What is the surface area produced by rotating #f(x)=2/x-1/x^2, x in [1,3]# around the x-axis?

Answer 1

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Answer 2

To find the surface area produced by rotating the function ( f(x) = \frac{2}{x} - \frac{1}{x^2} ) around the x-axis over the interval ( x ) in [1,3], we can use the formula for surface area of revolution:

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

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

[ f'(x) = -\frac{2}{x^2} + \frac{2}{x^3} ]

Now, we can plug in the values and integrate:

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

After evaluating this integral, you'll find the surface area.

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Answer from HIX Tutor

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.

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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