What is the surface area of the solid created by revolving #f(x) =2x^3-2x , x in [2,4]# around the x axis?
The area is given by the following integral :
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To find the surface area of the solid created by revolving the function ( f(x) = 2x^3 - 2x ) around the x-axis over the interval ([2, 4]), you can use the formula for surface area of revolution, which is given by:
[ S = \int_{a}^{b} 2\pi f(x) \sqrt{1 + \left( f'(x) \right)^2} , dx ]
First, find the derivative of ( f(x) ) which is ( f'(x) = 6x^2 - 2 ). Then, plug the function and its derivative into the formula and integrate over the given interval ([2, 4]) to find the surface area.
[ S = \int_{2}^{4} 2\pi (2x^3 - 2x) \sqrt{1 + (6x^2 - 2)^2} , dx ]
Compute this integral to find the surface area of the solid.
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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.
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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