# #f:RR->RR;f(x)=e^x-x-1;g:[0,1]->RR;g(x)=f(x)+x#.How to calculate the volume of the body obtained by rotating the graph of the function "g" axis OX?

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To calculate the volume of the body obtained by rotating the graph of the function ( g ) around the ( x )-axis, you can use the method of cylindrical shells.

The formula for the volume of the solid generated by rotating the graph of ( g(x) ) around the ( x )-axis from ( x = a ) to ( x = b ) is given by:

[ V = 2\pi\int_{a}^{b} x \cdot g(x) , dx ]

In this case, ( g(x) = f(x) + x ). So, you'll need to substitute ( f(x) + x ) for ( g(x) ) in the formula and integrate from ( x = 0 ) to ( x = 1 ), as ( g(x) ) is defined over the interval ([0, 1]).

[ V = 2\pi\int_{0}^{1} x \cdot (f(x) + x) , dx ]

The function ( f(x) = e^x - x - 1 ), so you need to substitute ( e^x - x - 1 ) for ( f(x) ).

[ V = 2\pi\int_{0}^{1} x \cdot (e^x - x - 1 + x) , dx ]

[ V = 2\pi\int_{0}^{1} x(e^x - 1) , dx ]

You can integrate ( x(e^x - 1) ) with respect to ( x ) from ( x = 0 ) to ( x = 1 ) using integration by parts or other appropriate techniques. Once you find the antiderivative, evaluate the definite integral to get the volume 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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