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7/21/2019 Integralni-23_1_2015_A (1)
http://slidepdf.com/reader/full/integralni-2312015a-1 1/14
7/21/2019 Integralni-23_1_2015_A (1)
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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)
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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)
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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 )
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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
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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)
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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 $unkcije:
y. =( ln
x+1
2 x+1 ).
=2 x+1
x+1 ( x+1
2 x+1 ).
=2 x+1
x+1
( x+1 ). (2 x+1 )−( x+1 ) (2 x+1 ).
(2 x+1 )2 =¿
1
x+1
(2 x+1 )−( x+1 ) (2 )2 x+1
=1
x+1
2 x
y. =
−1
( x+1 ) (2 x+1 )
≠ 0, nema ekstremne
Pre%ojne tačke& kon%eksnost'konka%nost:
y.. =( −1
( x+1 ) (2 x+1 ) ).
=0 ( x+1 ) (2 x+1 )−(−1 ) [ ( x+1 ) (2 x+1 ) ]
.
[ ( x+1 ) (2 x+1 ) ]2
=¿ ( x+1 ). (2 x+1 )+( x+1 ) (2 x+1 ).
[ ( x+1 ) (2 x+1 ) ]2
=(2 x+1 )+2 ( x+
[ ( x+1) (2 x+1 ) ]
y.. =
4 x+3
[ ( x+1 ) (2 x+1 ) ]2=0, 4 x+3=0, x≠−
3
4 , nema prevojne
−(−1−1
2+(
−1 −¿ −¿
( x+1 ) −¿ +¿
(2 x+1 ) −¿ +¿
y. −¿ −¿
tok ↘ ↘
7/21/2019 Integralni-23_1_2015_A (1)
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(ra$:
−(−1−3
4−
1
2+(
4 x+3 −¿ +¿
[ ( x+
1 ) (2 x+
1 ) ]
2 +¿ +¿
y.. −¿ +¿
ykonv / konk 0 ∪
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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
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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
¿
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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 ).
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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)
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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