Prove$f(x)=\int_{2}^{x^2}\int_{0}^{s}e^{-t^2}\ dtds$ is continuous and convex

Question

I want to show:

$$f(x)=\int_{2}^{x^2}\int_{0}^{s}e^{-t^2}\ dtds$$ is a continuous and convex function

Consider: $h(s)=\int_{0}^{s}e^{-t^2}\ dt$

Then I get: $f(x)=\int_{2}^{x^2}h(s)\ ds$

implies that

$f'(x)=2xh(x^2)\geq0 $ and $f''(x)=2h(x^2)+4x^2 e^{-x^4}\geq0 $

Is this true? thanks for any help.