Integralni-23_1_2015_A (1)

7/21/2019 Integralni-23_1_2015_A (1)

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7/21/2019 Integralni-23_1_2015_A (1)


1. Indukcijska baza: (n=2 )

¿

(cos x+isin x )2=cos x2 +i2cos x sin x−sin x

2,

adicijske formule :sin (a±b )=¿ sin acosb ± cos a sin bcos (a± b )=cos acosb∓sin

¿

x−¿ sin xsin x

cos x cos¿¿¿¿

x=¿

x−sin x22 +i 2cos xsin ¿cos¿

2. Indukcijska hipoteza: (n=k )

(cos x+isin x )k =cos kx+ isin kx,k ∈ Z

3. Dokaz: (n=k +1 )

(cos x+isin x )k +1=cos [ (k +1 ) x ]+i sin [ (k +1 ) x ] ,

x cos kx−¿sin xsin kx+i (sin xcos kx+cos x sin kx )=¿cos ( x+kx )+i sin ( x+kx )=cos [ ( k +1 ) x ]+isin [ (k +1 ) x ] ,tv(cos x+i sin x ) (cos x+i sin x )k

=(cos x+ isin x ) (cos kx+ isin kx )=¿cos ¿

7/21/2019 Integralni-23_1_2015_A (1)


Riješiti matričnu jednadžbu:

(2 1 1

1 1 −2

4 3 −2) ∙( x y

z)=(10

2) , A=(2 1 1

1 1 −2

4 3 −2) , X =( x y

z) ,B=(10

2) ,

A ∙ X =B A ∙ X =B/→ ∙ ( A−1 ) , s lijeve strane ,det ( A )≠ 0 A−1

∙ A ∙ X = A−1

∙B

X = A−1

∙ B

A−1=

1

det ( A ) adj ( A )

det ( A )=|2 1 1

1 1 −2

4 3 −2|2 1

1 1

4 3⏟

!arus

=+ (−4−8+3)−(4−12−2 )=−9+10=1≠ 0

A" =

(

2 1 1

1 1 −2

4 3 −2

)

"

=

(

2 1 4

1 1 3

1 −2 −2

)=

(

2 1 4

1 1 3

1 −2 −2

)

7/21/2019 Integralni-23_1_2015_A (1)


adj ( A )=(|2 1 4

1 1 31 −2 −2| |

2 1 4

1 1 31 −2 −2| |

2 1 4

1 1 31 −2 −2|# # #

|2 1 4

1 1 3

1 −2 −2| |2 1 4

1 1 3

1 −2 −2| |2 1 4

1 1 3

1 −2 −2|

# # #

|2 1 4

1 1 3

1 −2 −2| |2 1 4

1 1 3

1 −2 −2| |2 1 4

1 1 3

1 −2 −2|)$ovajdio sene pi%e ,

tu je demonstrativno =¿

¿

(

+| 1 3

−2 −2| −|1 3

1 −2| +|1 1

1 −2|−| 1 4

−2 −2| +|2 4

1 −2| −|2 1

1 −2|+|

1 4

1 3

| −|2 4

1 3

| +|2 1

1 1

|

)=¿

¿(+[ (−2 )−(−6 ) ] −[ (−2 )−(3 ) ] +[ (−2 )−(1 ) ]−[ (−2 )−(−8 ) ] +[ (−4 )−( 4 ) ] −[ (−4 )−(1 ) ]+ [(3 )−(4 ) ] −[ (6 )−(4 ) ] +[ (2 )−(1 ) ] )=¿

¿( 4 −(−5 ) −3

−6 −8 −(−5 )−1 −2 1 )=(

4 5 −3

−6 −8 5

−1 −2 1 )

A−1=

1

det ( A ) adj ( A )=

1

1 ( 4 5 −3

−6 −8 5

−1 −2 1 )=( 4 5 −3

−6 −8 5

−1 −2 1 )

7/21/2019 Integralni-23_1_2015_A (1)


X = A−1

∙ B=( 4 5 −3

−6 −8 5

−1 −2 1 )

3 x 3

∙(1

0

2)

3 x 1

=( 4 ∙ 1+5∙ 0+(−3 ) ∙2

(−6 )∙ 1+(−8 ) ∙0+5 ∙ 2

(−1 )∙ 1+(−2 )∙ 0+1 ∙2)

3 x 1

=¿(− 4+0−6

6−0+10

−1−0+2)

3 x 1

=(−2

4

1 )

3 x 1

X =( x y z)=(−2

4

1 ) , x=−2, y=4, z=1

y=ln x+1

2 x+1

7/21/2019 Integralni-23_1_2015_A (1)


Domen:

x+1

2 x+1>0,

&om ( f )={∀ x ∈ ' :( x ∈ ⟨−( ,−1 ⟩∪⟨−1

2 ,+(⟩)}

Parnost i periodičnost:

f (− x )=ln − x+1

−2 x+1=ln

−( x−1 )−

(2 x−1

)

=ln x−1

2 x−1, nije parnainije neparna

nematri)onometrijske elemente pa nije periodi*na

u!e i znak:

¿

ln x+1

2 x+1=0,

x+1

2 x+1=1,

x+1

2 x+1−1=0,

x+1−2 x−1

2 x+1 =0,

− x

2 x+1=0, x=0¿

¿

y + 0 za x ∈ ⟨−1

2 , 0 ]

y<0 za x∈ ⟨−( ,−1 ⟩∪ ⟨ 0,+( ⟩

"simptote:

vertikalne :

ln x+1

2 x+1= ## } ln {left ({0} over {1} right )} mtri! {

¿ =−(

x →−1

2+¿ ln

x+1

2 x+1= ## } ln {left ({ {1} over {2} } over {0} right )} mtri! {

¿=+(¿

x→−1−¿¿ lim¿¿

¿lim¿

¿

¿

−(−1−12+(

x+1 −¿ +¿ +¿

2 x+1 −¿ −¿ +¿

&om>0 +¿ −¿ +¿

−(−1−1

2 0+(

− x +¿ +¿ −¿

2 x+1 −¿ +¿ +¿

znak −¿ +¿ −¿

7/21/2019 Integralni-23_1_2015_A (1)


orizontalne:

## } ln {left ({ " } over {"} right )} mtri! {

ln x+1

2 x+1=

lim x →−(

¿ 1

2= ln

1

2

¿=ln lim x →−(

¿ x+1

2 x+1 -

. /ospital

⇒

ln¿

¿

ln x+1

2 x+1=

## } ln {left ({ " } over {"} right )} mtri! {

¿=ln lim x→+(

¿ x+1

2 x+1 -

. /ospital

⇒

ln lim x→+(

¿1

2=ln

1

2¿¿

lim x →−(

¿ lim x →+(

¿

¿

nemakosu, jer imaorizontalne za ( x→−( )i ( x→+( )

#kstremne tačke i tok $unkcije:

y. =( ln

x+1

2 x+1 ).

=2 x+1

x+1 ( x+1

2 x+1 ).

=2 x+1

x+1

( x+1 ). (2 x+1 )−( x+1 ) (2 x+1 ).

(2 x+1 )2 =¿

1

x+1

(2 x+1 )−( x+1 ) (2 )2 x+1

=1

x+1

2 x

y. =

−1

( x+1 ) (2 x+1 )

≠ 0, nema ekstremne

Pre%ojne tačke& kon%eksnost'konka%nost:

y.. =( −1

( x+1 ) (2 x+1 ) ).

=0 ( x+1 ) (2 x+1 )−(−1 ) [ ( x+1 ) (2 x+1 ) ]

.

[ ( x+1 ) (2 x+1 ) ]2

=¿ ( x+1 ). (2 x+1 )+( x+1 ) (2 x+1 ).

[ ( x+1 ) (2 x+1 ) ]2

=(2 x+1 )+2 ( x+

[ ( x+1) (2 x+1 ) ]

y.. =

4 x+3

[ ( x+1 ) (2 x+1 ) ]2=0, 4 x+3=0, x≠−

3

4 , nema prevojne

−(−1−1

2+(

−1 −¿ −¿

( x+1 ) −¿ +¿

(2 x+1 ) −¿ +¿

y. −¿ −¿

tok ↘ ↘

7/21/2019 Integralni-23_1_2015_A (1)


(ra$:

−(−1−3

4−

1

2+(

4 x+3 −¿ +¿

[ ( x+

1 ) (2 x+

1 ) ]

2 +¿ +¿

y.. −¿ +¿

ykonv / konk 0 ∪

7/21/2019 Integralni-23_1_2015_A (1)


7/21/2019 Integralni-23_1_2015_A (1)


1 =∫ e2 x

sin x dx=| u=e2 x

du=2e2 x

dx

dv=sin xdx v=−cos x |=e2 x (−cos x )−2∫ e

2 x (−cos x )dx⏟

1 1

1 1=∫ e2 x (−cos x ) dx=∫−e

2 xcos x dx=| u=−e

2 xdu=−2e

2 xdx

dv=cos x dx v=sin x |=¿−e2 x

sin x−∫−2e2 x

sin x dx=−e2 x

1 =e2 x (−cos x )−2 1 1=−e

2 xcos x−2 (−e

2 xsin x+2 1 )=−e

2 xcos x+2e

2 xsin x−4 1

1 =e2 x (−cos x+2sin x )−4 1 1 +4 1 =e

2 x (2sin x−cos x ) 5 1 =e2 x (2sin x−cos x )

1 =e

2 x

5(2sin x−cos x )+2

7/21/2019 Integralni-23_1_2015_A (1)


Izračunati površinu zatvorenu grafovima funkcijama:

y1=t) x y1=sin x x=0 x=1

zajednička tačka:

t) x=sin x

sin x

cos x=sin x

cos x=1⇒

x=0

povr%inanije isje*enai *itava senalazina podru*ju y + 0, pa inte)ralne trebamorazdvajatina vi%e dijelova zb

1

¿

1

¿

¿¿¿¿

0

=¿

¿|cos x|¿¿

¿

| z|¿¿¿¿

0

=−ln ¿

¿−1

z dz=−ln ¿

1 1=∫0

1

t)xdx=∫0

1sin x

cos x=|cos x= z

−sin x dx=dz|=∫01

¿

7/21/2019 Integralni-23_1_2015_A (1)


1

¿

¿¿¿¿

0

=−(cos1−cos0 )=−¿

¿ x¿¿

x dx=−cos ¿sin ¿

1 2=∫0

1

¿

3= 1 1− 1 2=−ln|cos1|−(−cos1+1 )=−ln|cos1|+cos 1−1=0,1559

Računati (trigonometrijske) u radijanima !

Ispitati monotonost:

y=arccos x−√ 1− x2

{ x∈ [−1,1 ] }0 {1− x2>0, x

2<1, ( x<10 x>−1 ) }

&om ( f )= ∀ x ∈ ' : ( x ∈ ⟨−1,1 ⟩ )}

y. =(arccos x ). −(√ 1− x

2 ). ,tabli*niizvod : (arccos x ). = −1

√ 1+ x2

y. = −1

√ 1− x2−

1

2

1

√ 1− x2(1− x

2 ).

7/21/2019 Integralni-23_1_2015_A (1)


y. = −1

√ 1− x2−

1

2

1

√ 1− x2

(−2 x )=−1+ x

√ 1− x2

y.

=−1+ x

√ 1− x2=0,−1+ x=0, x≠ 1

funkcija raste neprekidno

pr%i i dru)i iz%od:

y=ln ( x+√ 1+ x2 )

y. =[ ln ( x+√ 1+ x2 )]

.

y. =

1

x+√ 1+ x2( x+√ 1+ x2 ).

−(−11+(

−1− x +¿

√ 1− x2 +¿

y. +¿

tok ↗

7/21/2019 Integralni-23_1_2015_A (1)


y. =

1+ 1

2√ 1+ x2(1+ x

2 ).

x+√ 1+ x2

y. =

1+1

2√ 1+ x2(2 x )

x+√ 1+ x2=

2√ 1+ x2+2 x

2√ 1+ x2

x+√ 1+ x2

=2 (√ 1+ x

2+ x )2√ 1+ x2 ( x+√ 1+ x2 )

=1

√ 1+ x2

y.. =( 1

√ 1+ x2 ).

y.. =

0−(√ 1+ x2 ).

(√ 1+ x2 )

2 =

−(√ 1+ x2).

1+ x2

y.. =

− 1

2√ 1+ x2(1+ x

2 ).

1+ x2

= −1

2 (1+ x2 )√ 1+ x2

y.. =

− 1

2√ 1+ x2(2 x )

1+ x2

= −2 x

2 (1+ x2 )√ 1+ x2=

− x

( 1+ x2 )√ 1+ x2

Integralni-23_1_2015_A (1)

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