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

Options

A. 2
B. 2(e-1)
C. 0.2
D. e(e-1)

Answer

To find the minimum value of the function \( f(x) = \int_{0}^{2} e^{|x-t|} dt \), we will analyze the integral by considering different cases based on the value of \( x \).

### Step 1: Analyze the cases for \( x \)

1. **Case 1: \( x < 0 \)**

In this case, \( |x - t| = t - x \) for all \( t \) in the interval [0, 2]. Thus, we can rewrite the function as:
\[ f(x) = \int_{0}^{2} e^{t - x} dt = e^{-x} \int_{0}^{2} e^{t} dt \]
Now we compute the integral:
\[ \int_{0}^{2} e^{t} dt = [e^{t}]_{0}^{2} = e^{2} - 1 \]
Therefore, we have:
\[ f(x) = e^{-x} (e^{2} - 1) \]

2. **Case 2: \( 0 \leq x < 2 \)**

Here, we need to split the integral into two parts:
- From \( 0 \) to \( x \): \( |x - t| = x - t \)
- From \( x \) to \( 2 \): \( |x - t| = t - x \)

Thus, we can express \( f(x) \) as:
\[ f(x) = \int_{0}^{x} e^{x - t} dt + \int_{x}^{2} e^{t - x} dt \]
Calculating the first integral:
\[ \int_{0}^{x} e^{x - t} dt = e^{x} \int_{0}^{x} e^{-t} dt = e^{x} [-e^{-t}]_{0}^{x} = e^{x} (1 - e^{-x}) = e^{x} - 1 \]
Now for the second integral:
\[ \int_{x}^{2} e^{t - x} dt = e^{-x} \int_{x}^{2} e^{t} dt = e^{-x} [e^{t}]_{x}^{2} = e^{-x} (e^{2} - e^{x}) = e^{2 - x} - 1 \]
Therefore, we have:
\[ f(x) = (e^{x} - 1) + (e^{2 - x} - 1) = e^{x} + e^{2 - x} - 2 \]

3. **Case 3: \( x \geq 2 \)**

In this case, \( |x - t| = x - t \) for all \( t \) in the interval [0, 2]. Thus:
\[ f(x) = \int_{0}^{2} e^{x - t} dt = e^{x} \int_{0}^{2} e^{-t} dt = e^{x} [-e^{-t}]_{0}^{2} = e^{x} (1 - e^{-2}) = e^{x} (1 - \frac{1}{e^{2}}) = e^{x} \left(1 - \frac{1}{e^{2}}\right) \]

### Step 2: Finding the minimum value

Now we need to analyze the expressions we obtained for \( f(x) \):

- For \( x < 0 \): \( f(x) = e^{-x} (e^{2} - 1) \)
- For \( 0 \leq x < 2 \): \( f(x) = e^{x} + e^{2 - x} - 2 \)
- For \( x \geq 2 \): \( f(x) = e^{x} \left(1 - \frac{1}{e^{2}}\right) \)

To find the minimum, we can evaluate \( f(x) \) at critical points and endpoints.

1. At \( x = 0 \):
\[ f(0) = e^{0} + e^{2} - 2 = 1 + e^{2} - 2 = e^{2} - 1 \]

2. At \( x = 2 \):
\[ f(2) = e^{2} \left(1 - \frac{1}{e^{2}}\right) = e^{2} - 1 \]

3. Check the behavior as \( x \to -\infty \) and \( x \to \infty \):
- As \( x \to -\infty \), \( f(x) \to \infty \).
- As \( x \to \infty \), \( f(x) \to \infty \).

Thus, the minimum value occurs at both \( x = 0 \) and \( x = 2 \), giving us:
\[ \text{Minimum value of } f(x) = e^{2} - 1 \]