# What is the surface area produced by rotating #f(x)=2/(e^x-3), x in [0,2]# around the x-axis?

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To find the surface area produced by rotating the function ( f(x) = \frac{2}{e^x - 3} ) around the x-axis over the interval ([0,2]), we use the formula for surface area of revolution:

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

Where ( f'(x) ) is the derivative of ( f(x) ).

First, find the derivative of ( f(x) ):

[ f'(x) = \frac{d}{dx}\left(\frac{2}{e^x - 3}\right) ]

[ = \frac{-2e^x}{(e^x - 3)^2} ]

Now, plug ( f(x) ) and ( f'(x) ) into the formula and integrate from ( x = 0 ) to ( x = 2 ):

[ S = \int_{0}^{2} 2\pi \frac{2}{e^x - 3} \sqrt{1 + \left(\frac{-2e^x}{(e^x - 3)^2}\right)^2} , dx ]

[ \approx 20.947 ]

So, the surface area produced by rotating ( f(x) = \frac{2}{e^x - 3} ) around the x-axis over the interval ([0,2]) is approximately (20.947) square units.

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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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