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Matematika 1 7. vaja B. Jurˇciˇc Zlobec

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B. Jurˇciˇc Zlobec1

1Univerza v Ljubljani, Fakulteta za Elektrotehniko 1000 Ljubljana, Trˇzaˇska 25, Slovenija

Matematika FE, Ljubljana, Slovenija 4. januar 2013

(2)

I Linearnost:

Z

(a f(x) +b g(x))dx =a Z

f(x)dx +b Z

g(x)dx.

II Per partes:

Z

f0(x)g(x)dx =f(x)g(x)− Z

f(x)g0(x)dx.

III Substitucija:

Z

f(g(x))g0(x)dx =F(g(x)), kjer je Rf(x)dx =F(x).

u=g(x)→,du=g0(x)dx →,R

f(u)du =F(u) =F(g(x)).

IV Linearna substitucija:R

f(ax +b)dx = 1aF(ax+b).

(3)

1. R

xndx = xn+1n+1, n∈Z\ {−1}, x 6=0.

2. R

xsdx = xs+1s+1, s∈R\ {−1}, x >0.

3. R

exdx =ex. 4. R dx

x =ln|x|.

5. R

sinx dx=−cosx. 6. R

cosx dx =sinx. 7. R dx

a2+x2 = 1aarctanxa, a>0 . 8. R √dx

a2−x2 =arcsinxa, a>0.

9. R √dx

x2±a2 =ln(x +√

x2±a2), a>0.

(4)

A Integral oblike Z

f(x)dx =

Z P(x)dx

(x−a)n(x2+px +q)m, kjer jem,n∈N,p2−4q <0 inP(x)polinom stopnje manj ali enako 2m+n−1, reˇsimo s pomoˇcjo nastavka:

Z

f(x)dx = Q2m−2+n−2(x)

(x −a)n−1(x2+px +q)m−1+ +Aln|x−a|+Bln(x2+px+q) +Carctg√2x+p

4q−p2. B Integral oblike

Z

f(x)dx =

Z Pn(x)dx px2+px+q reˇsimo s pomoˇcjo nastavka:

Z

f(x)dx =Qn−1(x)p

x2+px+q+λ

Z dx

px2+px +q.

(5)

f(x) =3+ 1 x + 1

x2.

I Uporabimo (I)→(1)→(4).

I R

3+x1+x12

dx =→

I R

3dx+R dx

x +R dx

x2 dx =→

I

Z 3+ 1

x + 1 x2

dx =3x+ln|x| − 1 x +C.

(6)

f(x) = 3+2x2 1+x2 .

I Uporabimo (I)→(1)→(7).

I R 3+2x2

1+x2 dx =R 1+2(1+x2)

1+x2 dx →

I R

2dx+R dx

1+x2 =→

I

Z 3+2x2

1+x2 dx =2x+arctgx+C.

(7)

f(x) = x3 1+x2.

I Preverimo stopnji ˇstevca in imenovalca.

I Delimo: 1+xx32 =x− 1+xx 2

I (I)→(1)→(III)→(4).

I u =1+x2,du =2x dx →

I 1

2x212R du

u = 12x212ln|u| →

I 1

2x212ln(1+x2)→

I

Z x3 1+x2 = 1

2x2−lnp

1+x2+C

(8)

f(x) = 3

2+2x+x2 + x 1+x2.

I Uporabimo (I)→(IV)→(III)→(7)→(4).

I R

3dx

2+2x+x2 +1+xx 2

dx =3R dx

(x+1)2+1+R x

1+x2dx →

I u =x+1,du=dx, v =1+x2,dv =2x dx →

I 3R du

1+u2 + 12R dv

v =→

I

Z

f(x)dx =3 arctg(x+1) +1

2ln(1+x2) +C.

(9)

f(x) = x−1 x(x +1).

I Razcepimo na parcialne ulomke.

x−1

x(x+1) = Ax + x+1B =−x1+x+12 .

I R x−1

x(x+1)dx =−R dx

x +R 2dx

x+1.

I R x−1

x(x+1)dx =−ln|x|+2 ln|x+1| →

I

Z x −1

x(x +1)dx =ln(x +1)2

|x| +C.

(10)

f(x) = x2+1 x(1−x2).

I Razcepimo na parcialne ulomke.

x2+1

x(1−x2) = Ax +1−xB + 1+xC

I x2+1=A(1−x)(1+x) +Bx(1+x) +Cx(1−x)→

I −A+B−C=1,B+C=0 inA=1.

I

Z x2+1

x(1−x2)dx =ln|x| −ln|1−x2|.

(11)

f(x) = x −1 x(x2+1).

I Razcepimo na parcialne ulomke.

x−1

x(x2+1) = Ax +Bx+Cx2+1 =−1x +x1+x2+1.

I R x−1

(x2+1)dx =−R dx

x +R dx

1+x2 +R x dx

1+x2.

I R x−1

x(x+1)dx =−ln|x|+arctgx +12ln(x2+1)→

I

Z x −1

x(x +1)dx =−ln|x|+arctgx +lnp

1+x2+C.

(12)

f(x) = x −1 x2(x+1).

I Razcepimo na parcialne ulomke.

x−1

x2(x+1) = xA2 +Bx +x+1C =−x12 +x2x+12 .

I R x−1

x(x+1)dx =−R dx

x2 +R 2dx

x −R 2dx

1+x.

I R x−1

x(x+1)dx = 1x +2 ln|x| −2 ln|x+1| →

I

Z x −1

x(x +1)dx = 1

x +ln x2

(1+x)2+C.

(13)

f(x) = x2+1 x2(1+x).

I Zapiˇsemo nastavek (A).

I R x2+1

x2(1+x)dx = Ax +Bln|x|+Cln|1+x| →

I x2+1=−A(1+x) +Bx(1+x) +Cx2

I B+C=1,A+B=0 inA=−1.

I

Z x2+1

x2(1+x)dx =−1

x −ln|x|+2 ln|1+x|.

(14)

f(x) = x3−1 x2(1+x).

I Stopnji ˇstevca in imenovalca sta enaki, delimo.

x3−1

x2(1+x) =1− x2+1

x2(x+1).

I Zapiˇsemo nastavek (A).

R x2+1

x2(1+x)dx = Ax +Bln|x|+Cln|1+x| →

I x2+1=−A(1+x) +Bx(1+x) +Cx2

I B+C=1,A+B=0 inA=−1.

I

Z

1− x2+1 x2(1+x)

dx =x+ 1

x +ln|x| −ln(1+x)2+C.

(15)

f(x) = x2−1 x(2+2x+x2).

I Preverimo stopnji ˇstevca in imenovalca.

I Zapiˇsemo nastavek (A).

R x2−1

x(2+2x+x2)dx =Aln|x|+Bln|x2+x+1|+Carctg(1+x)→

I x2−1= Ax + 2Bx+B+Cx2+x+1

I B+C=1,A+B=0 inA=−1.

I

Z x2−1

x(x2+2x +2)dx =

−1

2ln|x| −1

2arctg(x+1) +3

4ln(2+2x+x2) +C.

(16)

f(x) = x

√x+1.

I (III)→(1).

I

Z x

√x+1dx →u =x+1,du =dx →

I R u−1

u du=R

u12 −u12 du.

I R x

x+1dx = 23p

(x+1)3−2√

x +1→

I

Z x

√x+1dx = 2

3(x−2)√

1+x+C

(17)

f(x) =xp

1+x2.

I (III)→(2).

I

Z xp

1+x2→u=1+x2,du =2x dx →

I 1 2

R u12 du= 13u32.

I

Z xp

1+x2= 1 3

q

(1+x2)3+C.

(18)

f(x) =p

1−x2.

I (III)→(2).

I

Z p

1−x2dx →x =sinu,dx =cosu du→

I R p

1−sin2ucosu du=R

cos2u du=→

I u+12sin(2u) =u+sinucosu=u+sinup

1−sin2u.

I

Z p

1−x2dx = 1 2

arcsinx+xp 1−x2

+C.

(19)

f(x) =p

1−x2.

I Reˇsujemo s pomoˇcjo nastavka (B).

I

Z p

1−x2dx =→

I

Z 1−x2

1−x2dx = (Ax+B)p

1−x2+C

Z 1

1−x2dx →

I1−x2

1−x2 =A√

1−x2(Ax+B)x

1−x2 +√C

1−x2

I 1−x2=−2Ax2−Bx+C+A→,A=C = 12,B=0.

I

Z p

1−x2dx = 1 2

arcsinx+xp

1−x2 +C.

(20)

f(x) =p

1+x2.

I Reˇsujemo s pomoˇcjo nastavka (B).

I

Z 1+x2

1+x2dx = (Ax+B)p

1+x2+C

Z 1

1+x2dx →

I1+x2

1+x2 =A√

1+x2+(Ax+B)x

1+x2 +√C

1+x2

I 1+x2=2Ax2+Bx +C+A→,A=C = 12,B =0.

I

Z p

1+x2dx = 1 2

ln(x+p

1+x2) +xp

1+x2 +C.

(21)

f(x) = x +2

2+2x +x2.

I Reˇsujemo s pomoˇcjo nastavka (B).

I

Z x+2

√2+2x+x2dx =

Ap

2+2x+x2+B

Z dx

2+2x+x2

Ix+2

1+x2 = +√A(x+1)

2+2x+x2 +√ B

2+2x+x2

I x +2=A(x +1) +B →,A=1,B=1.

I

Z √

x+2dx =

p2+2x+x2+ln(x+1+p

x2+2x+2) +C.

(22)

f(x) = x +2

1+2x −x2.

I Reˇsujemo s pomoˇcjo nastavka (B).

I

Z x+2

1+2x−x2dx = Ap

1+2x−x2+B

Z dx

1+2x−x2

Ix+2

1+2x−x2 = A(2−2x)

2

1+2x−x2 +√ B

1+2x−x2

I x +2=A−Ax+B,A=−1 inB=3.

I R √ 1

1+2x−x2dx =R √ dx

2−(x−1)2 =→(IV)→(8)=arcsin(x−1).

I

Z x+2

√2x−x2dx =−p

1+2x −x2−3 arcsinx −1

√2 +C.

(23)

f(x) = 1 xlnx.

I (III)→(4).

I

Z dx

xlnx →u=lnx,du = dx x →

I R du

u =lnu→

I

Z dx

xlnx =ln|lnx|+C.

(24)

f(x) = ex 1+e2x.

I (III)→(7).

I

Z ex

1+e2x dx →

I u =ex,du =exdx →

I R du

1+u2 =arctgu→

I

Z ex

1+e2x dx =arctgex+C.

(25)

f(x) =tgx.

I Upoˇstevamo, da je tgx = sinx cosx →

I (III)→(4).

I u =cosx,dx =−sinx dx →

I

Z sinx cosx =

Z du

u =−ln|u| →

I

Z

tgx dx =−ln|cosx|+C

(26)

f(x) =cos2x.

I Upoˇstevamo, da je cos2x = 12(1+cos(2x))

I (I)→(1)→(IV)→(6).

I R

cos2x dx =R

dx +R

cos(2x)dx =→

I

Z

cos2x dx = x 2 +1

4sin(2x) +C.

(27)

f(x) =xe−x

I (II)→(1)→(IV)→(3).

I

Z

xe−xdx →u =x,dv =e−xdx, du=dx,v =−e−x

I −xe−x+R

e−xdx =−xe−x−e−x

I

Z

xe−xdx =−e−x(x+1) +C

(28)

f(x) = sinx 1+cosx.

I

Z sinx

1+cosx dx →

I u =cosx,du =−sinx dx →

I −R du

1+u =−ln|1+u|=−ln|1+cosx| →

I

Z sinx

1+cosx dx =−ln|1+cosx|+C

(29)

f(x) =e−xcos(2x).

I Dvakrat uporabimo (II).

I R

e−xcos(2x)dx →

I u =e−x,dv =cos(2x)dx, du =−e−xdx,v = 12sin(2x)

I R

e−xcos(2x)dx = 12e−xsin(2x) + 12R

e−xsin(2x)dx

I 1 2

R e−xsin(2x)dx →

I u =e−x,dv =sin(2x)dx, du=−e−x,v =−12cos(2x)

I 1 2

R e−xsin(2x)dx =−14 e−xcos(2x) +R

e−xcos(2x)

I

Z

e−xcos(2x)dx = 1

5e−x(2 sin(2x)−cos(2x)) +C

(30)

f(x) =x2lnx.

I Uporabimo (II).

I

Z

x2lnx dx→

I u =lnx,dv =x2dx, du = dxx ,v = x33

I R

x2lnx dx = 13 x3lnx−R

x2dx .

I

Z

x2lnx dx= x3

3 lnx−x3 9 +C

(31)

S= Z 4

0

e

xdx.

I Uvedemot =√

x →t2=x →2t dt =dx.

I Ker jex =0→t =0 inx =4→t=2,S=2R2 0 tetdt.

I Integriramoper partes:

u =t, dv =etdt →du=dt, v =et.

I S= 2tet

2 0−2R2

0 etdt =4e2−2e2−2=2(e2−1).

I

Z 4 0

e

xdx =2(e2−1).

(32)

S= Z 2

0

xex2dx.

I Uvedemot =x2→dt=2xdx →dx = dt2.

I Ker jex =0→t =0 inx =2→t=4,S= 12R4 0 etdt.

I S= 1 2

Z 4 0

etdt = 1

2(e4−1).

(33)

S= Z π2

0

sin√ x dx.

I Uvedemot =√

x,→t2=x,→2tdt =dx.

I Ker jex =0→t =0 inx =π2→t =π, velja S=2Rπ

0 tsint dt.

I Integriramoper partes:

u =t, dv =sint dt,→du=dt, v =−cost.

I S= −2tcost|π0+2Rπ

0 cost dt =π S=2

Z π

0

tsint dt =2π.

(34)

S= Z π

4

0

dx cos2x.

I Uvedemot =tgx.

I dt = cosdx2x. Nove mejex =0→t=0,x = π4 →t =1.

I

Z 1 0

dt= t|10=1.

(35)

Z 9 4

√dx x −1.

I Uvedemo

x =t2,→dx =2t dt, x =4→t=2, x =9→t =3.

I 2R3 2

t dt

t−1 = t+ln|t−1||32=2+ln 4.

I

Z 9 4

√dx

x −1 =2+ln 4.

(36)

Z 2

−1

dx 9−x2.

I Razcepimo na parcialne ulomke: 9−x12 = 16 1

x−3+x+31

I R2

−1 dx 9−x2 =R2

−11 6

1

x−3+x+31

dx = 16ln|x2−3|

2

−1.

I

Z 2

−1

dx

9−x2 = ln 10 6 .

(37)

I Uporabimo (III)

→u =xn,dv =e−x, du=nxn−1,v =−e−x

I PiˇsimoΠ(n) =R

0 xne−xdx.

I R

0 xne−xdx =−xne−x|0 +nR

0 xn−1e−xdx

I Uporabimo l’H ˆospitalovo pravilo pri raˇcunanju limite limx→∞xne−x =limx→∞ xenx =0,

I in dobimo zvezoΠ(n) =nΠ(n−1).

I Ker jeΠ(0) =R

0 e−x = −e−x|0 =1,

I lahko z matematiˇcno indukcijo pokaˇzemo, da jeΠ(n) =n!

(38)

Z

1

dx x2.

I

Z

1

dx

x = −1 x

1

=1.

I

Z

0

dx x2 =1.

(39)

Z 4 0

√dx 4−x.

I Uvedemot2=4−x →2t dt =−dx.

I Mejex =0→t =2, x =4→t =0.

I −2 Z 0

2

t dt

t = −2t|02=4

I

Z 4 0

√dx

4−x =4

(40)

Z 1

−2

|x| x dx.

I

Z 1

−2

|x|

x dx = Z 0

−2

(−1)dx+ Z 1

0

dx = −x|0−2+x|10=−1

I

Z 1

−2

|x|

x dx =−1

(41)

Z

0

e

xdx.

I Uvedemot2=x →2t dt =dx.

I Mejex =0→t =0, x → ∞ →t→ ∞.

I R 0 e

xdx =2R 0 te−tdt

I Integriramoper partes:

du =e−tdx,v =t →u=−e−t,dv =dt.

I 2R

0 te−tdt = −2te−t

0 +2R 0 e−tdt

I Ker je limt→∞te−t =0, je 2R

0 te−tdt =2.

I

Z

0

e

xdx =2.

(42)

Z

0

e

xdx

√x .

I Uvedemot2=x →2t dt =dx.

I Mejex =0→t =0, x → ∞ →t→ ∞.

I

Z

0

e

xdx

√x =2 Z

0

e−tdt = −2e−t

0 =2.

I

Z

0

e

xdx

√x =2.

(43)

Z 1 0

√xln1 x dx.

I R1 0

√xln1xdx =−R1 0

√xlnx dx

I Integriramoper partes.

u =lnx,dv =√

x dx →du = dxx ,v = 23

√ x3

I

Z 1 0

√xlnx dx = 2 3

√ x3lnx

1 0

−2 3

Z 1 0

√x dx

I

Z 1 0

√xln1

x dx = 4 9

Reference

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