Integral of 1/2^x | Homework Solution

[Maddie1609]

Homework Statement

∫₁^∞ 1/2^x

Homework Equations

2^-x = (e^{ln 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

...

 

[HallsofIvy]

Homework Statement

∫₁^∞ 1/2^x

Homework Equations

2^-x = (e^{ln 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

...

 

[Samy_A]

Homework Equations

2^-x = (e^{ln 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.

 

[Maddie1609]

Quite relevant, try to build on that.

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

∫(1/2^x) dx = ∫(1/e^{x ln 2}) dx

u = ln 2 x → du = ln 2 dx

1/ln 2 ∫ du/e^u = ln e^u / ln 2 = ln e^{x ln 2} / ln 2 = x ln 2/ln2 = x

Okay, what did I do wrong?

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?

 

[Maddie1609]

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

Wouldn't that be e^{-x ln 2}?

 

[Samy_A]

∫(1/2^x) dx = ∫(1/e^{x ln 2}) dx

u = ln 2 x → du = ln 2 dx

1/ln 2 ∫ du/e^u = ln e^u / 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).

 

[Samy_A]

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.

 

[Maddie1609]

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.

 

[HallsofIvy]

Wouldn't that be e^{-x ln 2}?

Yes. Thanks.