Mathematical Reflections 2017, Issue 2 - Problem U407

Problem:
Prove that for every $\varepsilon>0$ $$\int_2^{2+\varepsilon} e^{2x-x^2} \ dx<\dfrac{\varepsilon}{1+\varepsilon}.$$

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
Observe that for all real numbers $t \geq 1$ we have $t^2 \leq e^{t^2-1}$. Indeed, put $f(t)=e^{t^2-1}-t^2$. Then, $f(1)=0$ and $f'(t)=2t(e^{t^2-1}-1) \geq 0$ for all $t \geq 1$. Now, put $t=x-1$. We have $(x-1)^2 \leq e^{x^2-2x}$ for all $x \geq 2$, i.e. $$e^{2x-x^2} \leq \dfrac{1}{(x-1)^2}.$$
So, $$\int_2^{2+\varepsilon} e^{2x-x^2} \ dx \leq \int_2^{2+\varepsilon} \dfrac{dx}{(x-1)^2}=\left[-\dfrac{1}{x-1}\right]_2^{2+\varepsilon}=\dfrac{\varepsilon}{1+\varepsilon}.$$