What is the surface area of the solid created by revolving #f(x) = x^2+e^x , x in [2,4]# around the x axis?
First, take a look at the graph
I've drawn vertical lines at the endpoints to show where the solid would revolve around the
The formula for find the volume of a shape like this is
In this case, you can plug in all the values and find the integral in a straight forward fashion.
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To find the surface area of the solid created by revolving ( f(x) = x^2 + e^x ) around the xaxis over the interval ([2, 4]):

Determine the formula for the surface area of revolution, which is given by: [ SA = 2\pi \int_{a}^{b} f(x) \sqrt{1 + \left(\frac{dy}{dx}\right)^2} , dx ]

Find the derivative of ( f(x) ) with respect to ( x ), which is: [ f'(x) = 2x + e^x ]

Use the derivative to find the expression for ( \frac{dy}{dx} ).

Substitute the expression for ( \frac{dy}{dx} ) into the formula for surface area.

Integrate the resulting expression from ( x = 2 ) to ( x = 4 ) to find the surface area of the solid.
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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