 5x
a) The optimal level of production occurs when M R − M C = 0, i.e. when
→
0 = 44 − 5x − 3x − 20
24
x =
=3
8
so x = 3 units need to be sold to optimize production.
Math 1100-6
November 7, 2008
Page 5
b) Note that:
Z
5
M R dx = 44x − x2
2
Z
3 2
C(x) = M C dx =
x + 20x + c
2
Set C(80) = 11400 = 9600 + 1600 + c
R(x) =
→
∴
c = 200
5
3
P = R − C = 44x − x2 − x2 − 20x − 200
2
2
= 24x − 4x2 − 200
c) The profit at the maximum level of production is: P (3) = 72 − 36 − 200 = −164.
10.
√
M C = 60 x + 1
→
C(0) = 340
3
2
C(x) = 40(x + 1) + c
Set C(0) = 340 = 40 + c
⇒
c = 300
3
2
C(x) = 40(x + 1) + 300
M R = 80x
R(x) = 40x2
3
P (x) = R(x) − C(x) = 40x2 − 40(x + 1) 2 − 300
⇒
P (3) = 360 − 320 − 300 = −260
P (8) = 2560 − 1080 − 300 = 1180
17.
dC
0.2
= 0.3 + √
C(0) = 8
dy
y
√
C(y) = 0.3y + 0.4 y + c
→
Set C(0) = 8 = 0 + 0 + c
√
C(y) = 0.3y + 0.4 y + 8
⇒
26.
→
dS
dy
dC
dy
= 0.2 + e−1.5y = 1 −
= 0.8 − e−1.5y
C(y) = 0.8y +
Set C(0) = 3 =
c =
⇒
2
+c
3
7
3
C(y) = 0.8y +
2 −1.5y
e
+c
3
2 −1.5y 7
e
+
3
3
dC
dy
C(0) = 3