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.

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