Can you simplify 2e^(ln12/2) and 2e^(ln6/2)?

[jrjack]

Homework Statement

The full problem is y=e^(x/2) and y=e^(-x) revolved around the x-axis, from [ln6,ln12].
The part I'm struggling with is evaluating 2e^(ln12/2) and 2e^(ln6/2).
I am just reviewing for my Calculus 1 final.

Homework Equations

The Attempt at a Solution

2e^{(ln12/2)}

2 e^{(1/2)(ln12)}

2 (e^{1/2}), e^{ln12}=12, and 2(12)e^{1/2}?

 

[rock.freak667]

Well you can easily show that e^(lnA) = A. So you can use that result as is.

 

[jrjack]

The problem is e^{(\frac{ln12}{2})}
not
e^{ln\frac{12}{2}}, that would be e^(ln6)=6

 

[jrjack]

I have a full equation that looks like this:
\pi((2e^{\frac{ln12}{2}}+e^{-12})-(2e^{\frac{ln6}{2}}+e^{-6}))

 

[SammyS]

Homework Statement

The full problem is y=e^(x/2) and y=e^(-x) revolved around the x-axis, from [ln6,ln12].
The part I'm struggling with is evaluating 2e^(ln12/2) and 2e^(ln6/2).
I am just reviewing for my Calculus 1 final.
...

I doubt that you have stated the full problem ?

What are you finding? Volume, Area, ... ?

What is being revolved around the x-axis ?

 

[jrjack]

Sorry, maybe I have made a mistake in my prior work. (this may need to be moved to the calculus section)
The problem states: Use the washer method to find the volume of the solid when the region bounded by y=e^(x/2) and y=e^(-x), x=ln6, x=ln12, when the region is revolved around the x-axis.

So I started with:
\int_{ln6}^{ln12}\pi(e^{\frac{x}{2}}-e^{-x})dx

\pi(2e^{x/2}+e^{-x})|_{ln6}^{ln12}

\pi((2e^{\frac{ln12}{2}}+e^{-ln12})-(2e^{\frac{ln6}{2}}+e^{-ln6}))

 

[Mentallic]

a^{\frac{1}{2}}=\sqrt{a}

and

\left(a^b\right)^c=a^{bc}

using these two rules for indices, you should be able to answer your question.

 

[jrjack]

Thanks, so...
2e^{\frac{ln12}{2}}=2\sqrt{e^{ln12}}=2\sqrt{12}
=4\sqrt{3} and
2\sqrt{e^{ln6}}=2\sqrt{6}
giving me:
\pi((4\sqrt{3}-\frac{1}{12})-(2\sqrt{6}-\frac{1}{6}))
\pi(\frac{1}{12}+4\sqrt{3}-2\sqrt{6})
But that was wrong, and I am unsure how to get the correct answer.
How does that become \frac{575\pi}{96}?

 

[Karamata]

giving me:
\pi((4\sqrt{3}-\frac{1}{12})-(2\sqrt{6}-\frac{1}{6}))

No, this isn't true.

-\frac{1}{12},-\frac{1}{6} here is mistake.

How does that become \frac{575\pi}{96}?

Hm, I checked this in one program, and it isn't true also.

 

[SammyS]

Sorry, maybe I have made a mistake in my prior work. (this may need to be moved to the calculus section)
The problem states: Use the washer method to find the volume of the solid when the region bounded by y=e^(x/2) and y=e^(-x), x=ln6, x=ln12, when the region is revolved around the x-axis.

So I started with:
\int_{ln6}^{ln12}\pi(e^{\frac{x}{2}}-e^{-x})dx
...

Using the washer method to find the volume of the solid when the region bounded by y=f(x) and y=g(x), x=a, x=b, is revolved around the x-axis, (Assumes f(x)>g(x) on the interval [a,b].) results in the following integral:

\displaystyle \int_a^b\pi\left((f(x))^2-(g(x))^2\right)\,dx​

You failed to square the two functions.

 

[jrjack]

Man do I feel like an idiot. Thanks for pointing that out.