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## Homework Statement

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$$y= \frac 1 {\sqrt{2\pi}}e^{ \frac{- x^2} 2},~y=0,~x=0,~x=1$$

## Homework Equations

$$Volume=2\pi\int_a^b p(x)h(x)dx$$

## The Attempt at a Solution

I understand how to do the problem, I'm just having trouble integrating.

##h(x)=\frac 1 {\sqrt{2\pi}}e^{ \frac{- x^2} 2},~ p(x)=x##

And you just plug that into the equation.

When you integrate ##\frac 1 {\sqrt{2\pi}}e^{ \frac{- x^2} 2}x## , you use the chain rule right? But first you move the constant ##\frac 1 {\sqrt{2\pi}}## out of the integral next to the ##2\pi##

So it would look like this, ##Volume=2\pi~\frac 1 {\sqrt{2\pi}}~\int_0^1~ xe^{ \frac{- x^2} 2}dx##

I know the anti derivative of ##e^{ \frac{- x^2} 2} is, e^{ \frac{- x^2} 2}## and then you have to do the anti derivative of ## { \frac{- x^2} 2}## as part of the chain rule right? Well, that is where I get stuck.