Cho $ m \in  N.$ Tìm giá trị nhỏ nhất của: $f(x) = \int\limits_{1}^{x}t^m.e^{2t} -2 \left ( \frac{x^{m+3}}{m+3} +\frac{x^{m+2}}{m+2}  \right ) , x\geq 1$

Xét $ g(x) = e^{2x} – 2 (x^2 +x), x\geq 0$
      $ g'(x) = 2e^{2x} – 2 (2x + 1) = 2 (e^{2x} -2x-1)$
      $ g”(x) = 2 (2e^{2x} -2 ) = 4 (e^{2x}-1) \geq 0, \forall x \geq 0 $
$\Rightarrow g’ $ tăng trên $[0;+\infty  )    \Rightarrow    g'(x) \geq g'(0) = 0 , \forall x \geq 0 $
$\Rightarrow g$ tăng trên $[0;+\infty  )    \Rightarrow g(x) \geq g(0) = 1$
$\Rightarrow e^{2x} \geq 2(x^2 + x) + 1, \forall x  \geq 0$
$\Rightarrow x^me^{2x} \geq 2(x^{m+2} + x^{m+1})+x^m, \forall x \geq 0$
$\Rightarrow \int\limits_{1}^{x} t^me^{2t} \geq  \int\limits_{1}^{x} 2( t^{m+2}+t^{m+1})+t^mdt$
                    $= 2 \left ( \frac{x^{m+3}}{m+3} + \frac{x^{m+2}}{m+2}  \right )+\frac{x^{m+1}}{m+1}  – 2 \left ( \frac{1}{m+3}+\frac{1}{m+2}   \right )-\frac{1}{m+1} , \forall x \geq 1$
            $\Rightarrow f(x) \geq -2 \left ( \frac{1}{m+3}+ \frac{1}{m+2}   \right ) $
$\Rightarrow f(x) \geq -2\left ( \frac{1}{m+3}+ \frac{1}{m+2}   \right ), \forall x \geq 1$
Dấu $”=”  \Leftrightarrow   x=1$
Vậy : $\min_{x\geq 1}f(x)  = -\frac{2(2m+5)}{(m+2)(m+3)}.$