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anonymous

  • 5 years ago

∫√3+2xdx

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  1. anonymous
    • 5 years ago
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    Is 3+2x all under the square root or is it just root 3?

  2. anonymous
    • 5 years ago
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    \[\sqrt{3+2x}\] ?

  3. anonymous
    • 5 years ago
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    yes

  4. anonymous
    • 5 years ago
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    Ok. We will use u-substitution. Set u equal to what's under the square root (3+2x) and then take the derivative of 3+2x to find du. So du = 2 dx

  5. anonymous
    • 5 years ago
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    ok

  6. anonymous
    • 5 years ago
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    So you are left with \[\int\limits \sqrt{u}/2 du\]

  7. anonymous
    • 5 years ago
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    The reason we divide by 2 is when you get du = 2 dx you want dx by itself, so you get du/2 = dx

  8. anonymous
    • 5 years ago
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    ok

  9. anonymous
    • 5 years ago
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    Since your dividing by 2, you can pull a 1/2 out in front of the integral, leaving you with the integral of root u, which can be rewritten as u^(1/2). Then to integrate we add 1 to the exponent (giving us u^(3/2)). We have to multiply by the reciprocal of our new exponent.

  10. anonymous
    • 5 years ago
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    So your final answer should be [2u^(3/2)]/3

  11. anonymous
    • 5 years ago
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    ok

  12. anonymous
    • 5 years ago
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    Since 2/3 times u^(3/2) just moves the u^(2/3) on top and multiplies it by 2.

  13. anonymous
    • 5 years ago
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    ok

  14. anonymous
    • 5 years ago
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    Any other questions?

  15. anonymous
    • 5 years ago
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    yes

  16. anonymous
    • 5 years ago
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    Ok what?

  17. anonymous
    • 5 years ago
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    ∫x+1/2x-x2+2 dx

  18. anonymous
    • 5 years ago
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    Oh well I meant about this problem. I'll look at it and see if I can help in the other problem.

  19. anonymous
    • 5 years ago
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    tthanks God bless you

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