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anonymous

  • 4 years ago

find the exact area of the regions bounded by y=x^3/(sqrt(4-x^2), y=0,x=0,x=sqrt(2)

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  1. anonymous
    • 4 years ago
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    \[\int\limits_{0}^{\sqrt{2}} x ^{3}/\sqrt{4-x ^{2}}\]

  2. TuringTest
    • 4 years ago
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    |dw:1328082060815:dw|all we're doing here is integrating up to the line x=sqrt2\[\int_{0}^{\sqrt2}\frac{x^3}{\sqrt{4-x^2}}dx\]which can be done with the substitution\[x=2\sin\theta\to dx=2\cos\theta d\theta\]

  3. anonymous
    • 4 years ago
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    Or you can put 4-x^2=t and then do it..

  4. TuringTest
    • 4 years ago
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    ...how would that work? you have x^3 on top

  5. anonymous
    • 4 years ago
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    Yeah so one x goes in dt and what is left is x^2 which is 4-t denominator is sqrt(t) so the net integral reduces to (4t^(-1/2)-t^(1/2))/2

  6. anonymous
    • 4 years ago
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    it's a trigonometric integral u substituion wont work

  7. anonymous
    • 4 years ago
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    And a minus sign outside.

  8. anonymous
    • 4 years ago
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    The substitution works pretty well only....

  9. anonymous
    • 4 years ago
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    if you use substitution your integral will be left in terms of both t and x, which will not work

  10. anonymous
    • 4 years ago
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    if it were only x on top not x^3 you could but not in this form

  11. anonymous
    • 4 years ago
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    Dude read what i have written if you put 4-x^2as t you are left with only x^2 in the numerator which is 4-t.

  12. anonymous
    • 4 years ago
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    First do it and see you'll get the answer...

  13. anonymous
    • 4 years ago
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    ok well what do you get by doing it your method?

  14. anonymous
    • 4 years ago
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    Whatever you get by trignometric substitution you'll get the same thing only...

  15. anonymous
    • 4 years ago
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    matt: Did you get what i'm trying to tell or should i elaborate further and solve the whole thing?

  16. anonymous
    • 4 years ago
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    It's okay I think what you are saying does make sense, I don't actually have pen or paper right now but i'll cease to doubt you

  17. anonymous
    • 4 years ago
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    \[\int\limits_{}^{}x^3/\sqrt{4-x ^{2}}\] put 4-x^2=t. -2x dx=dt. dx=dt/2x\[\int\limits_{}^{}-x ^{3}/\sqrt{t} 2x\] cancelling x you get x^2 on top which is 4-t\[\int\limits_{}^{}(4-t)/\sqrt{t} = (4/\sqrt{t})-\sqrt{t}\] This is easily integrable.

  18. anonymous
    • 4 years ago
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    ah k, yeah i see. Well thank you

  19. anonymous
    • 4 years ago
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    i forgot a minus sign. :D

  20. anonymous
    • 4 years ago
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    ahha it's alright I see what you are doing now

  21. TuringTest
    • 4 years ago
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    Never would have thought of that, cool.

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