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amyna

  • one year ago

find intergal: this is an interrogation by parts problem

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  1. amyna
    • one year ago
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    |dw:1441854137872:dw|

  2. amyna
    • one year ago
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    i got: u=x^3 du=3x^2 dv=e^-x^2 v= i don't know lol

  3. freckles
    • one year ago
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    \[u=x^2 \\ du=2x dx \\ \text{ now multiply both sides of } du=2x dx \text{ by } \frac{1}{2}x^2 \\ \text{ so we have } \\ \frac{1}{2}x^2 du=x^3 dx \\ \text{ but recall } x^2=u \\ \text{ so we have } \frac{1}{2}u du=x^3 dx\] once I make this substitution then I would do integration by parts

  4. freckles
    • one year ago
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    or you could have actually went with the sub u=-x^2 instead

  5. amyna
    • one year ago
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    ok i'll try and do this, and see what i get. thanks!

  6. amyna
    • one year ago
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    ya i don't know what I'm doing! can i use integration by parts first?

  7. Jhannybean
    • one year ago
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    this is an interrogation huh? ;)

  8. zepdrix
    • one year ago
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    buhaha XD

  9. freckles
    • one year ago
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    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 ?

  10. amyna
    • one year ago
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    ya where did you get the x^3 from?

  11. freckles
    • one year ago
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    \[\int\limits x^3 e^{-x^2} dx \\ \text{ replace } x^2 \text{ with } u \\ \text{ replace } x^3 dx \text{ with } \frac{1}{2} u du \\ \text{ giving you } \\ \int\limits \frac{1}{2} u e^{-u} du\]

  12. freckles
    • one year ago
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    did you read the third line ?

  13. freckles
    • one year ago
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    \[u=x^2 \\ du=2x dx \\ \text{ now multiply both sides of } du=2x dx \text{ by } \frac{1}{2}x^2 \\ \text{ so we have } \\ \frac{1}{2}x^2 du=x^3 dx \\ \text{ but recall } x^2=u \\ \text{ so we have } \frac{1}{2}u du=x^3 dx\] that third line said I multiplied both sides by 1/2x^2

  14. amyna
    • one year ago
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    oh okay! thats more clear! Thanks!

  15. freckles
    • one year ago
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    it was just a copy and paste of what I said earlier :p \[\frac{1}{2}x^2 \cdot 2x dx =\frac{2}{2} x^3 dx=x^3 dx\]

  16. Jhannybean
    • one year ago
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    where is this \(\dfrac{1}2x^2\) coming from...

  17. freckles
    • one year ago
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    we have to write x^3 dx in terms of u

  18. freckles
    • one year ago
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    that is where it comes from

  19. Jhannybean
    • one year ago
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    Oh nevermind, it's to eliminate the 2 and to obtain x\(^3\) on the right side. Nvm.

  20. jacobciezki
    • one year ago
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    Why are you girls still in her

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