Show that $ \int_0^2 e^{x^2-x} dx \in [2e^{-1/4},2e^2] $ 
Show that $ \int_0^2 e^{x^2-x} dx \in [2e^{-1/4},2e^2] $

If $f(x)\leq g(x)$ for $x\in[a,b]$ then $\int^b_af(x)dx\leq \int^b_ag(x)dx$
if $x\in [0,2]$ then $x^2-x\leq x$, so 
$$0 \leq  \int_0^2 e^{x^2-x} dx\leq  \int_0^2 e^{x} dx = e^2-1 <2e^2$$
Which integral should I choose on the left side?

Since $x^2-x=\left(x-\frac{1}{2}\right)^2-\frac{1}{4}$ we have, for $x\in[0,2]$,
$$-\frac{1}{4}\le x^2-x \le 2$$
Then
\begin{align*}
\int_0^2e^{-\frac{1}{4}}\,dx&\le\int_0^2e^{x^2-x}\,dx\le\int_0^2e^2\,dx\\
2e^{-\frac{1}{4}}&\le\int_0^2e^{x^2-x}\,dx\le 2e^2
\end{align*} 
