Solução_Calculo_Stewart_6e

F.
SECTION TX.10 16.5 APPLICATIONS OF DOUBLE INTEGRALS ET SECTION 15.5 ¤ 239
in the Table of Integrals): xe −0.5x dx = −2xe −0.5x − −2e −0.5x dx = −2xe −0.5x − 4e −0.5x = −2(x +2)e −0.5x .
Thus
μ 1 =0.1 lim
−2(x +2)e
−0.5x t
lim
t→∞ 0 −5e
−0.2y t
t→∞
0
=0.1 lim(−2) (t +2)e −0.5t − 2 lim
t→∞
=0.1(−2)
lim
t→∞
t +2
e 0.5t
− 2
(−5)(−1) = 2
t→∞
(−5) e −0.2t − 1
[by l’Hospital’s Rule]
The expected value of Y is given by
μ 2 = yf(x, y) dA = ∞
∞
y0.1e −(0.5+0.2y) dy dx
R 2 0 0
=0.1 ∞
0
e −0.5x dx ∞
0
ye −0.2y dy =0.1 lim
t→∞
t
0 e−0.5x dx lim
t→∞
t
0 ye−0.2y dy
To evaluate the second integral, we integrate by parts with u = y and dv = e −0.2y dy (oragainwecanuseFormula96in
the Table of Integrals) which gives ye −0.2y dy = −5ye −0.2y + 5e −0.2y dy = −5(y +5)e −0.2y .Then
μ 2 =0.1 lim
−2e
−0.5x t
lim
t→∞ 0 −5(y +5)e
−0.2y t
t→∞
0
=0.1 lim −2(e −0.5t − 1)
lim −5 (t +5)e −0.2t − 5
t→∞ t→∞
t +5
=0.1(−2)(−1) · (−5) lim
t→∞ e − 5 =5 [by l’Hospital’s Rule]
0.2t
31. (a) The random variables X and Y are normally distributed with μ 1 =45, μ 2 =20, σ 1 =0.5,andσ 2 =0.1.
1
The individual density functions for X and Y , then, are f 1 (x) =
0.5 √ 2π e−(x−45)2 /0.5 and
1
f 2 (y) =
0.1 √ 2π e−(y−20)2 /0.02 .SinceX and Y are independent, the joint density function is the product
f(x, y) =f 1 (x)f 2 (y) =
Then P (40 ≤ X ≤ 50, 20 ≤ Y ≤ 25) = 50
40
1
0.5 √ 2π e−(x−45)2 /0.5 1
0.1 √ 2π e−(y−20)2 /0.02 = 10
π e−2(x−45)2 −50(y−20) 2 .
25
20
f(x, y) dy dx =
10
π
50
25
40 20 e−2(x−45)2 −50(y−20) 2 dy dx.
Using a CAS or calculator to evaluate the integral, we get P (40 ≤ X ≤ 50, 20 ≤ Y ≤ 25) ≈ 0.500.
(b) P (4(X − 45) 2 +100(Y − 20) 2 ≤ 2) = 10
D π e−2(x−45)2 −50(y−20) 2 dA,whereD is the region enclosed by the ellipse
4(x − 45) 2 + 100(y − 20) 2 =2. Solving for y gives y =20± 1 10 2 − 4(x − 45)2 , the upper and lower halves of the
ellipse, and these two halves meet where y =20 [since the ellipse is centered at (45, 20)] ⇒ 4(x − 45) 2 =2 ⇒
x =45± 1 √
2
.Thus
D
√
45+1/ 2
20+ 1
10
−50(y−20) 2
π e−2(x−45)2 dA = 10
10
√2 − 4(x−45) 2
e −2(x−45)2 −50(y−20) 2 dy dx.
π
45−1/ √ 2 20−
10
1 √2 − 4(x−45) 2
Using a CAS or calculator to evaluate the integral, we get P (4(X − 45) 2 +100(Y − 20) 2 ≤ 2) ≈ 0.632.
33. (a) If f(P, A) is the probability that an individual at A will be infected by an individual at P ,andkdAis the number of
infected individuals in an element of area dA, thenf(P, A)kdAis the number of infections that should result from
exposure of the individual at A to infected people in the element of area dA. IntegrationoverD gives the number of
infections of the person at A due to all the infected people in D. In rectangular coordinates (with the origin at the city’s