# 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?

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

We can calculate this integral by parts:

so

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To find the surface area of the solid created by revolving the function ( f(x) = \frac{e^{x+1}}{x+1} ) over ( x ) in the interval ([0,1]) around the x-axis, follow these steps:

- Calculate the arc length of the curve from ( x = 0 ) to ( x = 1 ) using the formula for arc length:

[ S = \int_{a}^{b} 2\pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx ]

- Determine the derivative ( \frac{dy}{dx} ) of the function ( f(x) ).
- Evaluate ( y = f(x) ) and ( \frac{dy}{dx} ) at each point in the interval ([0,1]).
- Plug the values into the arc length formula and integrate over the interval ([0,1]).
- The result will be the surface area of the solid generated by revolving the function around the x-axis.

Let's go through the steps:

- Calculate the derivative of ( f(x) ):

[ f(x) = \frac{e^{x+1}}{x+1} ]

[ f'(x) = \frac{e^{x+1}(x+1) - e^{x+1}}{(x+1)^2} = \frac{e^{x+1}(x) - e^{x+1}}{(x+1)^2} ]

- Plug the function and its derivative into the arc length formula:

[ S = \int_{0}^{1} 2\pi \left(\frac{e^{x+1}}{x+1}\right) \sqrt{1 + \left(\frac{e^{x+1}(x) - e^{x+1}}{(x+1)^2}\right)^2} dx ]

- Evaluate this integral numerically to find the surface area.

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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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