minimum

Solution as suggested by @Ramesh Goenka

$\begin{aligned} I & = \int_0^3 e^{|x-t|} dt & \small \color{#3D99F6} \text{Since }|x-t| = \begin{cases} x - t & \text{for }x > t \\ t - x & \text{for }x < t \end{cases} \\ & = \int_0^x e^{x-t} dt + \int_x^3 e^{t-x} dt \\ & = e^{x-t} \big|_x^0 + e^{t-x} \big|_x^3 \\ & = e^x - 1 + e^{3-x} - 1 \\ & = {\color{#3D99F6} e^x + e^{3-x}} - 2 & \small \color{#3D99F6} \text{By AM-GM inequality: } e^x + e^{3-x} \ge 2 e^\frac 32 \\ & = {\color{#3D99F6} 2e^\frac 32} - 2 & \small \color{#3D99F6} \text{Equality occurs when }x = \frac 32 \\ & \approx \boxed{6.963} \end{aligned}$

break the limits from 0-2 then 2-3 and accordingly rewrite the integral removing the mod function. Then apply A-M, G-M Inequality!!