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anonymous

  • 5 years ago

how do i integrate dx/sqrt(4-x)

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  1. anonymous
    • 5 years ago
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    Have you tried u substition?

  2. anonymous
    • 5 years ago
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    substitution as in integration by parts?

  3. anonymous
    • 5 years ago
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    for your question, do you mean \[\int\limits_{}^{}dx/\sqrt{4-x}\]

  4. anonymous
    • 5 years ago
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    yes thats exactly it. i am studying improper integrals so i need to find the regular integral first

  5. anonymous
    • 5 years ago
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    Think of the square root on the bottom not as a square root, but as \[(4-x)^{1/2}\]

  6. anonymous
    • 5 years ago
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    You will, however, have to use a u substitution, which is different the integration by parts. Integration by parts is only necessary in situations with multiplication or division of things you cannot simplify

  7. anonymous
    • 5 years ago
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    More explicitly, think of it as \[\int\limits_{}^{}(4-x)^{-1/2}dx\]

  8. anonymous
    • 5 years ago
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    okay, so then when you say substitute you mean, take (4-x) and substitute u. and then perform \[1/n+1\times u ^ {n+1} \]

  9. anonymous
    • 5 years ago
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    sorry getting used to the equation typing

  10. anonymous
    • 5 years ago
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    essentially yes. \[u = (4-x)\], and \[du/dx = -1\]. Substitute in u and du, integrate the function I wrote above, and replace (4-x) when you're done.

  11. anonymous
    • 5 years ago
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    why is du/dx = -1?

  12. anonymous
    • 5 years ago
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    when doing u substitutions, you substitute u, then take the derivative of what you substituted, in this case (4-x), for x, which = -1 in this case.

  13. anonymous
    • 5 years ago
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    oh i see. so what i have then would be \[-1^{1/2}/{-1/2}\]

  14. anonymous
    • 5 years ago
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    and that becomes -(u^1/2)/1 * 2/1 which is -2sqrt u. and that becomes -2sqrt(4-x) is that right?

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