Rearranging of an equation

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  • #1
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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.
 
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  • #2
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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.
Is this what you mean?

$$W = -A_1B_1^c \int_{B_1}^{B_2} \frac{B^{-c + 1} dB}{-c + 1} $$
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.

If you right-click on the integral I wrote, you can see the LaTeX that creates it.
 
  • #3
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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]
 
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  • #4
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Are you sure that your antiderivative is correct? I'm thinking you might have made a mistake. What was the problem you started with?
 
  • #5
pasmith
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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.

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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