How do you find the volume bounded by #y=e^(2x)#, the y-axis and the line y=2 revolved about the x=axis?
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To find the volume bounded by ( y = e^{2x} ), the y-axis, and the line ( y = 2 ) revolved about the ( x )-axis, you can use the method of cylindrical shells. The formula for the volume of a solid generated by revolving the region bounded by ( f(x) ), the ( x )-axis, and the vertical lines ( x = a ) and ( x = b ) about the ( x )-axis is:
[ V = 2\pi \int_{a}^{b} x \cdot f(x) , dx ]
In this case, ( f(x) = e^{2x} ), ( a = 0 ), and ( b ) is the ( x )-value where ( y = e^{2x} ) intersects ( y = 2 ). So, you first need to solve ( e^{2x} = 2 ) to find ( b ).
[ e^{2x} = 2 ] [ 2x = \ln(2) ] [ x = \frac{1}{2} \ln(2) ]
Then, you plug these values into the formula and integrate:
[ V = 2\pi \int_{0}^{\frac{1}{2}\ln(2)} x \cdot e^{2x} , dx ]
After integrating, you'll have the volume of the solid.
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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.
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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