Решение задач

, x > 0.

, x £ 0

∞								ì	1	,	(x, y)Î D
= 4òe	− x	dx = -4e	− x		∞	= 4, ihwlhfm	p1,2	ï
									4
					0			(x, y) = í			(x, y)Ï D.
0								ï0,
								î

ì ex		dy				e		x
ï ò					=				,
		4
ï−ex						2
= í	− x		dy				e− x
ï e
ï ò					=
		4						2
î−e− x

ì ln(− y )

dx

ln (- y)

x £ 0

ï

ò

=

, -1 £

:gZeh]bqgh p

( y) =

ï− ln(− y ) 4

2

2

í

dx

ln y

ï ln y

, x > 0.

ï ò

=

, 0

< y £ 1.

4

2

î− ln y

p1 (x) = ò p1,2 (x, y)dy =

\

y < 0

b jZ\gh ijb hk-

	lZevguo agZq_gbyo y. p ( y / x) =	p1,2 (x, y )
		p1 (x)
	2 /1

ìe− x

ï

= ïí 2 ξ1 b ξ2 aZ\bkbfu gZijbf_j

ïex ïî 2

p	(0,1) =				1	¹		p (0) p				(1) =		1		×0 = 0.
1,2				4			1				2		2
											dx + +∞											+∞ e− x dx = 0;					dx + +∞
		Mξ		=		0		xex					xe− x			dx = -		1	0	ex dx +	1		Mξ 2	=	0	xe2 x		xe−2 x	dx =
			1			ò	2					ò	2				ò			ò			1		ò	2	ò	2
						−∞						0						4 −∞		4 0					−∞		0
= -	1	0	e2 x dx +					1	+∞ e−2 x dx = 0 Þ Dξ								= 0, ihwlhfm dhwnnbpb_gl dhjj_eypbb g_ hij_^_ezg b km-
	4 −∞ò						4 ò0									1

^blv h dhjj_ebjh\Zgghklb g_\hafh`gh

6.4. >\mf_jgZy kemqZcgZy \_ebqbgZ ξ1, ξ2 jZkij_^_e_gZ jZ\ghf_jgh \ h[eZklb D h]- jZgbq_gghc weebikhf k ihemhkyfb a, b b p_gljhf \ gZqZe_ dhhj^bgZl GZclb iehlghklv kh- \f_klgh]h jZkij_^_e_gby p1,2(x, y fZj]bgZevgmx iehlghklv j1(x b mkeh\gmx iehlghklv j2/1(y/x Y\eyxlky eb kemqZcgu_ \_ebqbgu ξ1 b ξ2 aZ\bkbfufb b dhjj_ebjh\Zggufb"

ì 1

, (x, y )Î D

πab,

ï

. p1 (x) = ò p1,2 (x, y)dy =

J_r_gb_:

S(D) =

ihwlhfm

p1,2 (x, y ) = íπ ab

ï0,

(x, y )Ï D

\

î

b 1−

x2

2 1-

x

2

ì

2

a2 - x2

ì

2

b2 - y2

a

2

£ a

£ b

1

a

2

ï

,

x

ï

,

y

= ò

dx

=

= í

π a2

. :gZeh]bqgh p2

( y) = í

π b2

.

π ab

π a

x2

ï

x

³ a

ï

³ b

−b 1−

î0,

î0,

y

a2

p	( y / x) = 2b				a2 - x2 : hij_^_e_gZ ijb |x| £ a. x1 b x2 aZ\bkbfu gZijbf_j p (0, 0) =		1	¹
2 /1						1,2	π ab
¹ p		(0) p		(0) =	4	. Fξ1 = Dξ1 kf ihwlhfm dhwnnbpb_gl dhjj_eypbb g_ hij_^_-
					π 2ab
1			2

ezg b km^blv h dhjj_ebjh\Zgghklb g_evay