# 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?

Nov 28, 2016

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

$\pi {\int}_{a}^{b} {f}^{2} \left(x\right) \mathrm{dx}$

#### Explanation:

$V = \pi {\int}_{0}^{1} {\left(\frac{{e}^{x + 1}}{x + 1}\right)}^{2} \mathrm{dx} = \pi {\int}_{0}^{1} {e}^{2 \left(x + 1\right)} / {\left(x + 1\right)}^{2} \mathrm{dx}$

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

$V = 2 \pi {\int}_{2}^{4} {e}^{t} / {t}^{2} \mathrm{dt}$

We can calculate this integral by parts:

$\int {e}^{t} / {t}^{2} \mathrm{dt} = - \int {e}^{t} d \left({t}^{- 1}\right) = - {e}^{t} / t + \int {e}^{t} / t$

$\int {e}^{t} / t \mathrm{dt}$ 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!))