How is the equation rearranged to end up with a different form?

[iScience]

i have a given equation of the following.

W= -A₁B₁^c[$\frac{1}{-c+1}$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=$\frac{A_1B_1}{c-1}${((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.

 

[Mark44]

i have a given equation of the following.

W= -A₁B₁^c[$\frac{1}{-c+1}$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=$\frac{A_1B_1}{c-1}${((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.

 

[iScience]

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 = $-A_1B_1^c \frac{B^{-c + 1}}{-c + 1}$

 

[Mark44]

Are you sure that your antiderivative is correct? I'm thinking you might have made a mistake. What was the problem you started with?

 

[pasmith]

i have a given equation of the following.

W= -A₁B₁^c[$\frac{1}{-c+1}$B^-c+1] where the part in brackets is the integrated function of B going from B₁ to B₂.

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

Now we pull a common factor of $B_1^{1-c}$ out of the bracket:
$$W = \frac{A_1 B_1^c B_1^{1-c}}{c - 1} \left( \frac{B_2^{1-c}}{B_1^{1-c}} - 1\right)<br /> = \frac{A_1 B_1}{c - 1}\left( \left(\frac{B_2}{B_1}\right)^{1-c} - 1\right)$$

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