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Rearranging of an equation

  1. Oct 9, 2013 #1
    i have a given equation of the following.

    W= -A1B1c[[itex]\frac{1}{-c+1}[/itex]B-c+1] where the part in brackets is the integrated function of B going from B1 to B2.

    then this equation somehow ends up like so

    W=[itex]\frac{A_1B_1}{c-1}[/itex]{((B1)/(B2))^(c-1)-1}

    (sorry for the messy format on the right hand side i couldn't get latex to work..)

    i don't know how they got from the first equation to the second equation. i don't know where to start because i don't know how certain terms were combined.
     
    Last edited: Oct 9, 2013
  2. jcsd
  3. Oct 9, 2013 #2

    Mark44

    Staff: Mentor

    Is this what you mean?

    $$W = -A_1B_1^c \int_{B_1}^{B_2} \frac{B^{-c + 1} dB}{-c + 1} $$
    If you right-click on the integral I wrote, you can see the LaTeX that creates it.
     
  4. Oct 9, 2013 #3
    thanks, however that wasn't what i meant. that function inside the bracket is 'already' integrated; what i was trying to say was that the limits just weren't taken.

    W = [itex]-A_1B_1^c \frac{B^{-c + 1}}{-c + 1}[/itex]
     
    Last edited by a moderator: Oct 9, 2013
  5. Oct 9, 2013 #4

    Mark44

    Staff: Mentor

    Are you sure that your antiderivative is correct? I'm thinking you might have made a mistake. What was the problem you started with?
     
  6. Oct 9, 2013 #5

    pasmith

    User Avatar
    Homework Helper

    So that's
    [tex]
    W = -A_1 B_1^c \frac{1}{-c+1}\left(B_2^{-c+1} - B_1^{-c + 1}\right)
    = \frac{A_1 B_1^c }{c-1}\left(B_2^{1-c} - B_1^{1-c}\right)
    [/tex]
    after tidying up some signs.

    Now we pull a common factor of [itex]B_1^{1-c}[/itex] out of the bracket:
    [tex]
    W = \frac{A_1 B_1^c B_1^{1-c}}{c - 1} \left( \frac{B_2^{1-c}}{B_1^{1-c}} - 1\right)
    = \frac{A_1 B_1}{c - 1}\left( \left(\frac{B_2}{B_1}\right)^{1-c} - 1\right)
    [/tex]

    Finally we flip the fraction in the bracket, remembering to multiply the exponent by -1 as we do so.
     
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