Volume of region roated about specified axi

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SUMMARY

The discussion focuses on calculating the volume of the region bounded by the curve y = e^(-x), the x-axis (y = 0), and vertical lines x = -1 and x = 0, when rotated about the line x = 1. The integral used for this calculation is 2π∫ (1-x)e^(-x) dx from -1 to 0, which simplifies to 2π(e) after applying integration by parts. The final result confirms the correctness of the approach and calculations presented.

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Homework Statement



y = e^(-x), y = 0, x = -1, x = 0, about x = 1

Homework Equations





The Attempt at a Solution


2pi∫ (1-x)(e^(-x) dx from -1 to 0

=2pi∫ e^(-x) -x*e^(-x) dx from -1 to 0
=2pi(-e^(-x) - ∫x*e^(-x) dx from -1 to 0
u=x
du=dx
dv=e^(-x)dx
v=-e^(-x)
2pi(-e^(-x)-(-xe^(-x) +∫e^(-x)dx))
2pi(-e^(-x) +xe^(-x) +e^(-x))
2pi(xe^(-x)) from -1 to 0
2pi(e)

is this right?
 
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Seems ok to me.
 

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