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

Answer 1

The surface area is approximately 105.75.

The surface area #S# is given by
#S=2piint_1^3xe^-x-e^xdx=2pi[int_1^3xe^-xdx-int_1^3e^xdx]#.
#intxe^-x=-e^-x(x+1)+C#

which can be obtained by integration by parts, so

#S=2pi[-e^-x(x+1)-e^x]# evaluated from 1 to 3.
#S=2pi[-e^-3*4-e^3+e^-1*2+e]~~-105.75#

To make this physically meaningful, we would just take the absolute value of this and say that

#S~~105.75#.
The volume, #V#, can be calculated from #piint_1^3(xe^-x-e^x)^2dx#.
#piint_1^3(xe^-x-e^x)^2dx=piint_1^3x^2e^(-2x)-2x+e^(2x)dx#
#=pi(int_1^3x^2e^(-2x)dx-2int_1^3xdx+int_1^3e^(2x)dx)#
The tough one here is #intx^2e^(-2x)dx#. This is done by integration by parts. If you need me to show you the steps for you, just indicate in the comments. For now, I will simply note that
#intx^2e^(-2x)dx=-1/4e^(-2x)(2x^2+2x+1)+C#

And so we have

#V=pi[-1/4e^(-2x)(2x^2+2x+1)-x^2+e^(2x)/2]# evaluated from 1 to 3.
#V=pi[-e^(-6)/4*25-9+e^6/2+e^(-2)/4*5+1-e^2/2]~~597.45#
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Answer 2

To find the surface area of the solid created by revolving the function ( f(x) = xe^{-x} - e^x ) over the interval ( x ) in ([1,3]) around the x-axis, you can use the formula for the 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 ) is the function being revolved, ( a ) and ( b ) are the interval endpoints, and ( \frac{dy}{dx} ) represents the derivative of ( y ) with respect to ( x ).

First, find the derivative of ( f(x) ):

[ f'(x) = e^{-x} - xe^{-x} - e^x ]

Then, calculate ( \left(\frac{dy}{dx}\right)^2 ):

[ \left(\frac{dy}{dx}\right)^2 = \left(e^{-x} - xe^{-x} - e^x\right)^2 ]

Now, substitute ( f(x) ), ( f'(x) ), and the interval ([1,3]) into the surface area formula:

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

This integral will give you the surface area of the solid created by revolving ( f(x) ) around the x-axis over the interval ([1,3]). You can evaluate this integral using numerical methods or software if an exact solution is not feasible.

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Answer from HIX Tutor

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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