anonymous
  • anonymous
integral (x^2 e^-x^2) dx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Integrate by parts once. The resulting integral will be a simple substitution.
anonymous
  • anonymous
That's not true.
anonymous
  • anonymous
Ah, missed the minus sign...

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anonymous
  • anonymous
is u=x^2 and dv=?
anonymous
  • anonymous
It's not the minus sign, is the fact that its \[ e^{x^2} \] it's called a Gaussian integral.
anonymous
  • anonymous
@blinn13 disregard my suggestion
anonymous
  • anonymous
oops, sorry, \[ e^{-x^2} \]
anonymous
  • anonymous
okay
experimentX
  • experimentX
the integral is not in closed form.
experimentX
  • experimentX
*cannot be expressed
anonymous
  • anonymous
What are the limits of integration?
anonymous
  • anonymous
0 to 4....im trying to find a centroid
experimentX
  • experimentX
hmm ... i would use calculator
anonymous
  • anonymous
i have already found the number i need to show work though :/
experimentX
  • experimentX
numerical methods?
anonymous
  • anonymous
Well then there's a mistake somewhere because it can't be done by hand.
anonymous
  • anonymous
i see...is (e^-x)^2 equal to e^-2x or e^-x^2 then?
experimentX
  • experimentX
e^(-2x) of course
anonymous
  • anonymous
thats where my problem was then...thank you
anonymous
  • anonymous
then it just becomes u-sub.
experimentX
  • experimentX
yes yes
anonymous
  • anonymous
Ah. Yep, there we go.
experimentX
  • experimentX
or by parts
anonymous
  • anonymous
by parts would be better
anonymous
  • anonymous
so u=x^2, du=2x dx, dv=e^-2x dx, v=-e^-2x/2..right?
anonymous
  • anonymous
So by parts was right, eh? hehe
experimentX
  • experimentX
how do you use IE by parts? here we use \[ \int u v dx = u \int v - \int (du/dx \int v ) \]
anonymous
  • anonymous
i have never seen that before
experimentX
  • experimentX
u = x^2 v = e^(-2x)
anonymous
  • anonymous
would i be able to do it with normal integration by parts?
experimentX
  • experimentX
yes .. twice
anonymous
  • anonymous
ok ill try that...my previous values for u and dv are correct?
experimentX
  • experimentX
i know that type of form exist ... but i don't use it often.
experimentX
  • experimentX
\[ \begin{align*} \int x^2 e^{-2x}dx dx &= x^2 \int e^{-2x}dx - \int \left( \frac{dx^2}{dx} \int e^{-2x}dx\right )dx \\ &= x^2 \frac{e^{-2x}}{-2 } - \int 2x \frac{e^{-2x}}{-2 }dx\\ &= - \frac{x^2 e^{-2x}}{2 } + \int x e^{-2x}dx \\ &= - \frac{x^2 e^{-2x}}{2 } + x \int e^{-2x}dx - \int \left( \frac{dx}{dx}\int e^{-2x} dx\right )dx\\ \end{align*} \]
anonymous
  • anonymous
we have not gotten that far but i think it is coming up soon
experimentX
  • experimentX
this is integration by parts
anonymous
  • anonymous
i learned it a different way then : \[uv -\int\limits vdu\]
experimentX
  • experimentX
yes yes i know that ... integration by parts is just opposite of product rule in differentiation
anonymous
  • anonymous
i know that
anonymous
  • anonymous
thank you for your help!
experimentX
  • experimentX
\[ d(uv) = u dv + v du \\ \int d(vu) = \int u dv + \int v du \\ uv = \int u dv + \int v du \\ \int v du = uv - \int u dv \] put v=g(x), u = \( \int f(x) dx \) you get \[ \int g d\left( \int f \right ) = g \int f dx - \int \left(\frac{dg}{dx} \int f dx\right ) dx \] I think is is easier to use.
experimentX
  • experimentX
\[ \int f g dx = g \int f dx - \int \left(\frac{dg}{dx} \int f dx\right ) dx \]

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