# What is the surface area of the solid created by revolving #f(x) =x^2ln(x)-x , x in [1,3]# around the x axis?

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To find the surface area of the solid created by revolving ( f(x) = x^2 \ln(x) - x ) around the x-axis over ( x ) in ([1, 3]), we can use the formula for surface area of revolution:

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

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

First, find ( f'(x) ), then use it and ( f(x) ) to calculate the integral.

After integrating, multiply the result by ( 2\pi ) to find the surface area of the solid.

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