`y = e^(-x^2/2)/sqrt(2pi) , y = 0 , x = 0 , x = 1` Use the shell method to set up and evaluate the integral that gives the volume of the solid...

Using the shell method we can find the volume of the solid generated by the given curves,

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

Using the shell method the volume is given as

`V= 2*pi int _a^b p(x) h(x) dx`

where p(x) is the function of the average radius  `=x`

and

h(x) is the function of height =  `e^(-x^2/2)/sqrt(2pi)`

and the range of x is given as...

Using the shell method we can find the volume of the solid generated by the given curves,

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

Using the shell method the volume is given as

`V= 2*pi int _a^b p(x) h(x) dx`

where p(x) is the function of the average radius  `=x`

and

h(x) is the function of height =  `e^(-x^2/2)/sqrt(2pi)`

and the range of x is given as 0 to 1

So the volume is  `= 2*pi int _a^b p(x) h(x) dx`

`= 2*pi int _0^1 (x) (e^(-x^2/2)/sqrt(2pi)) dx`

`=(2*pi)/(sqrt(2pi)) int _0^1 (x*e^(-x^2/2)) dx`

`=(2*pi)/(sqrt(2pi)) int _0^1 (x*e^(-x^2/2)) dx`

let us first solve

`int (x*e^(-x^2/2)) dx`

let `u = x^2/2`

`du = 2x/2 dx = xdx`

so ,

`int (x*e^(-x^2/2)) dx`

`= int  (e^(-u)) du`

`= -e^(-u) = -e^(-x^2/2)`

So,  `V=(2*pi)/(sqrt(2pi)) int _0^1 (x*e^(-x^2/2)) dx`

 `=(2*pi)/(sqrt(2pi)) [-e^(-x^2/2)]_0^1`

=`(2*pi)/(sqrt(2pi)) [[-e^(-(1)^2/2)]-[-e^(-0^2/2)]]`

=`(2*pi)/(sqrt(2pi)) [[-e^(-1/2)]-[-e^(0)]]`

=`(sqrt(2pi)) [1-[e^(-1/2)]] `

=` 0.986`

is the volume