Integral of 1/2^x | Homework Solution

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Homework Statement


1 1/2x

Homework Equations



2-x = (eln 2)-x

Could possibly be relevant, I don't know.

The Attempt at a Solution



y = 2-x

ln y = -x ln 2

1/y dy/dx = -ln 2 -x/2

...
 
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Maddie1609 said:

Homework Statement


1 1/2x

Homework Equations



2^-x = (eln 2)-x
Better is 2^{-x}= e^{ln(2^{-x})}= e^{x ln(2)}

Could possibly be relevant, I don't know.

The Attempt at a Solution



y = 2-x

ln y = -x ln 2

1/y dy/dx = -ln 2 -x/2

...
 
Maddie1609 said:

Homework Equations



2-x = (eln 2)-x

Could possibly be relevant, I don't know.
Quite relevant, try to build on that.

ETA: never mind, didn't see the previous post.
 
Last edited:
Samy_A said:
Quite relevant, try to build on that.

ETA: never mind, didn't see the previous post.

∫(1/2x) dx = ∫(1/ex ln 2) dx

u = ln 2 x → du = ln 2 dx

1/ln 2 ∫ du/eu = ln eu / ln 2 = ln ex ln 2 / ln 2 = x ln 2/ln2 = x

Okay, what did I do wrong?o_O

Edit: I was thinking of u'/ln u, my bad!

1/ln 2 ∫ e-u du = -e-u/ln 2 = -e-x ln 2/ln 2 = -2-x/ln2

Is this correct?
 
Last edited:
HallsofIvy said:
Better is 2^{-x}= e^{ln(2^{-x})}= e^{x ln(2)}
Wouldn't that be e-x ln 2?
 
Maddie1609 said:
∫(1/2x) dx = ∫(1/ex ln 2) dx

u = ln 2 x → du = ln 2 dx

1/ln 2 ∫ du/eu = ln eu / ln 2
You lost me in this last equality, I guess here something is wrong.

Also, don't forget to adapt the limits of your definite integral when you apply the substitution (not that I think that you really need that substitution).
 
Maddie1609 said:
Edit: I was thinking of u'/ln u, my bad!

1/ln 2 ∫ e-u du = -e-u/ln 2 = -e-x ln 2/ln 2 = -2-x/ln2

Is this correct?
Yes, this is correct for the indefinite integral.
 
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Samy_A said:
You lost me in this last equality, I guess here something is wrong.

Also, don't forget to adapt the limits of your definite integral when you apply the substitution (not that I think that you really need that substitution).
Yes I made an edit on my post, I was thinking of u'/ln u. I think I've got it now.

Screenshot_2015-10-25-14-23-21.png
 
Maddie1609 said:
Wouldn't that be e-x ln 2?
Yes. Thanks.
 
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