# What is the surface area of the solid created by revolving #f(x) = x^2-3x+24 , x in [2,3]# around the x axis?

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To find the surface area of the solid created by revolving ( f(x) = x^2 - 3x + 24 ) on the interval ( x ) in ([2,3]) around the ( x )-axis, we can use the formula for the surface area of a solid of revolution:

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

First, we need to find the derivative of ( f(x) ), which is ( f'(x) ).

[ f'(x) = 2x - 3 ]

Next, we need to find ( \sqrt{1 + (f'(x))^2} ).

[ \sqrt{1 + (f'(x))^2} = \sqrt{1 + (2x - 3)^2} ]

[ = \sqrt{1 + 4x^2 - 12x + 9} ]

[ = \sqrt{4x^2 - 12x + 10} ]

Now, we can set up the integral:

[ S = 2\pi \int_2^3 (x^2 - 3x + 24) \sqrt{4x^2 - 12x + 10} , dx ]

We integrate this expression over the interval ([2,3]) to find the surface area of the solid of revolution.

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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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