Solução_Calculo_Stewart_6e

F.
370 ¤ CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION
TX.10
8.4 Applications to Economics and Biology
1. By the Net Change Theorem, C(2000) − C(0) = 2000
0
C 0 (x) dx ⇒
C(2000) = 20,000 + 2000
(5 − 0.008x +0.000009x 2 ) dx =20,000 + 5x − 0.004x 2 +0.000003x 3 2000
0 0
=20,000 + 10,000 − 0.004(4,000,000) + 0.000003(8,000,000,000) = 30,000 − 16,000 + 24,000
=$38,000
3. If the production level is raised from 1200 units to 1600 units, then the increase in cost is
C(1600) − C(1200) = 1600
1200 C0 (x) dx = 1600
1200 (74 + 1.1x − 0.002x2 +0.00004x 3 ) dx
= 74x +0.55x 2 − 0.002
3
x 3 +0.00001x 4 1600
=64,331,733.33 − 20,464,800 = $43,866,933.33
1200
5. p(x) =10 ⇒ 450 =10 ⇒ x +8=45 ⇒ x =37.
x +8
37
37
450
Consumer surplus = [p(x) − 10] dx =
x +8 − 10 dx
0
0
= 450 ln (x +8)− 10x 37
=(450ln45− 370) − 450 ln 8
0
= 450 ln
45
8 − 370 ≈ $407.25
7. P = p S (x) ⇒ 400 = 200 + 0.2x 3/2 ⇒ 200 = 0.2x 3/2 ⇒ 1000 = x 3/2 ⇒ x =1000 2/3 = 100.
Producer surplus = 100
[P − p
0 S(x)] dx = 100
[400 − (200 + 0.2x 3/2 )] dx =
100
200 − 1 0 0
5 x3/2 dx
100
= 200x − 2
25 x5/2 =20,000 − 8,000 = $12,000
0
9. p(x) = 800,000e−x/5000
x +20,000
=16 ⇒ x = x 1 ≈ 3727.04.
Consumer surplus = x 1
[p(x) − 16] dx ≈ $37,753
0
11. f(8) − f(4) = 8
f 0 (t) dt = 8
√ 8 √
4 4 tdt=
2
3 t3/2 = 2 3 16 2 − 8 ≈ $9.75 million
4
13. N =
b
a
Similarly,
x
Ax −k −k+1 b
dx = A
=
−k +1 A
a
1 − k (b1−k − a 1−k ).
b
a
Thus, x = 1 N
x
Ax 1−k 2−k b
dx = A
2 − k
a
b
a
= A
2 − k (b2−k − a 2−k ).
Ax 1−k dx = [A/(2 − k)](b2−k − a 2−k )
[A/(1 − k)](b 1−k − a 1−k ) = (1 − k)(b2−k − a 2−k )
(2 − k)(b 1−k − a 1−k ) .
15. F = πPR4
8ηl
= π(4000)(0.008)4
8(0.027)(2)
≈ 1.19 × 10 −4 cm 3 /s