We have to show that ,

$1<\int_{0}^{1} e^{x^{2}} d x<e$

$ 0< x <1$

It implies, $0 < x^2 <1$

Now with this reduced form of the equation why don't you give it a try yourself, I am sure you can do it.

Thus, $ e^0 < e^{x^2} <e^1 $

i.e $1 < e^{x^2} <e $

So you are just one step away from solving your problem, go on.............

Therefore, Integrating the inequality with limits $0$ to $1$ we get, $\int\limits_0^1 \mathrm dx < \int\limits_0^1 e^{x^2} \mathrm dx < \int\limits_0^1e \mathrm dx$