Prove that $(1)\frac{1}{2}\leq\int_{0}^{2}\frac{dx}{2+x^2}\leq\frac{5}{6}$ $(2)2e^{-1/4}<\int_{0}^{2}e^{x^2-x}dx<2e^2$ Prove that $(1)\frac{1}{2}\leq\int_{0}^{2}\frac{dx}{2+x^2}\leq\frac{5}{6}$
$(2)2e^{-1/4}<\int_{0}^{2}e^{x^2-x}dx<2e^2$
I tried to prove it but my answer is not correct.
For first part,As $0\leq x\leq2\Rightarrow 2\leq x^2+2\leq6\Rightarrow\frac{1}{6}\leq\frac{1}{x^2+2}\leq\frac{1}{2}\Rightarrow\frac{1}{6}\int_{0}^{2}1dx\leq\int_{0}^{2}\frac{dx}{2+x^2}\leq\frac{1}{2}\int_{0}^{2}1dx$$\Rightarrow\frac{1}{3}\leq\int_{0}^{2}\frac{dx}{2+x^2}\leq1$
In the second part,$0\leq x\leq2\Rightarrow -2<x^2-x<4\Rightarrow e^{-2}<e^{x^2-x}<e^4\Rightarrow2e^{-2}<\int_{0}^{2}e^{x^2-x}dx<2e^4$
Where have i gone wrong?What is the correct method to solve it.

For the second problem, we note that 
$$-\frac14\le x^2-x\le2$$for $x\in[0,2]$. Note that we can easily find the minimum by taking the derivative and setting it to $0$, while we merely check the endpoints to find the maximum.  
Inasmuch as the exponential function is monotonically increasing, then
$$e^{-\frac14}\le e^{x^2-x}<e^2 \tag 1$$
Then, we see immediately that 
$$2\,e^{-\frac14}\le \int_0^2 e^{x^2-x}\,dx \le 2\,e^2$$
