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anonymous

  • 5 years ago

How do you take the antiderivative of (2x-1)^-1/2???

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  1. anonymous
    • 5 years ago
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    Is that (2x-1)^(-1/2) or (2x-1)^-1 which is divided by 2?

  2. anonymous
    • 5 years ago
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    (2x-1)^(-1/2)

  3. anonymous
    • 5 years ago
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    O.k. Then you would let u = 2x-1 Then du = 2 dx So your integral:\[\int\limits \frac{1}{\sqrt{2x+1}}dx\] becomes

  4. anonymous
    • 5 years ago
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    1/2 times \[\int\limits \frac{2dx}{\sqrt{2x+1}}\]

  5. anonymous
    • 5 years ago
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    \[\int\limits \frac{du}{\sqrt{u}}=\]

  6. anonymous
    • 5 years ago
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    \[\int\limits u ^{-1/2}du=\]

  7. anonymous
    • 5 years ago
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    \[\frac{u ^{-1/2 + 1}}{-1/2+1}+C\]

  8. anonymous
    • 5 years ago
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    \[\frac{(2x+1) ^{1/2}}{1/2}+C\]

  9. anonymous
    • 5 years ago
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    and dividing by 1/2 is the same as multiplying by the reciprocal 2/1

  10. anonymous
    • 5 years ago
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    What about x/(2x-1)^(1/2) with an upper of 5 and a lower of 1. My calculator is giving me a correcct answer of 5.333333. But I can work the problem out to be the same

  11. anonymous
    • 5 years ago
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    sorry, I "can't" work the problem out the same

  12. anonymous
    • 5 years ago
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    o.k. well, you would want to plug 5 into \[2*\sqrt{2x+1}\] and then subtract what you get when you plug 1 into it.

  13. anonymous
    • 5 years ago
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    that would be 2 times the result of \[\sqrt{2(5)+1} - \sqrt{2(1)+1} = \sqrt{11}-\sqrt{3}\]

  14. anonymous
    • 5 years ago
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    which turns out to be about 3.17

  15. anonymous
    • 5 years ago
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    Oh! I didn't notice you had an "x" in the numerator.

  16. anonymous
    • 5 years ago
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    You would still use the same substitution

  17. anonymous
    • 5 years ago
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    but the x in the numerator can be rewritten in terms of "u"

  18. anonymous
    • 5 years ago
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    right. x= (1+u)/2

  19. anonymous
    • 5 years ago
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    since u = 2x+1 u-1 = 2x and so you get that x = (.5)*(u - 1)

  20. anonymous
    • 5 years ago
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    so dividing that by \[\sqrt{u} = u^{1/2}\] you can use exponent rules to get

  21. anonymous
    • 5 years ago
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    \[\frac{.5*(u-1)}{u^{1/2}} \]

  22. anonymous
    • 5 years ago
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    which boils down to \[.5* (u^{1-1/2}-u^{-1/2})\]

  23. anonymous
    • 5 years ago
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    which is\[.5* (u^{1/2}-u^{-1/2})\]

  24. anonymous
    • 5 years ago
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    and you can use the power rule for integrals to antidifferentiate

  25. anonymous
    • 5 years ago
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    \[.5* (u^{3/2}-u^{1/2})\]

  26. anonymous
    • 5 years ago
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    or \[.5* ((2x+1)^{3/2}-(2x+1)^{1/2})\]

  27. anonymous
    • 5 years ago
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    does that make sense?

  28. anonymous
    • 5 years ago
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    Yes, but when I plug in the upper and lower I'm still not getting the correct answer.......

  29. anonymous
    • 5 years ago
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    let me check

  30. anonymous
    • 5 years ago
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    The (2x+1) is suppose to be (2x-1)

  31. anonymous
    • 5 years ago
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    right...it's 14.8511...

  32. anonymous
    • 5 years ago
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    Oh! :) Well, I can guarantee you if you use 2x-1 instead it should work but I'll check

  33. anonymous
    • 5 years ago
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    My graphing calculator is giving me 5.333333

  34. anonymous
    • 5 years ago
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    \[9^{3/2}-9^{1/2}-(1^{3/2}-1^{1/2})\]

  35. anonymous
    • 5 years ago
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    is \[3\sqrt{3}-3\]

  36. anonymous
    • 5 years ago
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    Do you have a graphing cal on you?

  37. anonymous
    • 5 years ago
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    Actually, you have to go back and change the substitution also so it is more work than this

  38. anonymous
    • 5 years ago
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    if u = 2x-1 then x=.5*(u+1)

  39. anonymous
    • 5 years ago
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    Thanks for your help. I'm in FLorida and it's about 1am. I need to hit the hay. Thanks again......

  40. anonymous
    • 5 years ago
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    Yeah, I did it by hand and it works out. Just use that substitution

  41. anonymous
    • 5 years ago
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    and also multiply by the reciprocal of the new exponent you get when you integrate. ;)

  42. anonymous
    • 5 years ago
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    Will do. Thanks!

  43. anonymous
    • 5 years ago
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    Your welcome...sorry I'm new to typing this out. It's a bit distracting ;) but I'll get used to it. Night!

  44. anonymous
    • 5 years ago
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    Yeah, it's a pain in the retrice Laters

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