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Homework Help: Simply Checking an Answer in Calculus

  1. Feb 27, 2005 #1
    Hello all, again.

    Simple question.

    Compute [tex]\int_{9}^{4} (x - \frac{1}{\sqrt{x}}) dx[/tex]

    I did this:

    [tex]\int_{9}^{4} (x - \frac{1}{x^{0.5}}) dx[/tex] = [tex]\int_{9}^{4} (x - x^{-0.5}) dx[/tex]

    [tex]\left[ \frac{1}{2} x^2 - 2 x^{0.5} \right]_{9}^{4}[/tex] = [tex]\left[ \frac{x^2}{2} - 2 \sqrt{x} \right]_{9}^{4}[/tex]

    [tex](\frac{81}{2} - 6) - (\frac{16}{2} - 4) = \frac{81 - 12}{2} - \frac{16 - 8}{2} = \frac{69}{2} - \frac{8}{2} = \frac{61}{2} = 30.5[/tex]

    2 months of calculus and and still I am not sure of what I have done wrong. The answers in my book gives [tex]10 \frac{2}{3}[/tex]

    So if it is right then can someone enlighten me to what I have done wrong please.

    Cheers

    The Bob (2004 ©)
     
  2. jcsd
  3. Feb 27, 2005 #2
    Third step...
    x - x^-0.5 gives 1/2x^2 - 2x^0.5... hmmm...
    Your work looks right to me.. maybe a book error?
    K.
     
    Last edited: Feb 27, 2005
  4. Feb 27, 2005 #3
    That is what I was hoping for. I know I make simply mistakes but now 4 people reckon it is right.

    Cheers. Still though, can anyone else see anything wrong at all???

    The Bob (2004 ©)
     
  5. Feb 27, 2005 #4

    dextercioby

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    Nope,both your and the books answers are incorrect.My answer is [tex] -\frac{61}{2} [/tex] and i'm sure of it,because even my old rusty Maple says it is so...:wink:

    Daniel.
     
  6. Feb 27, 2005 #5
    That's just weird.. even my calculator gives 30.5. :\
    And yeah, I get the negative answer too... only mistake in his work I see now.
     
  7. Feb 27, 2005 #6
    Oh nuts. I see what I did wrong. I have written the original limtis of the question the wrong way around. My fault (obviously). I'm not used to the Tex of calculus problems.

    It should have been [tex]\int_{4}^{9} (x - \frac{1}{\sqrt{x}}) dx[/tex] to give [tex]\frac{61}{2}[/tex]

    Cheers. At least I can go to college on Wednesday and now I am right. :biggrin:

    The Bob (2004 ©)
     
  8. Feb 27, 2005 #7

    dextercioby

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    It seemed pretty weird to me too,with the larger value being the inferior limit,but,hey,anything is possible in mathemetics,right...?:wink:

    Daiel.
     
  9. Feb 28, 2005 #8
    Most things are possible but I am yet to see a larger, inferior limit too, Daiel. :wink:

    The Bob (2004 ©)
     
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