# 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

#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#

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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.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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