The minimum value of the function f(x) = ∫ _ 0 ^ 2 e ^ |x - t| dt is

Given,

$f\left(x\right) = {∫}_{0}^{2}{e}^{\left|x - t\right|}dt$

Now For $x \le 0$

$f\left(x\right) = {∫}_{0}^{2}{e}^{t - x}dt = {e}^{- x}\left({e}^{2} - 1\right)$

And for $0 < x < 2$

$f\left(x\right) = {∫}_{0}^{x}{e}^{x - t}dt + {∫}_{x}^{2}{e}^{t - x}dt = {e}^{x} + {e}^{2 - x} - 2$

For $x \ge 2$

$f\left(x\right) = {∫}_{0}^{2}{e}^{x - t}dt = {e}^{x - 2}\left({e}^{2} - 1\right)$

Now for $x \le 0, f\left(x\right)$ is decreasing as ${e}^{- x}$ is decreasing function and $x \ge 2, f\left(x\right)$ is increasing as ${e}^{x - 2}$ is increasing function,

So, minimum value of $f\left(x\right)$ lies in $x \in \left(0, 2\right)$

Applying A.M $\ge$ G.M in ${e}^{x} + {e}^{2 - x}$we get,

$\dfrac{{e}^{x} + {e}^{2 - x}}{2} \ge \sqrt{{e}^{x} \times {e}^{2 - x}}$

$⇒ {e}^{x} + {e}^{2 - x} \ge 2e$

Hence, the minimum value of $f\left(x\right)$ is $2e - 2 = 2\left(e - 1\right)$