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?

Question

Christopher Allers · Accepted Answer

Area #= 17.01332# square units

#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

Charles Anctil · Answer

undefined To find the surface area of the solid created by revolving $ f(x) = \frac{e^{x^2-x}}{x+1} - e^x $ over $ x $ in the interval $[0,1] $ around the x-axis, you can use the formula for the surface area of revolution:

$$ S = 2\pi \int_{a}^{b} f(x) \sqrt{1 + \left( f'(x) \right)^2} \, dx $$

Where $ f'(x) $ denotes the derivative of $ f(x) $.

First, compute $ f'(x) $:
$$ f'(x) = \frac{e^{x^2 - x}((2x - 1)(x + 1) - e^x)}{(x + 1)^2} $$

Then, compute $ \sqrt{1 + \left( f'(x) \right)^2} $:
$$ \sqrt{1 + \left( f'(x) \right)^2} = \sqrt{1 + \left( \frac{e^{x^2 - x}((2x - 1)(x + 1) - e^x)}{(x + 1)^2} \right)^2} $$

Now, integrate $ f(x) \sqrt{1 + \left( f'(x) \right)^2} $ from $ x = 0 $ to $ x = 1 $:
$$ S = 2\pi \int_{0}^{1} \left( \frac{e^{x^2-x}}{x+1} - e^x \right) \sqrt{1 + \left( \frac{e^{x^2 - x}((2x - 1)(x + 1) - e^x)}{(x + 1)^2} \right)^2} \, dx $$

This integral represents the surface area of the solid of revolution. You can solve it using numerical methods or integrate it symbolically if possible.