Inequality involving $\int_0^t e^{\alpha s^2} ds$. How to justify it?

Let $\alpha>0$ and let $$G(t) =\int_0^t e^{\alpha s^2} ds.$$ My Professor said that it is possible to estimate $$ G(t)\le e^{\alpha t^2} +C,$$ where $C$ denotes a positive constant. Could someone please help me understand why? I think it could be something related to the monoticity of exp functioin, but I am not sure about that.

Moreover, he left (as an exercise) to find a similar estimate but for $$F(t) =\int_0^ t (e^{\alpha s^2}-1) ds.$$ Could someone please help me?

Thank you in advance!

Solution 1:

Hint. Note that $$G(t) =\int_0^{\frac{1}{2\alpha}} e^{\alpha s^2} ds+\int_{\frac{1}{2\alpha}}^t e^{\alpha s^2} ds\leq \int_0^{\frac{1}{2\alpha}} e^{\alpha s^2} ds+\int_{\frac{1}{2\alpha}}^t 2\alpha s e^{\alpha s^2} ds=\int_0^{\frac{1}{2\alpha}} e^{\alpha s^2} ds+[e^{\alpha s^2}]_{\frac{1}{2\alpha}}^t.$$ Moreover $F(t)=G(t)-t$.