Matematica - Integrale-1

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CAPITOLUL I I N T E G R A L E (recapitulare liceu) TABEL DERIVATE Funcţii elementare Funcţii compuse 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12) 9
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Matematica - Integrale-1

Transcript of Matematica - Integrale-1

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CAPITOLUL I

I N T E G R A L E (recapitulare liceu)TABEL DERIVATEFuncii elementareFuncii compuse

1)

2)

3)

4)

5)

6)

7)

8)

9)

10)

11)

12)

13)

14)

15)

16) F. F. important

Reguli de derivare1) 4)

2) 5)

3)

TABEL INTEGRALE

1)

12)

2)

13)

3)

14)

4)

15)

5)

16)

6)

17)

7)

18)

8)

19)

9)

20)

10)

21)

11)

5)

R i

a)

b)

c)

Proprieti

1)

2) .

1.1. Primitive (integrale)n clasa a XI-a se ddea i se cerea unde:

Deci, tim funcia i se cere s aflm tangentele la grafic n fiecare punct (domeniul funciei).

n clasa a XII-a tim i se cere s aflm .

Fig. 1

Altfel spus cunosc tangentele la grafic n fiecare punct al graficului i se cere s se afle funcia care are aceast proprietate. (Problema invers a tangentelor sau se mai numete i determinarea primitivelor).Definiie: Fie R. Se numete primitiv a funciei f, o funcie F cu proprietile:

a) F este derivabil

b) F(x)=f(x)Dac F(x)=f(x), atunci condiia este echivalent cu .Notaie: Primitiva care se mai numete i integral.

Observaii:

1) indic n raport cu care variabil are loc integrarea;

2) reprezint difereniala argumentului x.

Definiie: Difereniala unei funcii este: , unde (este o diferen).

Proprietile diferenialei

1) Dac

2) Dac

3) Dac

Proprietile primitivelor

a) Dac i sunt dou primitive ale aceleiai funcii f, atunci ele difer printr-o constant

b)

c)

Aplicaie: Folosind proprietile i tabelul de integrale s se calculeze:

Soluie:

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3 Deci:

Verificare c:

Dar

EMBED Equation.3

Tema 1.1 Folosind tabelul de integrale i proprietile primitivelor s se calculeze urmtoarele integrale:

1) 2)

3) 4)

5) 6)

7) 8)

9) 10)

11) 12)

13) ; 14)

15) 16)

17) 18)

19) 20)

21) 22)

23) 24)

1.2. Metoda schimbrii de variabil

A (Varianta clasic= Metoda a I-a din carte)

Fie: , dac fac schimbarea de variabil: , atunci ,

iar prin nlocuire:

formula schimbrii de variabil

Aplicaia 1: S se calculeze:

Soluie:

Fcnd schimbarea de variabil

atunci:

EMBED Equation.3 Din substituie: deci:

Observaie: Proprietile funciilor inverse trigonometrice:

Dac avem de calculat: atunci:

cu , devine:

Aplicaia 2. S se calculeze:

Soluie: Dac facem substituia , atunci , iar

Din substituia:

, iar

Tema 1.2.1.1) 2)

3) 4)

5) 6)

7) 8)

9) 10)

11) 12)

13) 14)

15)

Observaie: Schimbrile de variabil indicate trebuie s fie gsite de rezolvitor!!

Metoda schimbrii de variabilB (Varianta a II-a vezi metoda a II-a din carte)

Depistarea (gsirea) schimbrii de variabil se face cu ajutorul urmtoarei proprieti:

Proprietatea (*). Dac o parte are calitatea c face parte din funcia mpreun cu derivata ei , atunci:

n loc de demonstraie!!

Dac are proprietatea (*), atunci (vezi prima metod de schimbare de variabil) i atunci:

Observaie (F. F. important)

1)Dac are proprietatea (*), atunci nu este indicat s inversm substituia, ci s o difereniem direct . Cum funcia are proprietatea (*), atunci se nlocuiete direct.

ATENIE!!! Prin inversare se pierde timp!!

2) Proprietatea (*) ne nva!! ce substituie trebuie fcut pentru fiecare integral. Acest lucru se va deprinde din exemplele prezentate mai jos.

Aplicaia 1: S se calculeze:

Rezolvare. ntrebm ce funcie are derivata

Rspuns: care face parte din funcia f(x) Deci are calitatea c face parte din integrant mpreun cu derivata ei. Atunci:

(care se gsete n integral).

iar

Cum nu este indicat! Dac ghicesc substituia , atunci inversm relaia, iar:

EMBED Equation.3

Aplicaia 2: S se calculeze:

Rezolvare:

ntrebm ce funcie are derivata

Rspuns:

care are calitatea c face parte din funcia de sub integral. Atunci se face substituia

(care se gsete n componena integralei).

Atunci:

Cum nu este indicat!! Dac ghicesc substituia

,atunci inversez relaia

EMBED Equation.3

(vezi prima schimbare de variabil ). Atunci:

sau:

Calculez separat expresia:

EMBED Equation.3 .Atunci:

EMBED Equation.3 Deci, incomparabil prin inversarea substituiei se pierde timp!

Tema 1.2.2. S se calculeze urmtoarele integrale:1) 2)

3) 4)

5) 6)

7) 8)

9) 10)

11) 12)

13)

EMBED Equation.3 14)

15) 16)

17) 18) 19)

20) 21)

22) 23)

24) 25)

a) b) c) d) e)

1.3. Integrarea prin priFormula integrrii prin pri:

Dar: (integrala derivatei face funcia)

S-a obinut:

formula integralei prin pri.

Aplicaii: S se calculeze integralele urmtoare folosind metoda integrrii prin pri.

1)

Se alege:

Atunci:

2)

Se alege:

Notm:

Rezult:

. Atunci:

3)

EMBED Equation.3 .

sau:

(a)

Calculm prin pri:

sau:

(b)

nlocuind pe (b) n (a) se obine:

,

sau:

A)Calculul urmtoarelor integrale:

a) b) c)

a)

,

Atunci:

EMBED Equation.3 .

sau:

EMBED Equation.3 .

Deci:

care se gsete n tabelul de integrale.

Analog pentru b) i c).

b)

c)

Observaie. Integralele a), b), c) sunt ntlnite frecvent, motiv pentru care sunt trecute n tabelul de integrale. Important este ideea de a le calcula (artificiul), dar n probleme se utilizeaz direct din tabel.

B) Obinerea relaiilor de recuren pentru:

Obinerea unei relaii de recuren este indicat atunci cnd avem de calculat o integral care se preteaz de a fi determinat prin recuren.

Aplicaie. S se calculeze

Soluie. Deoarece I se calculeaz aplicnd metoda integrrii prin pri de patru ori, atunci considerm integrala general

care se calculeaz prin pri

,

Atunci:

EMBED Equation.3

care reprezint relaia de recuren.

Dar

;

Dar

Atunci:

Deci:

Tema 1.3.1

S se calculeze integralele urmtoare folosind relaiile de recuren:

1) 2)

3) 4)

5)

C) Metoda identificrii-derivrii pentru sau

Soluie:

Se identific:

Derivnd relaia se obine:

sau:

Prin identificare rezult:

.

Aplicaie. S se calculeze folosind metoda identificrii-derivrii.

Soluie.

.

Prin derivare rezult:

.

Sau:

EMBED Equation.3 Prin identificare se obine:

Atunci:

acelai rezultat ca atunci cnd folosim metoda recurenei.

Tema 1.3.2

Folosind metoda identificrii-derivrii s se calculeze urmtoarele integrale:

1) 2) 3)

4)

Tema 1.3.3.

S se calculeze integralele urmtoare folosind metoda integrrii prin pri:

1) 2)

3) 4)

5) 6)

7)

EMBED Equation.3 8)

9) 10)

11) 12)

13) 14)

15) 16)

17) 18)

19) 20)

21)

1.4. Tipuri de integrale

Tip I .

Substituia:

i (A nu se ine minte!!)

Atunci:

EMBED Equation.3

, unde pentru:

facem substituia

iar se calculeaz direct folosind tabelul.Tip II.

Substituia conduce la:

unde pentru se face substituia: , iar se calculeaz direct folosind tabelul.

Tip III. Substituia i conduce la a), b), c) din tabelul de integrale.

Tip IV. Substituia i conduce la tipul II).Tema 1.4.1.1) 2)

3) 4)

5) 6)

7) 8)

9) 10)

EMBED Equation.3 11) 12)

13) 14)

15) 16)

17) 18)

19) 20)

Tip V. Fracii raionale

A. Integrarea fraciilor simple Principalele fracii simple sunt:Integrarea fraciilor simple

substituia:

substituia:

substituia:

substituia: (conduce la 5)

Se face prin recuren:

4)

i A nu se ine minte!!

Atunci:

=

unde pentru se face substituia: , iar pentru

se face prin recuren (5).5)

Pentru calculul lui se pleac de la: i se integreaz prin pri:

atunci:

S-a obinut:

sau:

(A nu se ine minte!!)Aplicaie

Se pleac de la care se integreaz prin pri.

Atunci:

sau:

sau:

EMBED Equation.3 sau:

Dar:

EMBED Equation.3 Tema 1.4.21) 2)

3) 4)

5) 6)

7)

B. Integrarea fraciilor generale:

BDac grad grad (teorema mpririi)

cu grad grad atunci:

B Dac grad grad, atunci se aplic teoremele: .

Dac , atunci:

Dac atunci:

Pentru fraciile i se aplic recursiv i .

Practic:

se descompune n factori ireductibili de gradul I i II; pentru orice factor de gradul I se aplic pn la epuizarea lor; pentru orice factor de gradul II se aplic pn la epuizarea lor;Aplicaie: S se descompun n fracii simple fracia:

Soluie.

Descompunem pe n factori ireductibili de gradul I i II.

Atunci avem descompunerea n fracii simple:

Tema 1.4.31) S se descompun n fracii simple:

C. Determinarea coeficienilor Pentru determinarea coeficienilor se pot folosi dou metode:- Metoda identificrii

- Metoda valorilor.

Tema 1.4.4.1) ; 2)

3) 4)

5)

Tip VI. Integrale iraionale de forma:

Observaie: .

1) Se consider fraciile:

2) Se determin numitorul lor comun notat cu q

3) Se face substituia:

4) Se calculeaz:

5) Rezult integrala raional.

Tema 1.4.5.1) 2)

3) 4)

5) 6)

7) 8)

EMBED Equation.3 9) 10)

Aplicaie: S se calculeze integrala

Soluie: La prima vedere aceast integral pare c nu este de acest tip.

Dar: , care este de tipul VI. Apar radicalii:

Avem fraciile Numitorul comun

Fac substituia

Dar:

VII. Integrale iraionale de forma:

Substituiile Euler

1) Dac , atunci

Prin ridicare la ptrat se obine:

i

Rezult integral raional:

2) Dac , atunci

Prin ridicare la ptrat rezult:

,

atunci:

i

Rezult integral raional.

3) Dac

Se face substituia

Prin ridicare la ptrat se obine:

Rezult integral raional.

Aplicaie: S se calculeze folosind substituiile lui Euler:

.

Soluie:

Cum facem substituia

Ridicm la ptrat i rezult:

nlocuind n substituie rezult:

.

Atunci:

Facem descompunerea n fracii simple:

Tema 1.4.6

1) - prin toate substituiile;

2) 3)

4) ; 5)

VIII. Integrale iraionale de forma:

, unde polinom de gradul n.

Metoda identificrii-derivrii.

Se identific integrala cu un polinom de grad n-1 de forma:

unde este un polinom de grad n-1 iar .

Derivnd relaia de mai sus se obine:

.

innd cont c:

se obine:

EMBED Equation.3 Rezult:

Prin identificare rezult i

Rmne de calculat doar:

care este de tipul II.

Aplicaie: S se calculeze: .

Soluie:

Dac aplicm metoda identificrii-derivrii rezult:

Prin derivare se obine:

sau:

EMBED Equation.3 Atunci:

Tema 1.4.6.1) 2) 3)

IX. Integrale iraionale de forma:

. Se face substituia

Conduce la tipul VIII

Aplicaie: S se calculeze

Soluie:Facem substituia iar,

Calculm:

EMBED Equation.3

EMBED Equation.3 . Prin nlocuire n integral rezult:

.

Prin derivare:

Fcnd identificarea rezult:

EMBED Equation.3 Atunci:

EMBED Equation.3 Tema 1.4.71) 2)

3) 4)

5) 6)

X. Integrale trigonometrice de forma:

Substituie standard:

EMBED Equation.3 .

Prin nlocuire rezult integrala raional:

Aplicaie: S se calculeze:

Soluie:Facem substituia

;

Atunci:

EMBED Equation.3

Tema 1.4.8.1) 2)

3) 4)

5) 6)

XI. Integrale trigonometrice de forma:

Se rezolv folosind formulele de transformare a produsului n sume:

a)

b)

c)

Aplicaie: S se calculeze:

Tema 1.4.9.1) 2)

3) 4)

5) 6)

7) 8)

9)

XII. Integrale trigonometrice - Substitutii speciale pentru:

1) Dac (impar n ),

atunci se face substituia

2) Dac (impar n ),

atunci se face substituia .

3) Dac (par),

atunci se face substituia

i de reinut formulele!!!:

Prin nlocuire n integral rezult integrala raional.

Aplicaie: S se calculeze

Soluie. Cum , stabilim care substituie este cea mai avantajoas?!!1)

EMBED Equation.3

2)

Facem substituia

Atunci:

3)

Facem substituia

atunci:

Abandon!!!!

Altfel:

EMBED Equation.3 Facem substituia

atunci:

Tem 1.4.10.1) 2)

3) 4)

5) 6)

7) 8)

9)

XIII. Integrala general

a)Dac (impar n cos x) , deci se face substituia

iar

EMBED Equation.3 Se dezvolt dup binomul lui Newton, apoi aplicm tabelul.b) Dac (par n cos x) ATENIE!! (Nu se face substituia )

Se folosesc formulele de trecere la unghi dublu:

Atunci:

Se dezvolt dup binomul lui Newton i se aplic recursiv (a) i (b).

Tema 1.4.11.1) 2)

3) 4)

5) 6)

7)

XIV. Integrala general de forma: .a) Dac (impar n sin x), deci se face substituia

iar

Se aplic binomul lui Newton, apoi se aplic tabelul.b) Dac (par) ATENIE!! (Nu se face substituia )

Se folosesc formulele de trecere la unghi dublu:

Atunci:

Se dezvolt dup binomul lui Newton i se aplic recursiv (a) i (b).

Tema 1.4.12.1) 2) 3)

4) 5) 6)

XV. Integrala general de forma:

Se face substituia special (3)

i rezult:

funcie raional.

Tema 1.4.13.1) 2) 3)

4) 5)

XVI.Integrala general de forma:

a) Dac i (impar n i n ).

Facem substituia sau

b) Dac i (impar n

iar

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3 c)Dac i (impar n ) Substituia este

EMBED Equation.3

EMBED Equation.3

d)Dac i (pare).ATENIE !! (Nu se face substituia ) .Se folosesc formulele de trecere la unghi dublu:

,

atunci:

EMBED Equation.3

Se aplic binomul lui Newton i se aplic recursiv (a), (b), (c) si (d).

Tema 1.4.14.1) 2)

3) 4)

5)

XVII. Integrale generale de forma:

Substituia:

i

Atunci:

Se dezvolt dup binomul Newton, apoi se mparte n n integrale.

Aplicaie: S se calculeze:

Soluie:Facem substituia

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3 Tema 1.4.15.1) 2) 3) 4) 5) 6) .

XVIII. Integrale generale de forma:

Observaie: ,

atunci facem substituia:

iar

Vezi tipul XVII.

Tema 1.4.16.1) 2) 3) 4) 5) 6)

XIX. Integrale generale de forma:

Se folosete metoda recurenei:

, ,

atunci:

EMBED Equation.3 Deci s-a obinut relaia de recuren:

(A nu se ine minte!)

Aplicaie: S se calculeze:

Soluie:

Pentru a evita integrarea prin pri de 4 ori, se pleac de la integrala general:

pentru care se obine o relaie de recuren.

,

atunci:

EMBED Equation.3 ,Unde:

EMBED Equation.3

EMBED Equation.3

Atunci:

Observaie: Astfel de integrale se pot face i prin metoda identificrii care a fost prezentat la metoda integrrii prin pri.

Tema 1.4.17.

1) 2)

XX. Integrale generale de forma:

Se folosete metoda recurenei pentru ambele integrale. Vom exemplifica doar pentru:

pentru care se obine relaia de recuren.

, ,

atunci:

se integreaz prin pri.

, ,

atunci:

atunci:

relaie de recuren. (A nu se ine minte!!)

Aplicaie: S se calculeze: i .

Plecnd de la integrala general

creia i determinm o relaie de recuren

atunci:

a)

se face prin pri:

,

b)

atunci:

i

Observaie: Analog se procedeaz i pentru .

Tema 1.4.181) 2)

3) 4)

XXI. Substituii trigonometrice pentru:a)

b)

c)

Prin aceste substituii integrala se transform ntr-o integral trigonometric.

Aplicaie: S se calculeze:

Soluie:

Facem substituia :

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3 ,

impar n sin x, deci se face substituia

atunci:

integral raional.

Facem descompunerea n fracii simple:

(*)x

y

O

EMBED Equation.3

y=f(x)

(T)

PAGE 58

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