Calculate normilization constant for f(x) = e^ [1.5(x^2)]
I think you mean $\displaystyle e^{-1.5x^2}$ because the former does not convergene when we take the integral.
We need to find a constant ("normalization") $\displaystyle k$ so that:
$\displaystyle k\int_{-\infty}^{\infty} e^{-1.5x^2} dx = 1$
In order to turn it into a probability density function.
Thus,
$\displaystyle k\int_{-\infty}^{\infty} e^{-(\sqrt{1.5} x)^2} dx = 1$
Let $\displaystyle t=\sqrt{1.5}x$ use this substitution to get,
$\displaystyle \frac{k}{\sqrt{1.5}}\int_{-\infty}^{\infty} e^{-t^2} dt = 1$
Thus,
$\displaystyle k \cdot \sqrt{1.5}^{-1} \cdot \sqrt{\pi} = 1$
Thus,
$\displaystyle k = \frac{\sqrt{3}}{\sqrt{2\pi}}$