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

[Jovy]

Homework Statement

[/B]

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

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.

 

[Buzz Bloom]

Hi Jovy:

I suggest you investigate the substitution u = (x^2)/2, du = x dx

Hope this helps.

Regards,
Buzz

 

[Ray Vickson]

Homework Statement

[/B]

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

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.

No, the antiderivative of $e^{-x^2/2}$ is a non-elementary function. In other words, it is not possible to write down the antiderivative in a finite number of terms involving the standard functions. Even if you try to write it down in 1 billion pages of horrible algebra, you cannot do it.

Fortunately, you don't need the antiderivative of $e^{-x^2/2}$, but of $x e^{-x^2/2}$ instead, and that is a much different story.

 

[Mark44]

Homework Statement

[/B]

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

What exactly is the problem? All you have given us is an equation in x and y, but you haven't said what it is you're supposed to do?

From your attempt below, it would seem that you're trying to find the volume of a surface of revolution, but there is no clue above that that's what you need to do.

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.