# What is the surface area of the solid created by revolving f(x)=e^(x^2-x)/(x+1)-e^x over x in [0,1] around the x-axis?

Area $= 17.01332$ square units
$A = 2 \pi {\int}_{0}^{1} f \left(x\right) \mathrm{ds}$
A=2 pi* int_0^1 ((e^(x^2-x))/(x+1)-e^x) (sqrt(1+(((2x^2+x-2)(e^(x^2-x)))/(x+1)^2-e^x)^2) dx
$A = - 17.01332$ square units