IMC 2010 · Day 1 · P1

Let $f(x) = \int_0^x (t^2 + 1) e^{-t^2} dt$ and let $g(x) = -e^{-x^2}$; both functions are increasing.

By Cauchy's Mean Value Theorem, there exists a real number $x \in (a, b)$ such that $$\frac{f(b) - f(a)}{g(b) - g(a)} = \frac{f'(x)}{g'(x)} = \frac{(x^2 + 1) e^{-x^2}}{2x e^{-x^2}} = \frac{1}{2} \left( x + \frac{1}{x} \right) \ge \sqrt{x \cdot \frac{1}{x}} = 1.$$ Then $$\int_a^b (x^2 + 1) e^{-x^2} dx = f(b) - f(a) \ge g(b) - g(a) = e^{-a^2} - e^{-b^2}.$$

$$\int_a^b (x^2 + 1) e^{-x^2} dx \ge \int_a^b 2x e^{-x^2} dx = \left[ -e^{-x^2} \right]_a^b = e^{-a^2} - e^{-b^2}.$$