Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

SECTION 1505 APPLICATIONS OF DOUBLE INTEGRALS 0 265
(b) (i) No restriction is placed on X, so
P(Y ~ 1) = f~oo I 1
00
f (x,y) dy dx = I oo Jt 0°1e-(OoSx+0. 2 y) dydx
= o ."1 roo e- 0 ·5"'
dx f 00 e- 0·211 dy = oo1 tim rt e - o.s., dx lim It e- 0· 211 dv
Jo 1 t-+oo Jo t-oo 1
= 001 lim [-2e- 0 ·5"']t lim [-5e- 0·211 ]t = 001 lim [- 2(e- 0·5 ' - 1)] lim [- 5(e- 0 ·2 ' - e- 0 ·2 )]
t-oo 0 t-oo 1 t-+oo t-oo
(0.1) 0 ( -2)(0- 1) 0 ( - 5)(0 - e- 0·2 ) = e- 0 ·2 ~ 008187
(ii) P(X $2, Y $ 4) = I~ oo I~ oo
(c) The expected value of X is given by
f (x,y) dyd.x = I 02
J 0
4 Oo1e-<
0·5"'+ 0·2Y) dydx
= 001 1~ 2 e- o.sx dx 1~ 4 e- 0 ·2 ll dy = 0.1 [ -2e- 0 ·5"']
~ [.- 5e - 0 · 211 ]~
= (0.1) · (-2)(e- 1 -1) · (- 5)(e- 0·8 -1)
= (e- 1 - 1)(e- 0 ·8 - 1) = 1 + e-J.B - e- o.s - e- 1 ~ 0.3481
Mt = IIR2 x f(x, y) dA = 1~ 00 1~ 00 :z; [oo1e-(o.sx,+o. 2 v)J dydx
= 0.1 ]; 00 xe-O.ux dx ]; 00 e- 002 !1 dy = 001 lim ];t xe- 005 "' dx lim rt e- 0 ·2 Y dy
0 0 t~oo o t-oo ./o
To evaluate the first inteirat, we integrate by parts with u. = x and dv = e- 0·5"' dx (or we can use Formula 96
in the Table of integrals): I xe- 0 ·5"'
dx = - 2xe- 0·5"'- J - 2e- 0·5"' do'l = -2xe- 0·5"' -
4e- 0 ·5 "' = - 2(x + 2)e- 0 ·5"'0
Thus
JJ. 1
= 0°1 lim [-2(x + 2)e- 0·5 "'] '· lim [- 5e- 0·2ll] t
t-+oo 0 t -+oo 0
= 001 lim ( -2) [ (t + 2)e- 0 ·5t
- 2] lim ( - 5) [e- 0·2 t - 1]
t -...oo
t-+oo
The expected value ofY is given by
= 0.1(-2)(lim t +. 2 - 2)(- 5)(- 1) = 2 [by !'Hospital's Rule]
t -+oo eO.at
J1.2 = Ifa2 y f(x , y) dA = Io 00 fa"" v [oo1e-(0.5+0. 2 y)] dy dx
= 001 ]; 00 e- 0 ·5"'
dx {; 00 ye- 0 ·211 dy = 001 lim {;t e-o.sx dx lim rt ye- 0 · 2Y dy
0 . 0 t-oo. o t-oo _Jo
To evaluate the second integral, we integrate by parts with u = y and dv = e - 0·2 !1 dy (or again we can use Formula 96 in
the Table oflntegrals) which gives I ye- 0·211 dy = - 5ye- 0·2ll + .r se- 0·2 11 dy = -5(y + 5)e- 0 ·2U
0 Then
JJ. 2 = 0.1 lim [- 2e- 0 · 0·"] t lim [- 5(y + 5)e- 0 ·2 Y] '
t-oo 0 t-+oo 0
= Oollim [- 2(e-o.st _ 1)] lim (- 5[(t + 5)e- 0 ·2t- 5])
t-+oo
t -oo
0
=Oo1(- 2)(- 1)o (-5) lim -
t+5 )
( 0
.,t - 5 = 5
t- oo e ....
[Py )'Hospital's Rule]
310 (a) The random variables X and Y are normally distributed with JJ. 1 = 45, JJ. 2 = 20, 0'1 = 005, and a 2 = Ool.
The individual density functions for X and Y, then, are !J (x) =
h (y) =
~ e-