Integral

Ko tudi učitelj ne more pomagati...
Tasko
Posts: 37
Joined: 10.3.2008 9:49

Re: Integral

Post by Tasko » 15.5.2011 13:35

No, in se je zataknilo pri tem primeru:

Image
Image
Mathematica izračuna rezultat \(\frac{512}{9009}\)

Kje je napaka?

User avatar
shrink
Posts: 14571
Joined: 4.9.2004 18:45

Re: Integral

Post by shrink » 15.5.2011 17:10

\(\Gamma(\frac{3}{2})=\frac{1}{2}\Gamma(\frac{1}{2})\)

\(\Gamma(6+\frac{3}{2})=\Gamma(\frac{15}{2})=\frac{13}{2}\cdot\frac{11}{2}\cdot\frac{9}{2}\cdot\frac{7}{2}\cdot\frac{5}{2}\cdot\frac{3}{2}\cdot\frac{1}{2}\Gamma(\frac{1}{2})\)

Tasko
Posts: 37
Joined: 10.3.2008 9:49

Re: Integral

Post by Tasko » 15.5.2011 18:14

shrink wrote:\(\Gamma(\frac{3}{2})=\frac{1}{2}\Gamma(\frac{1}{2})\)

\(\Gamma(6+\frac{3}{2})=\Gamma(\frac{15}{2})=\frac{13}{2}\cdot\frac{11}{2}\cdot\frac{9}{2}\cdot\frac{7}{2}\cdot\frac{5}{2}\cdot\frac{3}{2}\cdot\frac{1}{2}\Gamma(\frac{1}{2})\)
Hvala.

User avatar
sniper
Posts: 231
Joined: 30.10.2006 13:08

Re: Integral

Post by sniper » 17.5.2011 3:14

Kako pa bi tule izračunal površino med krivuljama ? Problem imam, ker nevem kako bi dobil površino med -4 do -2 ter z osjo x da jo odštejem..?

Image

User avatar
Aniviller
Posts: 7263
Joined: 15.11.2004 18:16

Re: Integral

Post by Aniviller » 17.5.2011 8:37

1) Poisci presecisci krivulj (od tukaj dobis meje integrala)
2) Odstej krivulji in integriraj (ali integriraj in potem odstej, ni vazno)

Tisti koscek, o katerem govoris, sploh ne igra vloge.

User avatar
sniper
Posts: 231
Joined: 30.10.2006 13:08

Re: Integral

Post by sniper » 17.5.2011 19:46

aja to se da tako, fino :)

hvala

naty11
Posts: 6
Joined: 2.5.2010 14:28

Re: Integral

Post by naty11 » 18.5.2011 13:23

Ali mi lahko kdo pomaga rešiti naslednji integral:
Brez naslova.jpg
Brez naslova.jpg (2.21 KiB) Viewed 1719 times
Bi lepo prosila za postopek reševanja! Hvala...

Zajc
Posts: 1099
Joined: 26.6.2008 19:15

Re: Integral

Post by Zajc » 18.5.2011 13:37

Uvedem \(u=\sin{x}\) in dobim \(\int\frac{dx}{\cos{x}}=\int\frac{\cos{x}dx}{\cos^2{x}}=\int\frac{du}{1-u^2}=\frac{1}{2}(\int\frac{du}{1-u}+\int\frac{du}{1+u})=\) \(\frac{1}{2}(\log{(1+u)}-\log{(1-u)})+C=\) \(\log\sqrt{\frac{1+u}{1-u}}+C=\log\frac{1+u}{\sqrt{1-u^2}}+C=\log\frac{1+\sin{x}}{\cos{x}}+C\).

naty11
Posts: 6
Joined: 2.5.2010 14:28

Re: Integral

Post by naty11 » 18.5.2011 16:05

Hvala za tvoj odgovor.

Me zanima, kako si dobil 1/2 za tretjim enačajem in kako si potem naredil vsoto dveh integralov?

User avatar
Aniviller
Posts: 7263
Joined: 15.11.2004 18:16

Re: Integral

Post by Aniviller » 18.5.2011 17:15

Razcep na parcialne ulomke.

naty11
Posts: 6
Joined: 2.5.2010 14:28

Re: Integral

Post by naty11 » 23.5.2011 0:38

Hvala.

Kako pa je za četrtim enačajem dobil razliko dveh logaritmov, če je bila prej vsota?

User avatar
Aniviller
Posts: 7263
Joined: 15.11.2004 18:16

Re: Integral

Post by Aniviller » 23.5.2011 8:32

Integrirat moras. In en izmed integrandov ima - pred u, kar po integraciji prinese dodaten minus.

User avatar
sniper
Posts: 231
Joined: 30.10.2006 13:08

Re: Integral

Post by sniper » 24.5.2011 11:34

Kaj pa tule:
Izračunaj ploščino lika med krivuljama \(y=x^2-3x\) in \(y=|x-1|\)
Krivulji med sabo odštejem in dobim: \(3x-x^2+|x-1|\)

Meji integrala sta pri \(1- \sqrt2\) in \(2+ \sqrt{3}\)

Nevem pa kako integrirat tole: \(3x-x^2+|x-1|\), ker je notri abs vrednost ?

User avatar
Aniviller
Posts: 7263
Joined: 15.11.2004 18:16

Re: Integral

Post by Aniviller » 24.5.2011 14:53

Razbij na vsoto integralov, enega do 1 in enega od 1 naprej. Vsak kos tako lahko izgubi absolutno vrednost.

User avatar
sniper
Posts: 231
Joined: 30.10.2006 13:08

Re: Integral

Post by sniper » 24.5.2011 20:10

Hm neki ni OK. Vsota obeh integralov do 1 in od 1 naprej pride različno kot če vse skupaj:


1 del
Image

2 del

Image

Skupaj pride malo več kot 6




Image

Tukaj pa 8

Post Reply