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anonymous

  • one year ago

Need Help! Evaluate the integral (1-x/x)^2 *dx

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  1. anonymous
    • one year ago
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    |dw:1439271323768:dw|

  2. ganeshie8
    • one year ago
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    whats stopping you from expanding and working each term ?

  3. anonymous
    • one year ago
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    So I did it, but I got stuck on one part

  4. anonymous
    • one year ago
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    Let me draw it

  5. anonymous
    • one year ago
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    |dw:1439271526217:dw|

  6. anonymous
    • one year ago
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    What happens after this...because the solution manual is saying something else

  7. ganeshie8
    • one year ago
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    looks good! expand that out using the identity \[(a-b)^2=a^2-2ab+b^2\]

  8. ganeshie8
    • one year ago
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    |dw:1439271609165:dw|

  9. ganeshie8
    • one year ago
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    \[\left(\frac{1}{x}-1\right)^2=?\]

  10. anonymous
    • one year ago
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    OH thats what they did

  11. anonymous
    • one year ago
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    okay, I totally forgot about the identity and was wondering how they came up with the middle term

  12. ganeshie8
    • one year ago
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    it helps to review basic algebraic and trig identities while doing integrals

  13. anonymous
    • one year ago
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    I would honestly love doing integrals....but theres just so much algebra

  14. ganeshie8
    • one year ago
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    Exactly! calculus is always easy, its just the algebra part that is hard

  15. ganeshie8
    • one year ago
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    its not hard as such, its just pain

  16. anonymous
    • one year ago
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    lol tell me about it, I have my final in a few days..and im freaking out Either way, how is the middle term -2/x , im getting +2/x

  17. ganeshie8
    • one year ago
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    \[(a-b)^2=a^2\color{red}{-}2ab+b^2\]

  18. anonymous
    • one year ago
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    So it would be -2*(1/x)*(-1), right?

  19. ganeshie8
    • one year ago
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    \[\left(\frac{1}{x}-1\right)^2=\frac{1}{x^2}-\frac{2}{x}+1\]

  20. anonymous
    • one year ago
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    But the middle part is -2*(1/x)*(-1), so wouldnt that make it +2/x instead of -2/x?

  21. ganeshie8
    • one year ago
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    \[\left(\frac{1}{x}-1\right)^2=(\frac{1}{x})^2-2(\frac{1}{x})(1)+1^2\]

  22. anonymous
    • one year ago
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    why dont we multiply by negative, because isnt that the b value?

  23. ganeshie8
    • one year ago
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    \(a = \frac{1}{x}\) \(b=1\)

  24. anonymous
    • one year ago
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    kk, I get it

  25. ganeshie8
    • one year ago
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    :)

  26. ganeshie8
    • one year ago
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    don't forget to integrate the terms

  27. anonymous
    • one year ago
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    btw are you in Uni?

  28. anonymous
    • one year ago
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    Also, how do we integrate 1/x^2?

  29. Astrophysics
    • one year ago
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    \[\frac{ 1 }{ x^2 } = x^{-2} ~~~~~\int\limits \frac{ 1 }{ x^2 } dx = \int\limits x^{-2} dx = \frac{ x^{-1} }{ -1 }+C = - \frac{ 1 }{ x }+C\]

  30. anonymous
    • one year ago
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    nvm, I watched this video https://www.youtube.com/watch?v=ZFZRid582Lk, and even though its in spanish...I figured it out, thanks though astro

  31. ganeshie8
    • one year ago
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    here is a trick question, \[\int \frac{1}{x}~dx = ?\] @mary_am

  32. anonymous
    • one year ago
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    lnIxI + c

  33. ganeshie8
    • one year ago
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    nvm that video is a spoiler

  34. Astrophysics
    • one year ago
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    lol

  35. anonymous
    • one year ago
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    k im not dumb.. I knew what ∫1x dx was

  36. anonymous
    • one year ago
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    *THAT

  37. ganeshie8
    • one year ago
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    Ofcourse you're not, I see :)

  38. Astrophysics
    • one year ago
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    I am dumb, and yes I am xD

  39. Astrophysics
    • one year ago
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    Sounds about right

  40. Astrophysics
    • one year ago
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    Eihsenag university

  41. Astrophysics
    • one year ago
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    We enjoy helping people, it's a good way to learn ourselves :-)

  42. Astrophysics
    • one year ago
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    They intertwine

  43. Astrophysics
    • one year ago
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    Thank you and you to!

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