Try this problem from ISI-MSQMS 2015 which involves the concept of Integral Inequality.
Show that $1<\int_{0}^{1} e^{x^{2}} d x<e$
Real Analysis
Inequality
Numbers
ISI - MSQMS - B, 2015, Problem 7b
"INEQUALITIES: AN APPROACH THROUGH PROBLEMS BY BJ VENKATACHALA"
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, $\int\limits_0^1 \mathrm dx < \int\limits_0^1 e^{x^2} \mathrm dx < \int\limits_0^1e \mathrm dx$
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