Hi there can someone pls help me with part (i) of this question. It is the integration I'm unsure about. It's probably quite basic but I can't work it out. The question requires you to integrate f(x) = ax.exp{-x^2} for 0<x<2 so I thought the way to do it is integration by parts where I set u=x, du=1 dv=exp{-x^2}, v=(exp{-x^2})/(-2x) and then carry out uv - integral (vdu), leaving out a for the moment as it's s constant. but then I realised I have to keep doing integration by parts on the second part. Also the solutions just gives a(exp{-x^2})/(-2)) can someone pls help, am I overlooking something? Thanks loads! a(exp{-x^2})/(-2x
They make use of the trick that integral of f'(x) Exp{ f(x)} dx= Exp {f(x)} +c where f'(x) = d/dx{f(x)} To see this diffentiate the RHS. Your integration is flawed as you can't simply integrate exp(-x^2) (The normal distribution pdf is effectively like this (with some extra constants) and we resort to numerical approximations and tables to calculate normal probabilities.)
f(x) = ax.exp{-x^2} for 0<x<2 What about this approach: Let, t=e{-x^2} dt=-2x.e{-x^2}dx x.exp{-x^2}dx=-dt/2 The integral becomes: Int(-at/2)dt
Substitution is a good method to use here, but you have a slight mistake. LHS is exactly the same as what you want to find except for the a, so the integral becomes Integral {-a/2 dt} = [-a/2 t ] (no t term in LHS) I would change it back to x before applying the limits but if you really want you could find the equivalent values of t and use those.
The solution in the revision notes for (iii) assumes C=2. Why make an assumption instead of doing actual calc to determine the max. of f(x)/h(x) ?
