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Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the given curves about the specified axis. y = 7e^x, y = 7e−x, x = 1; about the y-axis

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I don't get what I did wrong in my work. Let me post it.
Did I set up the integral properly?

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Other answers:

What did you get for your points of intersection? y = 7e^x vs x = 1 is obviously (1,7e) y = 7e−x vs x = 1 is obviously (1,7e-1) But what about y = 7e^x vs y = 7e−x ? Since this cannot be solved by direct methods, we get (0.9488478,18.0791250) Maybe the problem statement is not written correctly?
Let me send a pic.
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I used the e points of intersection so I used 0 to 1.
x not e.
\(y = 7e^{x}\) -- You wrote this one okay. \(y = 7e^{-x}\) -- This is quite a different story. Please learn to code LaTeX or to use the Equation Editor. It will be a great benefit to you now and in the future. You're f(x) is not correct. You just picked one of the curves. You need the distance between them.
Ohh I can write in latex. It just takes too long lol.
It would have avoided much time and error. 7e-x is VERY different from 7e^(-x) or \(7e^{-x}\)
So I wrote my integral as: \[14\pi \int\limits_{}^{}xe^x dx\]
How do I incorporate the e^-x then?
\(e^{x}\) is no good. You need \(e^{x} - e^{-x}\).
Huh? Why?
Nevermind. I see why.
The limits of intergation should still be the same right?
Did you draw a picture? The height of your cyllinders is the difference of the two curves.
Yeah I did, One moment.
Something like that.
That's it. And the final answer is?
I see why it's the difference though.
\[\frac{ 28\pi }{ e}\] is what I got.
I like it. Now, your punishment for writing one of the equations incorrectly in the beginning is to find the point of intersection for what you wrote. Find the value of x such that \(7e^{x} = 7e - x\). It's a wonderful exploration.
Eww... Natural logs...
No, no. It's far worse than that. :-) When you are ready...
Ahh I see... Hehe...
Yeah... I have to use a calculator lol.

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