amyna
  • amyna
find intergal: this is an interrogation by parts problem
Mathematics
schrodinger
  • schrodinger
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amyna
  • amyna
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amyna
  • amyna
i got: u=x^3 du=3x^2 dv=e^-x^2 v= i don't know lol
freckles
  • freckles
\[u=x^2 \\ du=2x dx \\ \text{ now multiply both sides of } du=2x dx \text{ by } \frac{1}{2}x^2 \\ \text{ so we have } \\ \frac{1}{2}x^2 du=x^3 dx \\ \text{ but recall } x^2=u \\ \text{ so we have } \frac{1}{2}u du=x^3 dx\] once I make this substitution then I would do integration by parts

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freckles
  • freckles
or you could have actually went with the sub u=-x^2 instead
amyna
  • amyna
ok i'll try and do this, and see what i get. thanks!
amyna
  • amyna
ya i don't know what I'm doing! can i use integration by parts first?
Jhannybean
  • Jhannybean
this is an interrogation huh? ;)
zepdrix
  • zepdrix
buhaha XD
freckles
  • freckles
you could try integration by parts but as you wrote earlier it is a little confusing to do a substitution would make the integration by parts less confusing -- what do you mean you don't know what you are doing? like you couldn't do the substitution I gave you above ?
amyna
  • amyna
ya where did you get the x^3 from?
freckles
  • freckles
\[\int\limits x^3 e^{-x^2} dx \\ \text{ replace } x^2 \text{ with } u \\ \text{ replace } x^3 dx \text{ with } \frac{1}{2} u du \\ \text{ giving you } \\ \int\limits \frac{1}{2} u e^{-u} du\]
freckles
  • freckles
did you read the third line ?
freckles
  • freckles
\[u=x^2 \\ du=2x dx \\ \text{ now multiply both sides of } du=2x dx \text{ by } \frac{1}{2}x^2 \\ \text{ so we have } \\ \frac{1}{2}x^2 du=x^3 dx \\ \text{ but recall } x^2=u \\ \text{ so we have } \frac{1}{2}u du=x^3 dx\] that third line said I multiplied both sides by 1/2x^2
amyna
  • amyna
oh okay! thats more clear! Thanks!
freckles
  • freckles
it was just a copy and paste of what I said earlier :p \[\frac{1}{2}x^2 \cdot 2x dx =\frac{2}{2} x^3 dx=x^3 dx\]
Jhannybean
  • Jhannybean
where is this \(\dfrac{1}2x^2\) coming from...
freckles
  • freckles
we have to write x^3 dx in terms of u
freckles
  • freckles
that is where it comes from
Jhannybean
  • Jhannybean
Oh nevermind, it's to eliminate the 2 and to obtain x\(^3\) on the right side. Nvm.
jacobciezki
  • jacobciezki
Why are you girls still in her

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