Solução_Calculo_Stewart_6e

F.
17. From (3), F =
T
0
A 6
=
c (t) dt 20I ,where
I =
10
0
TX.10
10
te −0.6t 1
dt =
(−0.6) 2 (−0.6t − 1) e−0.6t
6(0.36)
Thus, F =
20(1 − 7e −6 ) = 0.108 ≈ 0.1099 L/s or 6.594 L/min.
1 − 7e−6 0
integrating
by parts
SECTION 8.5 PROBABILITY ¤ 371
= 1
0.36 (−7e−6 +1)
19. AsinExample2,wewillestimatethecardiacoutputusingSimpson’sRulewith∆t =(16− 0)/8 =2.
16
c(t) dt ≈ 2 [c(0) + 4c(2) + 2c(4) + 4c(6) + 2c(8) + 4c(10) + 2c(12) + 4c(14) + c(16)]
0 3
Therefore, F ≈
≈ 2 [0 + 4(6.1) + 2(7.4) + 4(6.7) + 2(5.4) + 4(4.1) + 2(3.0) + 4(2.1) + 1.5]
3
= 2 (109.1) = 72.73mg· s/L
3
A
72.73 = 7 ≈ 0.0962 L/s or 5.77 L/min.
72.73
8.5 Probability
1. (a) 40,000
f(x) dx is the probability that a randomly chosen tire will have a lifetime between 30,000 and 40,000 miles.
30,000
(b) ∞
f(x) dx is the probability that a randomly chosen tire will have a lifetime of at least 25,000 miles.
25,000
3. (a) In general, we must satisfy the two conditions that are mentioned before Example 1—namely, (1) f(x) ≥ 0 for all x,
and (2) ∞
3
f(x) dx =1.For0 ≤ x ≤ 4,wehavef(x) = x √ 16 − x
−∞ 64 2 ≥ 0,sof(x) ≥ 0 for all x. Also,
∞
−∞ f(x) dx = 4
0
= − 1
64
Therefore, f is a probability density function.
(b) P (X