Question

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?

Answer 1

Area `= 17.01332` square units

Explanation:

`A=2 pi int_0^1 f(x) 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