`y = e^(x/2) + e^(-x/2) , y = 0 , x = -1 , x = 2` Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis.

Expert Answers

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Given

`y=e^(x/2)+e^(-x/2), y=0 x=-1,x=2`

so the solid of revolution about x-axis is given as

`V = pi * int _a ^b [R(x)^2 -r(x)^2] dx`

here

`R(x) = e^(x/2)+e^(-x/2)`

`r(x)=0` and the limits are `a=-1 ` and` b=2`

so ,

`V = pi * int _a ^b [R(x)^2 -r(x)^2] dx`

= `pi * int _-1 ^2 [(e^(x/2)+e^(-x/2))^2 -0^2] dx`

=`pi * int _-1 ^2 [(e^(x/2)+e^(-x/2))^2 ] dx`

=`pi * int _-1 ^2 [(e^(x/2)+e^(-x/2))^2 ] dx`

=`pi * int _-1 ^2 [e^x+e^(-x)+2 ] dx`

=`pi * [e^x -e^(-x)+2x]_-1 ^2 `

=`pi * [[e^2 -e^(-2)+4]-[e^(-1) -e^(1)+2(-1)]] `

=`pi*[[11.253]-[-4.350]]`

=`pi*[15.603] `

=`49.018`

is the volume

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