It is known that ∫e^(-x^2 )dx=sqrt(x)/2 from 0 to ∞ Using this, find the value of ∫x^2 e^(-x^2 ) dx from 0 to∞

Use integration by parts.

Let u = x, dv = xe^(-x^2) dx
du = dx, v = (-1/2)e^(-x^2).

So, ∫(x = 0 to ∞) x^2 e^(-x^2) dx
= (-1/2)xe^(-x^2) {for x = 0 to ∞} - ∫(x = 0 to ∞) (-1/2) e^(-x^2) dx
= 0 + (1/2) ∫(x = 0 to ∞) e^(-x^2) dx, via L'Hopital's Rule
= (1/2)(√(π) / 2), via corrected hint (which has a typo)
= √(π)/4.

I hope this helps!