36_integralele Euleriene (Rezumat)
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INTEGRALE EULERIENE BREVIAR TEORETIC • Integrala ga mma: ( ) ∫ ∞ − − > = Γ 0 1 0 ; a dx e x a x a . Propriet ăţi: 1) ( ) 1 1 = Γ . 2) ( ) ( ) ( ) ( ) 1 , 1 1 > ∀ − Γ − = Γ a a a a . 3) ( ) ( ) ( ) N n n n ∈ ∀ − = Γ , ! 1 . 4) π = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ Γ 2 1 . • Integrala beta : ( ) ( ) ∫ > > − = − − 1 0 1 1 0 , 0 ; 1 , b a dx x x b a b a β Propriet ăţi: 1) ( ) ( ) 0 , , , , > ∀ = b a a b b a β β 2) ( ) ( ) ( ) ( ) 0 , , , > ∀ + Γ Γ Γ = b a b a b a b a β . 2) ( ) ( ) ∫ ∞ + − + = 0 1 1 , dx x x b a b a a β . 3) Dacă 1 = + b a , atunci ( ) π π β a b a sin ) , ( = . 1
description
Integrale
Transcript of 36_integralele Euleriene (Rezumat)
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INTEGRALE EULERIENE BREVIAR TEORETIC
Integrala gamma: ( ) >=0
1 0; adxexa xa .
Proprieti: 1) ( ) 11 = . 2) ( ) ( ) ( ) ( ) 1,11 >= aaaa . 3) ( ) ( ) ( ) Nnnn = ,!1 . 4) =
21 .
Integrala beta: ( ) ( ) >>= 10
11 0,0;1, badxxxba ba Proprieti: 1) ( ) ( ) 0,,,, >= baabba 2) ( ) ( ) ( )( ) 0,,, >+
= babababa .
2) ( ) ( )
+
+= 01
1, dx
xxba ba
a .
3) Dac 1=+ ba , atunci ( )a
basin
),( = .
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PROBLEME REZOLVATE S se calculeze urmtoarele integrale:
1. +
+=
1
11 dxexI x .
Rezolvare: Folosim schimbarea de variabil dtdxtxtx ===+ 11 . Intervalul de integrare se modific dup cum rezult din tabelul de mai jos: x 1 t 0 Obinem: dtetI t
=0
21
. Prin identificare cu formula de definiie a
integralei gamma, rezult 23
211 == aa , prin urmare
( ) ( ) 21212123 ===I .
2. + =0
25 dxexI x .
Rezolvare: Folosim schimbarea de variabil dtdxtxtx 2
1212 === .
x 0 t 0 Obinem: ( )
815
2!56
21
21
21
2 6605
60
5====
=
dtetdtetI tt .
3. +
= dxexI x 26 .
2
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Rezolvare: Deoarece funcia care trebuie integrat este par, rezult c
+ =0
6 22 dxexI x .
Folosim schimbarea de variabil: dttdxtxtx 21
21
212 === .
x 0 t 0
8
1521
21
23
25
272
00213 2521 =
=
===
+ + dtetdttetI tt .
4. xdxxI 31
0ln= .
Rezolvare: Folosim schimbarea de variabil: dtedxextx tt ===ln x 0 1 t 0
==0
30
3 232 dtetdteteItt t
Facem transformarea: dydtytyt 32
32
23 ===
t 0 y 0
( ) ( ) ( )27324
8116
81160
0
3323
32 ====
dyeydyeyI yy .
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5. =0
2
dxeI x (integrala Euler-Poisson).
Rezolvare: Folosim schimbarea de variabil: dttdxtxtx 2
121
212 === .
x 0 t 0
221
21
021
021 2121 =
===
dtetdtteI tt .
6. 1,ln
1>
adx
xx
a .
Rezolvare:
Folosim schimbarea de variabil: dtedxextx tt ===ln . x 1 t 0
( ) ==0
1
0dtetdteetI tatat .
Folosim schimbarea de variabil: ( ) dydtytyta aa 11111 === . t 0 y 0
( ) ( ) ( ) ( )222 1111011 2
=== aaya dyeyI .
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7. Integrala dxeI xx
+=
1
15,0 2 are forma b
ake
2 . S se
determine valorile parametrilor reali k , a i b . Rezolvare:
Avem c: ===
+ +
11
1 2 1222
21
dxedxeIxxxx
+
+++ ==
1
21
1
23
212
2
23
2
dxeedxexxx
. Folosim schimbarea de variabil:
dtdxtxtx 21221 ===+ .
x 1 t 0
=0
22
23
dteeI t . Folosind faptul c 20
2 = dte t (integrala Euler-Poisson), obinem c
21
23
23
222
== eeI , prin urmare
valorile cutate ale celor trei parametri sunt: 21,
23,1 === bak .
S se calculeze urmtoarele integrale:
8. ( ) =1
0 3 2 1 xx
dxI .
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Rezolvare:
( ) ( ) =
=
1
0
1
0 3 231
32
11
dxxxxx
dxI . Prin identificare cu formula
de definiie a integralei beta, obinem:
31
321 == aa ; 32311 == bb , prin urmare, avnd n
vedere definiia i proprietatea 3 pentru integrala beta, rezult: ( )3
2sin
,3
32
31 ===I .
9. ( ) = 10
38 1 dxxxI .
Rezolvare:
Facem schimbarea de variabil dttdxtxtx 32
31
313 === .
x 0 1 t 0 1
( ) ( ) ( )121
)5()2()3(
312,311
1
031
1
0
231
31 3238 =
==== dtttdttttI .
10. ( ) dxxxI = 10
5,123 1 .
Rezolvare:
Facem schimbarea de variabil: dttdxtxtx 21
21
212 === .
x 0 t 0
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Prin urmare, ( ) ( ) === 10
1
0
5,12 21236131 1211 dttttdxxxI
( )
== 25,
32
211
21 1
0
23
31 dttt .
11. S se calculeze: a) ( )
+= 0 61dx
xxI ; b)
+= 0 61
dxx
xI .
Rezolvare: a) Prin identificare cu a doua formul de definiie a integralei beta (proprietatea 2), obinem: 211 == aa ; 46 ==+ bba , prin urmare ( ) ( ) ( )( ) 20
16
424,2 === I .
b) Facem schimbarea de variabil dttdxtxtx 65
61
616 === .
x 0 t 0
( ) ===+=+= 0 332311
061
93
sin61,
61
161
161 3
2
656
1 ttdtt
ttI .
12. Integrala ( ) ( ) = 20
6,04,1 cossin
dxxxI are forma ),( qpk , unde 0,;,, > qpRqpk . S se afle valorile paramertilor qpk ,, . Rezolvare: Folosim schimbarea de variabil: dtxdxxtx == cossin2sin2 . x 0 2
t 0 1 Transformm funcia care trebuie integrat astfel:
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== 2
0
6,14,0 cossin2)(cos)(sin21
xdxxxxI
=2
0
8,022,02 cossin2)(cos)(sin21
xdxxxx . Obinem:
( )2,0;2,121)1(
21 1
0
8,02,0 == dtttI , deci 2,0;2,1;21 === qpk .
13. S se calculeze integrala: ( )( ) +=3
4 6 534 xx
dxI .
Rezolvare:
Integrala se poate scrie: ( ) ( )
+=3
4
65
61
34 dxxxI .
ncercm s facem schimbarea de variabil dtdxtxtx ===+ 44 .
x 4 3 t 0 7 Se observ c intervalul de integrare devine ( )7,0 , prin urmare, pentru a ajunge la intervalul ( )1,0 , vom folosi schimbarea de variabil dtdxtxtx 747
74 ===+ .
x 4 3 t 0 1
Obinem: ( ) ( ) ( ) === 10
1
0
65
61
65
61
65
61
17777777 dtttdtttI
( ) ( ) 2sin,, 665616165 ==== .
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PROBLEME PROPUSE S se calculeze valoarea urmtoarelor integrale:
1. 0
36 dxex x R: 24380 2.
0
7 2 dxex x R: 3;
3. ( ) dxxx 10
52 R: 27721 4. +
dxex x24
R: 43
5. 1
0
2dxxx R: 8
6. +
dxe x 2 R:
7. ( )
+1
151 dxex x R: 8. ( ) dxxx
+0
1
32 1 R: 601
9.
05 dxex x R: 120 10.
+0 2 23 dx
exx
x R: -1
11. ( ) 10
6314 1 dxxx R: 69301
12. ( ) 1
0 3 2 1
1 dxxx
R: 3
32
13. dxxx 2
0
22 4 R: 14. ( )
+0 64
1dx
xx R: 5
1
15. ( ) dxxx 10
42 R: 6301 16. ( )
1
0 6 5 1
1 dxxx
R: 2
17. ( )10
5ln dxxx R: 8
15 18. 0,0
222 > adxxaxa
R: 164a
19.
+0 411 dxx
R:22
20.
++
225)2( dxex x R: 120
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21.( )
1
0 4 3 1
1 dxxx
R: 2 22.
0
2
2
dxex
R:22
23. 0;0
> ndxe nx R: ( )nn 11
24. 0,;0
> nmdxex nxm R: ( )nmn 11 +
25. ( ) 2
272 dxex x R: !7 26.
0
dxe x R: 2
27. 2/
0
53 cossin
dxxx R: 121 28.
+ 0
7 5 7 dxex x R: !117
29. dxxx
0
3
24 9 R: 32729
30. +
dxe x2
2
R: 2 31. ( )
+0 3210
21dx
x
x R: 2
32. dxx1
0
1ln R: 2
33.( ) ( ) +
1
3 6 5 13 xx
dx R: 2
34. Nndxex xn
;2 R: 0 , dac n impar; ( ) ( )
22!!1
21
n
nn = + , dac n par
35. ( )
++1
131 dxex x R: -3! 36. ( )( ) 3
1 13dx
xxdx R:
37. e dxxxx143 )ln1(ln1 R: 280
1 38. +0 64
1dx
xx
R: 3
10
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s
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39. a
dxxax0
224 R:
32
6a
40. + +1
422 dxe xx R: 32e
41.
+
0
24
27
21 dxxx R:
524
42. dxxx 3
0
25 9 R: 355832 43. Nndxex
nxn ;0
2 R: ( )nnn 131+
44. 0
13 dxex x R: e6
45. ( ) 0;ln10
11 > pdxpx R: ( )p 46. ( )
+0 234
21dx
x
x R:
2723 3
47.
+
1
322 dxe xx R: 2
4 e 48. ( ) ( ) 1
151 dxexnx R:1
49. Nndxex xn ;0
2
R: 50.
+0 83
1dx
xx
R: 8
51. dxxx
0
4
26 16 R: 1280
52. ( ) 10
435 1 dxxx R: 901 53.
2/
0
24 cossin
dxxx R:
54. ( )
+0 3 11 dx
xxR:
32 55. ( ) 1
0
438 1 dxxx
11
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56.
+0 611 dxx
R: 3 57. ( ) +0 23 11 dxxx R: 3
58. ( )
+0 242
1dx
x
xR:
28
59. Nnmdxxx nm ,;cossin2/
0
1212
R:( ) ( )( )!12
!1!1+
nmnm
60.
++ dxe xx 12 2 R:
2
89 e
61. *2
;12
1
Nnn
xn
+
+
62.
++ dxe xx 142 2 R:
223 e
63.
+0 42
1dx
xx
R:4
2
64. 2
22 dxex x R: 2 65.
1
13 dxex x R:16
66. ( ) 10
523 1 dxxx R: 841
67. ( )
+0 324
21dx
x
xR:
12823
68. Integrala dxeI xx
+=
1
563 2 are forma bake , unde Rbak ,, . S se afle valorile parametrilor bak ,, .
R: 21
63 ,8, === bak .
69. Integrala =2/
0
42 cossin
dxxxI are forma ak unde
12
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r stu
dent
s
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Rak , . S se determine valorile parametrilor k i a . R: 1;32
1 == ak . 70. Integrala )(
0
45,2 3 badxexI x == , unde 0;, > bRba . S
se determine valorile parametrilor a i b .
71. Integrala ( ) == 10
8,436,3 ),(1 qpkdxxxJ , unde
0,;,, > qpRqpk . S se determine valorile parametrilor qpk ,, . 72. S se calculeze 0,0,
)1()1()1(1
12
1212>>+
+= +
nmdx
xxxT nm
nm.
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r stu
dent
s
CAPITOLUL 9 - CALCUL INTEGRAL9.1. INTEGRALE GENERALIZATE9.1.1. INTEGRALE CU LIMITE INFINITE9.1.2. INTEGRALE DIN FUNCII NEMRGINITE9.1.3. INTEGRALE EULERIENE
9.2. INTEGRALE DUBLE
