# What is the surface area of the solid created by revolving #f(x) = x^2-x , x in [2,7]# around the x axis?

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To find the surface area of the solid created by revolving ( f(x) = x^2 - x ) around the x-axis over the interval ( x \in [2, 7] ), you can use the formula for the surface area of a solid of revolution:

[ A = 2\pi \int_{a}^{b} f(x) \sqrt{1 + \left(\frac{df}{dx}\right)^2} , dx ]

Where ( a = 2 ) and ( b = 7 ), and ( f(x) = x^2 - x ).

First, find ( \frac{df}{dx} ):

[ \frac{df}{dx} = \frac{d}{dx} (x^2 - x) = 2x - 1 ]

Now, plug the function and its derivative into the formula:

[ A = 2\pi \int_{2}^{7} (x^2 - x) \sqrt{1 + (2x - 1)^2} , dx ]

[ = 2\pi \int_{2}^{7} (x^2 - x) \sqrt{1 + 4x^2 - 4x + 1} , dx ]

[ = 2\pi \int_{2}^{7} (x^2 - x) \sqrt{4x^2 - 4x + 2} , dx ]

This integral may require techniques like substitution or other advanced methods to solve, as it involves a square root. Once you find the antiderivative and evaluate it over the interval ([2, 7]), you will have 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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