1. We need to minimize $f(x)=\int_0^2 e^{|x-t|}\,dt.$

Because of the absolute value, split into cases depending on the position of $x$ relative to the interval $[0,2]$.

2. Case 1: $0\le x\le 2$

Then

• for $0\le t\le x$, $|x-t|=x-t$,
• for $x\le t\le 2$, $|x-t|=t-x$.

So $f(x)=\int_0^x e^{x-t}\,dt+\int_x^2 e^{t-x}\,dt.$

Now evaluate each part: $\int_0^x e^{x-t}\,dt=e^x\int_0^x e^{-t}\,dt=e^x(1-e^{-x})=e^x-1,$

and $\int_x^2 e^{t-x}\,dt=e^{-x}\int_x^2 e^t\,dt=e^{-x}(e^2-e^x)=e^{2-x}-1.$

Hence $f(x)=e^x+e^{2-x}-2.$

To minimize this on $[0,2]$, differentiate: $f'(x)=e^x-e^{2-x}.$

Set $f'(x)=0$: $e^x=e^{2-x}\implies x=2-x\implies x=1.$

Also, $f''(x)=e^x+e^{2-x}>0,$ so $x=1$ gives a minimum.

Then $f(1)=e+e-2=2e-2=2(e-1).$

3. Case 2: $x\le 0$

For all $t\in[0,2]$, we have $t\ge x$, so $|x-t|=t-x$. Thus $f(x)=\int_0^2 e^{t-x}\,dt=e^{-x}\int_0^2 e^t\,dt=e^{-x}(e^2-1).$

Since $x\le 0$, $e^{-x}\ge 1$, so the minimum in this region occurs at $x=0$: $f(0)=e^2-1.$

4. Case 3: $x\ge 2$

For all $t\in[0,2]$, we have $x\ge t$, so $|x-t|=x-t$. Thus $f(x)=\int_0^2 e^{x-t}\,dt=e^x\int_0^2 e^{-t}\,dt=e^x(1-e^{-2}).$

Since $x\ge 2$, this is increasing in $x$, so the minimum in this region occurs at $x=2$: $f(2)=e^2-1.$

5. Compare the candidate minima:

• On $[0,2]$, minimum is $2(e-1)$.
• For $x\le 0$, minimum is $e^2-1$.
• For $x\ge 2$, minimum is $e^2-1$.

Now $e^2-1-(2e-2)=e^2-2e+1=(e-1)^2>0,$ so $2(e-1)<e^2-1.$

Therefore the global minimum value is $\boxed{2(e-1)}.$

So the correct option is B.