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

Volume :

graph{x^2e^(2x) [-5, 5, -10000, 10000]}

Now,

#=1/4(x^2e^(4x)--int e^(4x)d(x^2))

Substituting the limits for the difference, the value here is

So, the volume is

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

[ S = \int_{a}^{b} 2\pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} , dx ]

where ( y = f(x) ) and ( \frac{dy}{dx} ) is the derivative of ( f(x) ) with respect to ( x ).

First, let's find ( \frac{dy}{dx} ):

[ f(x) = xe^{2x} ] [ \frac{dy}{dx} = e^{2x} + 2xe^{2x} ]

Now, we can plug these values into the formula and integrate over the interval ( x \in [2,3] ):

[ S = \int_{2}^{3} 2\pi (xe^{2x}) \sqrt{1 + (e^{2x} + 2xe^{2x})^2} , dx ]

After integrating this expression, we obtain 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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