What is the surface area of the solid created by revolving #f(x)=e^(x+1)/(x+1)# over #x in [0,1]# around the x-axis?

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Answer 1 · 2016-11-28T22:15:19

The area of the solid of revolution around the x-axis of a curve is calculated as:

#pi int_a^b f^2(x)dx#

#V= pi int_0^1((e^(x+1))/(x+1))^2dx = pi int_0^1e^(2(x+1))/(x+1)^2dx#

Substitute #t=2(x+1); dx=dt/2#

#V=2pi int_2^4e^t/t^2dt#

We can calculate this integral by parts:

#int e^t/t^2dt = -int e^t d(t^(-1)) = -e^t/t + int e^t/t #

# int e^t/tdt # is not easily solved, you can find it on manuals:

# int e^t/tdt =log|x| + sum_0^oox^n/(n*n!)#,

so

#V=2pi(-e^4/4+e^2/2+log 4 -log 2 +sum_0^oo4^n/(n*n!)-sum_0^oo2^n/(n*n!))#