Recall the following property of integrals: if f ( t ) ≥ g ( t ) for all t ≥ 0 , then for all x ≥ 0 , ∫ x 0 f ( t ) d t ≥ ∫ x 0 g ( t ) d t 4 the inequality e t ≥ 1 holds for t ≥ 0 because e > 1. use (4) to prove that e x ≥ 1 + x for x ≥ 0 then prove, by successive integration, the following inequalities (for x ≥ 0 ): e x ≥ 1 + x + 1 2 x 2 e x ≥ 1 + x + 1 2 x 2 + 1 6 x 3.