Odredeni integrali

Osnovne osobine odredenog integrala:

10

∫ a

af(x)dx = 0 ,

20

∫ b

af(x)dx = −

∫ a

bf(x)dx ,

30

∫ b

af(x)dx =

∫ c

af(x)dx+

∫ b

cf(x)dx ,

40

∫ b

a

(αf(x) + βg(x)

)dx = α

∫ b

af(x)dx+ β

∫ b

ag(x)dx ,

50

∫ b

af(x)dx = F (x)

∣∣∣b

a= F (b)− F (a), gde je F ′(x) = f(x),

60 Ako je f parna funkcija(f(x) = f(−x), ∀x ∈ R), tada je

∫ a

−af(x)dx = 2

∫ a

0f(x)dx,

70 Ako je f neparna funkcija(f(x) = −f(−x), ∀x ∈ R), tada je∫ a

−af(x)dx = 0,

80 Ako je f periodicna funkcija sa periodom T(f(x+ T ) = f(x), ∀x ∈ R),

tada je∫ a+T

af(x)dx =

∫ T

0f(x)dx.

Cuvajmo drvece. Nemojte stampati ovaj materijal, ukoliko to nije neophodno.

1

2

Osnovne metode integracije:

METOD SMENE

∫ b

af(x)dx =

∣∣∣∣x = u(t) t = ϕ(x)α = ϕ(a) β = ϕ(b)

∣∣∣∣ =∫ β

αf(u(t))u′(t)dt.

METOD PARCIJALNE INTEGRACIJE

∫ b

audv = uv

∣∣∣b

a−∫ b

avdu.

Primena odredenog integrala:

POVRSINA RAVNE FIGURE

Neka je f(x) ≥ 0 za x ∈ [a, b]. Povrsinakrivolinijskog trapeza ogranicenog krivomy = f(x), pravama x = a, x = b i x−osom,iznosi

S =∫ b

af(x)dx.

a b

f HxL

S

Ukoliko je f(x) ≤ 0 za x ∈ [a, b], tada je

S = −∫ b

af(x)dx.

f HxL

S

a b

S =∫ b

a|f(x)|dx.

f HxL

Sa b

3

DUZINA LUKA KRIVE

Duzina luka krive f(x) od tacke na grafikusa apscisom a do tacke na grafiku sa apsci-som b, iznosi

l =∫ b

a

√1 +

(f ′(x)

)2dx.

f HxL

a b

l

POVRSINA I ZAPREMINA ROTACIONOG TELA

Povrsina i zapremina tela nastalog rotacijom dela luka krive f(x) oko ko-ordinatnih osa izracunavaju se sa:

Vx = π

∫ b

af(x)2dx,

Px = 2π∫ b

af(x)

√1+(f ′(x)

)2dx.

Vy = 2π∫ b

axf(x)dx

= π

∫ d

cx(y)2dy,

Py = 2π∫ d

cx(y)

√1+(x′(y)

)2dy.

Zadaci

1. Odrediti sledece integrale (a > 0) :

10

∫ 1

0

3√x(x+ 1)3dx 20

∫ 3

1

dx

x230

∫ 1

0

dx

1 + x2

40

∫ 65

2

dx3√x− 1

(6√x− 1 +

√x− 1

) 50

∫ a

0

√a2 − x2 dx 60

∫ 4

2

√x+ 2x− 1

dx

4

Resenje: 10

∫ 1

0

3√x(x+ 1)3dx =

∫ 1

0x1/3(x3 + 3x2 + 3x+ 1)dx

=∫ 1

0(x10/3 + 3x7/3 + 3x4/3 + x1/3)dx =

(x13/3

13/3+ 3 · x

10/3

10/3+ 3 · x

7/3

7/3+x4/3

4/3

)∣∣∣1

0

=313

+910

+97

+34.

20

∫ 3

1

dx

x2= −1

x

∣∣∣3

1= −1

3+ 1 =

23.

30

∫ 1

0

dx

1 + x2= arctanx

∣∣∣1

0=π

4.

40

∫ 65

2

dx3√x− 1

(6√x− 1 +

√x− 1

) =∣∣∣∣t6 = x− 1 dx = 6t5dt2 7→ 6

√2− 1 = 1 65 7→ 6

√65− 1 = 2

∣∣∣∣

= 6∫ 2

1

t2

1 + t2dt = 6

∫ 2

1

t2 + 1− 11 + t2

dt = 6∫ 2

1dt− 6

∫ 2

1

dt

1 + t2

= 6t∣∣∣2

1− 6 arctan t

∣∣∣2

1=

3π2

+ 6(1− arctan 2).

50

∫ a

0

√a2−x2 dx =

∣∣∣∣x=a sin t dx=a cos t dt0 7→ 0 a 7→ π/2

∣∣∣∣=a

∫ π/2

0

√a2−a2 sin2 t cos t dt

= a2

∫ π/2

0cos2 t dt=a2

∫ π/2

0

1+cos 2t2

dt=a2

2t∣∣∣π/2

0− a

2

4sin 2t

∣∣∣π/2

0=a2π

4.

5

60

∫ 4

2

√x+ 2x− 1

dx =

∣∣∣∣∣t =

√x+2x−1 dx = − 6t dt

(t2−1)2

2 7→ 2 4 7→ √2

∣∣∣∣∣ = −6∫ √2

2

t2

(t2 − 1)2dt

= 6∫ 2

√2

t2

(t2 − 1)2dt =

∣∣∣∣∣u = t dv = t dt

(t2−1)2

du = dt v = − 12(t2−1)

∣∣∣∣∣

= − 3tt2 − 1

∣∣∣2

√2

+ 3∫ 2

√2

dt

t2 − 1= 3√

2− 2 +32

log∣∣∣ t− 1t+ 1

∣∣∣∣∣∣∣2

√2

= 3√

2− 2 + 3 log√

2 + 1√3

.

2. Odrediti sledece integrale (n ∈ N) :

10

∫ π

0sinx sin 3x dx 20

∫ π/2

0cosx sin

x

2dx 30

∫ π/3

π/6

dx

sin2 x cos4 x

40

∫ π/6

0(2x−1) sin 3x dx 50

∫ π

−πx2 cosnx dx 60

∫ π/4

0

sinx cosx dx2 sin2 x−5 cos2 x

Resenje: 10

∫ π

0sinx sin 3x dx=

12

∫ π

0(cos 2x− cos 4x)dx

=14

sin 2x∣∣∣π

0− 1

8sin 4x

∣∣∣π

0= 0.

20

∫ π/2

0cosx sin

x

2dx=

12

∫ π/2

0

(sin

3x2−sin

x

2

)dx=cos

x

2

∣∣∣π/2

0− 1

3cos

3x2

∣∣∣π/2

0

=23

(√

2− 1) .

30

∫ π/3

π/6

dx

sin2 x cos4 x=∫ π/3

π/6

sin2 x+ cos2 x

sin2 x cos2 x

dx

cos2 x=

∣∣∣∣∣1

cos2 x= 1 + tan2 x

1sin2 x

= 1+tan2 xtan2 x

∣∣∣∣∣

=∫ π/3

π/6

(2 + tan2 x+

1tan2 x

)d(tanx)

=(

2 tanx+13

tan3 x− 1tanx

)∣∣∣π/3

π/6=

80√

327

.

6

40

∫ π/6

0(2x− 1) sin 3x dx =

∣∣∣∣u = 2x− 1 dv = sin 3xdu = 2dx dx v = −1

3 cos 3x

∣∣∣∣

=1− 2x

3cos 3x

∣∣∣π/6

0+

23

∫ π/6

0cos 3x dx = −1

3+

29

sin 3x∣∣∣π/6

0= −1

9.

50

∫ π

−πx2 cosnx dx = 2

∫ π

0x2 cosnx dx =

∣∣∣∣u = x2 dv = cosnx dxdu = 2x dx v = 1

n sinnx

∣∣∣∣

=2nx2 sinnx

∣∣∣π

0− 4n

∫ π

0x sinnx dx = − 4

n

∫ π

0x sinnx dx

=∣∣∣∣u=x dv=sinnx dxdu=dx v=− 1

n cosnx

∣∣∣∣=4n2x cosnx

∣∣∣π

0− 4n2

∫ π

0cosnx dx

= (−1)n4πn2− 4n3

sinnx∣∣∣π

0= (−1)n

4πn2.

60

∫ π/4

0

sinx cosx dx2 sin2 x− 5 cos2 x

=∣∣∣∣t=2 sin2 x−5 cos2 x dt=14 sinx cosx dx0 7→ −5 π/4 7→ −3/2

∣∣∣∣

=114

∫ −3/2

−5

dt

t=

114

log |t|∣∣∣−3/2

−5= − 1

14log

103.

3. Odrediti sledece integrale:

10

∫ π

0sin2 x dx 20

∫ π

0

sin2 x

2(1+cosx)dx

30

∫ π

0

√sinx−sin3x dx 40

∫ π/2

−π/2

√cosx−cos3x dx

50

∫ π/2

0

dx

(2+cosx)(3+cosx)60

∫ 1007π

0

√1−cos 2x dx

Resenje: 10

∫ π

0sin2 x dx =

∫ π

0

1−cos 2x2

dx =12x∣∣∣π

0− 1

4sin 2x

∣∣∣π

0=π

2.

7

20

∫ π

0

sin2 x

2(1+cosx)dx=

∫ π

0

sin2 x2 cos2 x

2

cos2 x2

dx=∫ π

0sin2x

2dx=

∫ π

0

1−cosx2

dx=π

2.

30

∫ π

0

√sinx−sin3x dx =

∫ π

0

√sinx(1−sin2x) dx =

∫ π

0

√sinx| cosx| dx

=∫ π/2

0

√sinx cosx dx−

∫ π

π/2

√sinx cosx dx=

∣∣∣∣t=sinx dt=cosx dx0 7→ 0, π/2 7→ 1, π 7→ 0

∣∣∣∣

=∫ 1

0

√t dt−

∫ 0

1

√t dt = 2

∫ 1

0

√t dt = 2

t3/2

3/2

∣∣∣1

0=

43.

40

∫ π/2

−π/2

√cosx−cos3x dx =

43.

50

∫ π/2

0

dx

(2+cosx)(3+cosx)=∫ π/2

0

(3+cosx)−(2+cosx)(2+cosx)(3+cosx)

dx

=∫ π/2

0

dx

2+cosx−∫ π/2

0

dx

3+cosx=∣∣∣∣t = tan x

2 dx = 2dt1+t2

0 7→ 0 π/2 7→ 1

∣∣∣∣

= 2∫ 1

0

dt

3 + t2− 2

∫ 1

0

dt

4 + 2t2=

2√3

arctant√3

∣∣∣1

0− 1√

2arctan

t√2

∣∣∣1

0

=π

3√

3− 1√

2arctan

1√2.

60

∫ 1007π

0

√1−cos 2x dx=

∫ 1007π

0

√2 | sinx|dx = 1007

√2∫ π

0sinx dx

=−1007√

2 cosx∣∣∣π

0= 2014

√2 .

8

4. Odrediti sledece integrale:

10

∫ π/4

0

3 tanx+ 13− tanx

dx 20

∫ π/4

0tan2 x dx

30

∫ 1

0

x log(x+√

1 + x2)√1 + x2

dx 40

∫ 6

4x log

2 + x

2− x dx

50

∫ log 5

0

ex√ex − 1

ex + 2dx 60

∫ 1

0

ex√ex + 3

3√ex + 3

dx

Resenje: 10

∫ π/4

0

3 tanx+ 13− tanx

dx =∣∣∣∣t = tanx dx = dt

1+t2

0 7→ 0 π/4 7→ 1

∣∣∣∣

=∫ 1

0

3t+ 1(3− t)(1 + t2)

dt =∫ 1

0

t2 + 1 + t(3− t)(3− t)(1 + t2)

dt

=∫ 1

0

dt

3− t +∫ 1

0

t dt

t2 + 1= − log |3− t|

∣∣∣1

0+

12

log |t2 + 1|∣∣∣1

0

= log3√2.

20

∫ π/4

0tan2 x dx =

∫ π/4

0

1− cos2 x

cos2 xdx = (tanx− x)

∣∣∣π/4

0= 1− π

4.

30

∫ 1

0

x log(x+√

1 + x2)√1 + x2

dx =

∣∣∣∣∣∣u = log(x+

√1 + x2) dv = x dx√

1+x2

du = dx√1+x2

v =√

1 + x2

∣∣∣∣∣∣

=√

1 + x2 log(x+√

1 + x2)∣∣∣1

0−∫ 1

0dx =

√2 log(1 +

√2)− 1.

40

∫ 6

4x log

x+2x−2

dx=

∣∣∣∣∣u=log x+2

x−2 dv=x dx

du= 4dx4−x2 v= x2

2

∣∣∣∣∣=x2

2log

x+2x−2

∣∣∣6

4+2∫ 6

4

x2

x2−4dx

= 18 log 2− 8 log 3 + 2∫ 6

4

x2 − 4 + 4x2 − 4

dx

= 18 log 2− 8 log 3 + 2x∣∣∣6

4− 2 log

x+ 2x− 2

∣∣∣6

4= 16 log 2− 6 log 3 + 4.

9

50

∫ log 5

0

ex√ex − 1

ex + 2dx =

∣∣∣∣t =√ex − 1 2t dt = exdx

0 7→ 0 log 5 7→ 2

∣∣∣∣ = 2∫ 2

0

t2

t2 + 3dt

= 2∫ 2

0

t2 + 3− 3t2 + 3

dt = 2(t−√

3 arctant√3

)∣∣∣2

0= 4− 2

√3 arctan

2√3.

60

∫ log 2

0

ex√ex + 3

3√ex + 3

dx=∣∣∣∣t6 = ex + 3 6t5 dt = exdx

0 7→ 6√

4 log 2 7→ 6√

5

∣∣∣∣ = 6∫ 6√5

6√4t6dt

=67t7∣∣∣

6√5

6√4=

67(5 6√

5− 4 6√

4).

5. Izracunati vrednost integrala

I1 =∫ +∞

−∞f(x) cosx dx, I2 =

∫ +∞

−∞f(x) sinx dx,

pri cemu je funkcija f(x) data sa

f(x) ={

1− |x|, −1 ≤ x ≤ 1,0, x < −1 ∨ x > 1.

Resenje: Kako je f(x) = 0 za x /∈ [−1, 1], to je onda i

f(x) cosx = 0, f(x) sinx = 0 za x /∈ [−1, 1].

Integral I1 tada glasi

I1 =∫ +∞

−∞f(x) cosx dx =

∫ 1

−1f(x) cosx dx =

∫ 1

−1

(1− |x|) cosx dx

= 2∫ 1

0

(1− x) cosx dx = 2

∫ 1

0cosx dx− 2

∫ 1

0x cosx dx

= 2 sinx∣∣∣1

0− 2

∫ 1

0x cosx dx =

∣∣∣∣u = x dv = cosx dxdu = dx v = sinx

∣∣∣∣

= 2 sin 1− 2(x sinx

∣∣∣1

0−∫ 1

0sinx dx

)= −2 cosx

∣∣∣1

0= 2(1− cos 1).

10

S obzirom da je f(x) sinx neparna funkcija, to je I2 = 0.

6. Izracunati integrale∫ π

−πcosnx cosmxdx, i

∫ π

−πsinnx sinmxdx,

gde su n,m ∈ N i vazi

a) n = m = 0; b) n = m 6= 0; c) n 6= m.

Resenje: Oznacimo

In,m =∫ π

−πcosnx cosmxdx, Jn,m =

∫ π

−πsinnx sinmxdx.

a) I0,0 =∫ π

−πdx = 2π, J0,0 = 0.

b) In,n=∫ π

−πcos2 nx dx = 2

∫ π

0cos2 nx dx =

∫ π

0(1 + cos 2nx)dx

=(x+

12n

sin 2nx)∣∣∣π

0= π.

Jn,n=∫ π

−πsin2 nx dx = 2

∫ π

0sin2 nx dx =

∫ π

0(1− cos 2nx)dx

=(x− 1

2nsin 2nx

)∣∣∣π

0= π.

c) In,m=∫ π

−πcosnx cosmxdx =

∫ π

0

(cos(n−m)x+ cos(n+m)x

)dx

=1

n−m sin(n−m)x∣∣∣π

0+

1n+m

sin(n+m)x∣∣∣π

0= 0.

Jn,m=∫ π

−πsinnx sinmxdx =

∫ π

0

(cos(n−m)x− cos(n+m)x

)dx

=1

n−m sin(n−m)x∣∣∣π

0− 1n+m

sin(n+m)x∣∣∣π

0= 0.

11

7. Dat je integral In =∫ e

1x(log x)n dx, n ∈ N0.

a) Izracunati I0 i I1.

b) Odrediti an i bn u relaciji In = anIn−1 + bn.

c) Naci vrednost integrala I3.

Resenje: a) I0 =∫ e

1xdx =

x2

2

∣∣∣e

1=e2 − 1

2.

I1 =∫ e

1x log x dx =

∣∣∣∣u = log x dv = x dx

du = dxx v = x2

2

∣∣∣∣ =x2

2log x

∣∣∣e

1− 1

2

∫ e

1x dx

=e2 + 1

4.

b) In =∫ e

1x(log x)n dx =

∣∣∣∣u = (log x)n dv = x dx

du = n(log x)n−1 dxx v = x2

2

∣∣∣∣

=x2

2(log x)n

∣∣∣e

1− n

2In−1 =

e2

2− n

2In−1,

an =−n2, bn =

e2

2.

c) I3 =e2

2− 3

2I2 =

e2

2− 3

2

(e2

2− I1

)=e2 + 3

8.

8. Odrediti koeficijente an i bn u relaciji In+2 = anIn+1 + bnIn, n ∈ N0, gde je

In =∫ 1

0(1− x2)n/2dx, n ∈ N0.

Izracunati vrednost integrala I3.

12

Resenje: In+2 =∫ 1

0(1− x2)

n+22 dx =

∫ 1

0(1− x2)(1− x2)

n2 dx

=In−∫ 1

0x2(1− x2)

n2 dx =

∣∣∣∣∣u=x dv=x(1−x2)

n2 dx

du=dx v=− 1n+2(1−x2)

n+22

∣∣∣∣∣

=In +x

n+ 2(1− x2)

n+22

∣∣∣1

0− 1n+ 2

In+2 = In − 1n+ 2

In+2.

n+ 3n+ 2

In+2 = In =⇒ In+2 =n+ 2n+ 3

In

an = 0, bn =n+ 2n+ 3

.

I1 =∫ 1

0

√1− x2 dx =

∣∣∣∣x = sin t dx = cos t dt0 7→ 0 1 7→ π/2

∣∣∣∣ =∫ π/2

0cos2 t dt

=12

∫ π/2

0(1 + cos 2t)dt =

π

4.

I3 =34I1 =

3π16.

9. Dati geometrijsku interpretaciju odredenog integrala, a zatim izracunatipovrsinu figure ogranicene lukom krive f(x) = e

√x − e−

√x, pravom x = 1 i

x−osom.

Resenje:

1

f HxL

S

S =∫ 1

0

(e√x−e−

√x)dx=

∣∣∣∣t =√x 2t dt = dx

0 7→ 0 1 7→ 1

∣∣∣∣

= 2∫ 1

0t(et−e−t)dt=

∣∣∣∣u= t dv=

(et−e−t)dt

du=dt v=et+e−t

∣∣∣∣

= 2t(et+e−t

)∣∣∣1

0− 2

∫ 1

0

(et+e−t

)dt

= 2e+2e− 2(et−e−t)

∣∣∣1

0=

4e.

10. Izracunati povrsinu figure ogranicene linijama y = log x, y = 1, x = 4,kao i zapreminu tela nastalog rotacijom figure oko x−ose.

13

Resenje:

S =∫ 4

1

(f(x)− 1

)dx =

∫ 4

1

(log x− 1

)dx =

∫ 4

1log x dx−

∫ 4

1dx

=∣∣∣∣u = log x dv = dx

du = dxx v = x

∣∣∣∣ = x log x∣∣∣4

1− 3− x

∣∣∣4

1= 4 log 4− 6.

Vx = π

∫ 4

1

(f(x)−1

)2dx=π

∫ 4

1

(log x−1

)2dx=

∣∣∣∣u=log x−1 dv=(log x−1)dxdu= dx

x v=x(log x−2)

∣∣∣∣

= πx(log x− 1)(log x− 2)∣∣∣4

1− π

∫ 4

1(log x− 2)dx = 4(log 4− 2)2π − 1.

11. Odrediti povrsinu figure ogranicene lukom krive

f(x) =1√

x2 + x− 2, x ∈ [2, 4]

i x−osom, a zatim izracunati zapreminu tela nastalog rotacijom tog luka okox−ose.

Resenje:

S =∫ 4

2

dx√x2 + x− 2

=∫ 4

2

dx√(x+ 2)(x− 1)

=

∣∣∣∣∣t(x+2)=

√(x+2)(x−1) dx= 6t dt

(1−t2)2

2 7→ 1/2 4 7→ 1/√

2

∣∣∣∣∣

= 2∫ 1/

√2

1/2

dt

1− t2 = log1 + t

1− t

∣∣∣∣1/√

2

1/2

= log(

1+2√

23

).

14

Vx = π

∫ 4

2

dx

x2 + x− 2=π

3

∫ 4

2

x+ 2− (x− 1)(x+ 2)(x− 1)

dx

=π

3

∫ 4

2

dx

x− 1− π

3

∫ 4

2

dx

x+ 2=π

3log

x− 1x+ 2

∣∣∣∣4

2

=π

3log 2.

12. Data je funkcija

f(x) =x

1 + x2.

Izracunati povrsinu figure ogranicene krivom f(x), x−osom i pravom x = 1.Izracunati zapreminu tela nastalog rotacijom date figure oko x−ose.

Resenje:

S =∫ 1

0

x dx

1 + x2=

12

log(1 + x2)∣∣∣∣1

0

=12

log 2.

Vx = π

∫ 1

0

x2

(1 + x2)2dx =

∣∣∣∣∣u = x dv = x dx

(1+x2)2

du = dx v = −12(1+x2)

∣∣∣∣∣

=−π x

2(1 + x2)

∣∣∣1

0+π

2

∫ 1

0

dx

1 + x2

=−π4

+π

2arctanx

∣∣∣1

0=π2

8− π

4.

13. Izracunati povrsinu figure ogranicene lukovima krivih f1(x) = x2 − 2x +2, f2(x) = x2 + 4x + 5 i y = 1. Izracunati zapreminu tela nastalog rotacijomoko x−ose ove figure.

Resenje: Odredimo najpre presecnu tacku grafika krivih f1 i f2 :

x2 − 2x+ 2 = x2 + 4x+ 5 ⇐⇒ x = −1/2.

Primetimo da je slika simetricna u odnosu na pravu x = −1/2.

15

S =∫ −1/2

−2f2(x)dx+

∫ 1

−1/2f1(x)dx

= 2∫ 1

−1/2

(x2 − 2x+ 2

)dx

= 2(x3

3− x2 + 2x

)∣∣∣1

−1/2=

214.

Vx = 2π∫ 1

−1/2f1(x)2dx = 2π

∫ 1

−1/2

(x2 − 2x+ 2

)2dx

= 2π∫ 1

−1/2

((x− 1)2 + 1

)2dx =

843π80

.

14. Izracunati zapreminu tela nastalog rotacijom luka krive y(x) = cos2 x, x ∈[−π/2, π/2] oko x−ose.

Resenje:

Vx = π

∫ π/2

−π/2cos4 x dx = 2π

∫ π/2

0cos4 x dx =

π

2

∫ π/2

0(1 + cos 2x)2dx

=π

2

∫ π/2

0

(1 + 2 cos 2x+

1 + cos 4x2

)dx =

3π2

8.

15. Izracunati zapreminu tela nastalog rotacijom oko x−ose figure ogranicenelukom krive f(x) = x log x i x−osom.

Resenje: Kako je

limx→0

x log x = limx→0

log x1/x

= limx→0

1/x−1/x2

= 0,

i f(1) = 0, to posmatramo deo luka krive f za x ∈ (0, 1].

Vx = π

∫ 1

0x2 log2 x dx =

2π27

.

16. Izracunati zapreminu tela koje nastaje rotacijom figure ogranicene de-lovima krivih y = x2 + 1 i y = 2 oko x−ose.

16

Resenje: Vx = π

∫ 1

−1

(22 − (x2 + 1)2

)dx =

64π15

.

17. Izracunati zapreminu tela nastalog rotacijom oko x−ose figure ogranicenelinijama

y = e3x + ex, x = 0, y = e3 + e.

Resenje:

Vx = π

∫ 1

0

((e3 + e)2 − (e3x + ex)2

)dx

=16

(7 + 3e2 + 9e4 + 5e6)π .

18. Izracunati duzinu luka krive L date sa y(x) = log(1− x2), x ∈ [0, 1/2].

Resenje: y′(x) =2x

x2 − 1.

f HxL

1�2

l =∫ 1/2

0

√1 +

4x2

(x2 − 1)2dx =

∫ 1/2

0

x2 + 11− x2

dx

=−∫ 1/2

0dx+ 2

∫ 1/2

0

dx

1− x2= log 3− 1

2.

19. Izracunati duzinu luka krive y(x) =√

81− x2 za 3 ≤ x ≤ 9, kao i povrsinurotacione povrsi nastale rotacijom tog luka oko x−ose.

17

Resenje: y′(x) = − x√81− x2

,√

1 + y′(x)2 =9√

81− x2.

l =∫ 9

3

√1 + y′(x)2 dx = 9

∫ 9

3

dx√81− x2

=∣∣∣∣x = 9 sin t dx = 9 cos t dt3 7→ arcsin 1

3 9 7→ π2

∣∣∣∣

= 9(

arcsin13− π

2

)= 9 arccos

13.

Px = 2π∫ 9

3y(x)

√1 + y′(x)2 dx = 18π

∫ 9

3dx = 108π.