Answer
Converges to $125$
Work Step by Step
Let us consider that $I_n=\int_{0}^{n} 5x e^{-0.02x^2} \ dx$
Suppose $a=-0.02x^2 \implies da=-0.04x \ dx$
$I_n=-125\int_{0}^{-0.02n^2} e^{a} \ da=125 (1-e^{-0.02n^2})$
We define the given integral as a limit of $I_n$ as: $I=\lim\limits_{n \to \infty} I_n=\int_{0}^{\infty} 5x e^{-0.02x^2} \ dx \\=\lim\limits_{n \to \infty} 125 (1-e^{-0.02n^2})=125$
This means that the limit exist and the given integral converges to $125$.