How do you get from one to the other

[SMOF]

Hello.

I am going through some worked examples for a class I have, and there is one step I don't understand, and I hope someone can help me with that.

It goes from

$\frac{1}{D_pB}\int e^{x/d}(-de^{-W_B/d} + de^{-x/d}) dx$

to

$\frac{1}{D_pB}\int (1 - e^{x-W_B/d}) dx$

The limits are 0 to W_B

I have tried to work through a few things, such as multiplying out the brackets and all, but I am not getting it right, so I am doing something wrong.

Thanks.

Seán

 

[Mark44]

Hello.

I am going through some worked examples for a class I have, and there is one step I don't understand, and I hope someone can help me with that.

It goes from

$\frac{1}{D_pB}\int e^{x/d}(-de^{-W_B/d} + de^{-x/d}) dx$

to

$\frac{1}{D_pB}\int (1 - e^{x-W_B/d}) dx$

The limits are 0 to W_B

I have tried to work through a few things, such as multiplying out the brackets and all, but I am not getting it right, so I am doing something wrong.

Show us what you have done.

Thanks.

Seán

 

[I like Serena]

Hi SMOF!

Multiplying the brackets out is the way to go.
What did you get?
Did you consider to use that $e^x e^y = e^{x+y}$?

Btw, there is a typo in your example.
It appears you dropped a factor "d".

 

[Dick]

Hello.

I am going through some worked examples for a class I have, and there is one step I don't understand, and I hope someone can help me with that.

It goes from

$\frac{1}{D_pB}\int e^{x/d}(-de^{-W_B/d} + de^{-x/d}) dx$

to

$\frac{1}{D_pB}\int (1 - e^{x-W_B/d}) dx$

The limits are 0 to W_B

I have tried to work through a few things, such as multiplying out the brackets and all, but I am not getting it right, so I am doing something wrong.

Thanks.

Seán

It looks like in going from the first expression to the second they just made a factor of $d$ disappear. I don't think that's right.

 

[SMOF]

Hello.

Sorry, there was a typo, the second line should have been

$\frac{d}{D_pB}\int e^{x/d}(-de^{-W_B/d} + de^{-x/d}) dx$

Working out the brackets, I get

$e^{x/d}(-de^{-W_B/d} + de^{-x/d}) dx$

goes to (I think)

$-de^{x-W_B/d} + de^{d}$

I am not sure about the de^d part.

Seán

 

[I like Serena]

Not quite.

You should get:
$$-de^{(x/d)-(W_B/d)} + de^{(x/d) - (x/d)}$$
Perhaps you can simplify that?

 

[SMOF]

Hello, thanks for the reply!

Right, am I right in saying that the de^(x-d)-(x-d) comes to e^0 = 1?

But the why is the other other exponential all over d, and not d - d?

Thanks.

Seán

 

[I like Serena]

I do not understand why you are using minus signs now instead of division slashes.
Any particular reason?

Anyway, you need to evaluate $e^{x/d}e^{-W_B/d}$.


Considering that the power formula is $\displaystyle e^a e^b = e^{a+b}$.
Can you substitute $a=x/d$ and $b=-W_B/d$ in this formula?

 

[SMOF]

Any particular reason?

Yes, two. One, I am a bit of an idiot, and two, I should have been asleep hours ago! Sorry about that

Yes, I think I have it now.

Thanks for everyones help! And I shall make sure I read over what I have written before I submit it in future!

Again, many thanks.

Seán