Solução_Calculo_Stewart_6e

F.
238 ¤ CHAPTER 16 MULTIPLE INTEGRALS ET CHAPTER 15
TX.10
27. (a) f(x, y) is a joint density function, so we know R 2 f(x, y) dA =1.Sincef(x, y) =0outside the
rectangle [0, 1] × [0, 2], we can say
Then 2C =1 ⇒ C = 1 2 .
(b) P (X ≤ 1,Y ≤ 1) = 1
−∞
R 2 f(x, y) dA = ∞
−∞
= 1
0
∞ f(x, y) dy dx = 1 2
Cx(1 + y) dy dx
−∞ 0 0
= C 1
0 x y + 1 2 y2 y=2
y=0 dx = C 1
0 4xdx= C 2x 2 1
0 =2C
1 f(x, y) dy dx = 1 1
−∞ 0 0
1
x y + 1 y2 y =1
dx = 1 1
x 3
2 2 y =0 0 2 2
1
x(1 + y) dy dx
2
dx =
3 1 x2 1
= 3 or 0.375
4 2 0 8
(c) P (X + Y ≤ 1) = P ((X, Y ) ∈ D) where D is the triangular region shown in
the figure. Thus
P (X + Y ≤ 1) = f(x, y) dA = 1
D 0
= 1
0
= 1 1
4 0
1 − x
0
1
x(1 + y) dy dx
2
1
x 2
y + 1 y2 y =1−x
2
dx = 1 1
x 1
y =0 0 2
2 x2 − 2x + 3 2
dx
x 3 − 4x 2 +3x
dx = 1 x 4
− 4 x3
4 4 3 +3x2 2
= 5
48 ≈ 0.1042
29. (a) f(x, y) ≥ 0,sof is a joint density function if R 2 f(x, y) dA =1. Here, f(x, y) =0outside the first quadrant, so
R 2 f(x, y) dA = ∞
0
∞
0
0.1e −(0.5x +0.2y) dy dx =0.1 ∞
0
=0.1 lim
t→∞
t
0 e−0.5x dx lim
t→∞
t
0 e−0.2y dy =0.1 lim
t→∞
−2e
−0.5x t
0 lim
t→∞
∞
0
1
0
e −0.5x e −0.2y dy dx =0.1 ∞
e −0.5x dx ∞
e −0.2y dy
0 0
−5e
−0.2y t
0
=0.1 lim
t→∞
−2(e −0.5t − 1) lim
t→∞
−5(e −0.2t − 1) =(0.1) · (−2)(0 − 1) · (−5)(0 − 1) = 1
Thus f(x, y) is a joint density function.
(b) (i) No restriction is placed on X,so
P (Y ≥ 1) = ∞
−∞
∞
1
(ii) P (X ≤ 2,Y ≤ 4) = 2
−∞
f(x, y) dy dx = ∞
0
=0.1 ∞
e −0.5x dx ∞
e −0.2y dy =0.1 lim
0 1
=0.1 lim
t→∞
−2e
−0.5x t
0 lim
t→∞
(c) The expected value of X is given by
∞
1
0.1e −(0.5x+0.2y) dy dx
t
t
t→∞ 0 e−0.5x dx lim
t→∞ 1 e−0.2y dy
−5e
−0.2y t
=0.1 lim
1 t→∞
(0.1) · (−2)(0 − 1) · (−5)(0 − e −0.2 )=e −0.2 ≈ 0.8187
4 f(x, y) dy dx = 2 4
−∞ 0 0 0.1e−(0.5x+0.2y) dy dx
=0.1 2
0 e−0.5x dx 4
0 e−0.2y dy =0.1 −2e −0.5x 2
0
=(0.1) · (−2)(e −1 − 1) · (−5)(e −0.8 − 1)
−2(e −0.5t − 1) lim
t→∞
−5(e −0.2t − e −0.2 )
−5e
−0.2y 4
0
=(e −1 − 1)(e −0.8 − 1) = 1 + e −1.8 − e −0.8 − e −1 ≈ 0.3481
μ 1 = R 2 xf(x, y) dA = ∞
0
∞
x0.1e −(0.5x+0.2y) dy dx
0
=0.1 ∞
0
xe −0.5x dx ∞
0
e −0.2y dy =0.1 lim
t→∞
t
0 xe−0.5x dx lim
t→∞
t
0 e−0.2y dy
To evaluate the first integral, we integrate by parts with u = x and dv = e −0.5x dx (or we can use Formula 96