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Dido525

  • 3 years ago

Evaluate the intergral:

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  1. Dido525
    • 3 years ago
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    |dw:1355802290129:dw|

  2. Dido525
    • 3 years ago
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    I thought I could use substitution but I can't ...

  3. winterfez
    • 3 years ago
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    is this right? \[\int\limits_{0}^{4}\frac{ x }{ \sqrt{1+2x} }\]

  4. Dido525
    • 3 years ago
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    yep :) .

  5. eliassaab
    • 3 years ago
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    put u= 1 + 2 x

  6. Mimi_x3
    • 3 years ago
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    \( u = 1+ 2x => du/dx = 2\) \( 1+2x = u => 2x = u-1 => x = (u-1)/2 \)

  7. Dido525
    • 3 years ago
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    I got to that step. Then I got stuck :P .

  8. Mimi_x3
    • 3 years ago
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    Maybe:\[\int\limits\frac{x}{\sqrt{1+2x}} dx => \int\limits \frac{(u-1)/2}{\sqrt{u}} *2du \int\limits\frac{u-1}{2} *\frac{1}{\sqrt{u}} 2du => 2\int\limits\frac{u-1}{2} *\frac{1}{\sqrt{u}} \]

  9. eliassaab
    • 3 years ago
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    Here what wolfram alpha gave

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  10. Mimi_x3
    • 3 years ago
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    Would my step work tho?

  11. zepdrix
    • 3 years ago
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    @Dido525 do you understand what mimi did? those are the correct steps to take.

  12. Dido525
    • 3 years ago
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    Yeah I actually do. I just need to process it.

  13. Dido525
    • 3 years ago
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    @Mimi_x3 : Wouldn't you write du/2 instead of 2*du ?

  14. Dido525
    • 3 years ago
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    Because we said du/2 was dx.

  15. zepdrix
    • 3 years ago
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    Oh yes :o good call.

  16. Mimi_x3
    • 3 years ago
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    Yeah typo lol sorry! that was where i went wrong :(

  17. Dido525
    • 3 years ago
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    It's okay :) .

  18. Dido525
    • 3 years ago
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    So so far I have: \[\frac{ 1 }{ 4 }\int\limits_{1}^{9} \frac{ u-1 }{ \sqrt{u} } du\]

  19. Mimi_x3
    • 3 years ago
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    then break it up!

  20. Dido525
    • 3 years ago
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    Ohh... Right :P .

  21. Mimi_x3
    • 3 years ago
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    integrate it; then don't forget to sub back in \(u\)

  22. Dido525
    • 3 years ago
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    Thanks! :) .

  23. Mimi_x3
    • 3 years ago
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    :)

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