Question:

Step 1: To find the minimum value of the function $f(x)=\displaystyle\int_0^2 e^{|x-t|}\,dt$, we first need to evaluate the integral. The integral involves the absolute value function $|x-t|$, which requires us to consider different cases based on the value of $x$. Step 2: We can split the integral into two parts: one from $0$ to $x$ and the other from $x$ to $2$. For $0 \leq t \leq x$, $|x-t| = x-t$, and for $x \leq t \leq 2$, $|x-t| = t-x$. So, the integral becomes $f(x) = \displaystyle\int_0^x e^{x-t}\,dt + \displaystyle\int_x^2 e^{t-x}\,dt$. Step 3: Now, we evaluate each integral separately. For the first integral, $\displaystyle\int_0^x e^{x-t}\,dt$, let $u = x-t$, then $du = -dt$. When $t = 0$, $u = x$, and when $t = x$, $u = 0$. So, this integral becomes $\displaystyle\int_x^0 e^u \cdot -du = \displaystyle\int_0^x e^u\,du = e^u \Big|_0^x = e^x - 1$. Step 4: For the second integral, $\displaystyle\int_x^2 e^{t-x}\,dt$, let $v = t-x$, then $dv = dt$. When $t = x$, $v = 0$, and when $t = 2$, $v = 2-x$. So, this integral becomes $\displaystyle\int_0^{2-x} e^v\,dv = e^v \Big|_0^{2-x} = e^{2-x} - 1$. Step 5: Combining the results of the two integrals, we get $f(x) = (e^x - 1) + (e^{2-x} - 1) = e^x + e^{2-x} - 2$. To find the minimum value, we need to find the critical points by taking the derivative of $f(x)$ with respect to $x$ and setting it equal to zero. Step 6: The derivative of $f(x)$ is $f'(x) = e^x - e^{2-x}$. Setting $f'(x) = 0$ gives $e^x - e^{2-x} = 0$, which simplifies to $e^x = e^{2-x}$. This implies $x = 2-x$, so $x = 1$. Step 7: To verify that $x = 1$ corresponds to a minimum, we can use the second derivative test. The second derivative of $f(x)$ is $f''(x) = e^x + e^{2-x}$. Since $f''(1) = e^1 + e^1 = 2e > 0$, $x = 1$ indeed corresponds to a minimum. Step 8: Substituting $x = 1$ into the expression for $f(x)$ gives $f(1) = e^1 + e^{2-1} - 2 = e + e - 2 = 2e - 2 = 2(e-1)$. Therefore, the minimum value of the function $f(x)$ is $2(e-1)$, which corresponds to Option 1. The final answer is: $2(e-1)$, which matches Option 1.