Intergration by parts

Attached I have a picture of an integration I am struggling with.

I am trying to use integration by parts, making u equal to "e to the power of minus x squared" and dv/dx equal to x but I keep getting the former (e to the power of minus x squared) in the integral which I can't/don't know how to integrate.

I can't seem to find any other way around it.

Please help.

Thank you.

Try a substitution: $u = x^{2}$

Attached I have a picture of an integration I am struggling with.

I am trying to use integration by parts, making u equal to "e to the power of minus x squared" and dv/dx equal to x but I keep getting the former (e to the power of minus x squared) in the integral which I can't/don't know how to integrate.

I can't seem to find any other way around it.

Please help.

Thank you.

Try a substitution: $u = x^{2}$

By using integration by substitution (substituition u= minus x squared), I've got the answer of 0.5 (i.e. a half). Hoping this is correct now.

Use polar coordinates.

You don't need polar coordinates for this integral - it's a simple substitution that's required

(you may be thinking of the normal probability integral which is easiest to do by squaring and transforming to polar coordinates)

Yea, I forgot there was the x otherwise

$\displaystyle I=\int_{- \infty}^{\infty} e^{-x^2} dx$

$\displaystyle I^2= (\int_{- \infty}^{\infty} e^{-x^2} dx)^2=(\int_{- \infty}^{\infty} e^{-x^2} dx)(\int_{- \infty}^{\infty} e^{-x^2} dx)$

$\displaystyle I^2= (\int_{- \infty}^{\infty} e^{-x^2} dx)(\int_{- \infty}^{\infty} e^{-y^2} dy)$

$\displaystyle I^2= \int_{- \infty}^{\infty}\int_{- \infty}^{\infty} e^{-(x^2+y^2)} dxdy$ Switching to polar coordinates $r=x^2+y^2$



$\therefore I^2= \pi \Rightarrow I=\sqrt{\pi}$.



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