The minimum value of `f(x)=int_0^4e^(|x-t|)dt` where `x in [0,3]` is : (A) `2e^2-1` (B) `e^4-1` (C) `2(e^2-1)` (D) `e^2-1`

Options

A. `2e ^(2) -1`
B. ` e ^(4) -1`
C. ` 2 (e ^(2) -1)`
D. `e ^(2) -1`

Answer

To find the minimum value of the function \( f(x) = \int_0^4 e^{|x-t|} dt \) where \( x \in [0, 3] \), we can break down the solution into several steps.

### Step 1: Break the Integral into Two Parts
The absolute value function \( |x - t| \) can be split into two cases based on the value of \( t \) relative to \( x \):
- For \( t \leq x \), \( |x - t| = x - t \)
- For \( t > x \), \( |x - t| = t - x \)

Thus, we can express the integral as:
\[ f(x) = \int_0^x e^{x-t} dt + \int_x^4 e^{t-x} dt \]

### Step 2: Evaluate Each Integral
Now we evaluate each part separately.

1. **First Integral**:
\[ \int_0^x e^{x-t} dt = e^x \int_0^x e^{-t} dt = e^x \left[-e^{-t}\right]_0^x = e^x \left[-e^{-x} + 1\right] = e^x(1 - e^{-x}) = e^x - 1 \]

2. **Second Integral**:
\[ \int_x^4 e^{t-x} dt = e^{-x} \int_x^4 e^t dt = e^{-x} \left[e^t\right]_x^4 = e^{-x} (e^4 - e^x) = e^{4-x} - 1 \]

### Step 3: Combine the Results
Now we combine the results of both integrals:
\[ f(x) = (e^x - 1) + (e^{4-x} - 1) = e^x + e^{4-x} - 2 \]

### Step 4: Find the Derivative
To find the minimum value, we differentiate \( f(x) \):
\[ f'(x) = e^x - e^{4-x} \cdot (-1) = e^x + e^{4-x} \]

### Step 5: Set the Derivative to Zero
Setting the derivative to zero to find critical points:
\[ e^x = e^{4-x} \]
Taking the natural logarithm of both sides:
\[ x = 4 - x \implies 2x = 4 \implies x = 2 \]

### Step 6: Evaluate \( f(x) \) at the Critical Point
Now we evaluate \( f(2) \):
\[ f(2) = e^2 + e^{4-2} - 2 = e^2 + e^2 - 2 = 2e^2 - 2 \]

### Step 7: Check the Endpoints
We also need to check the endpoints \( x = 0 \) and \( x = 3 \):
1. For \( x = 0 \):
\[ f(0) = e^0 + e^{4-0} - 2 = 1 + e^4 - 2 = e^4 - 1 \]
2. For \( x = 3 \):
\[ f(3) = e^3 + e^{4-3} - 2 = e^3 + e - 2 \]

### Step 8: Compare Values
Now we compare the values:
- \( f(2) = 2e^2 - 2 \)
- \( f(0) = e^4 - 1 \)
- \( f(3) = e^3 + e - 2 \)

### Conclusion
The minimum value occurs at \( x = 2 \):
\[ \text{Minimum value of } f(x) = 2e^2 - 2 \]

Thus, the answer is \( \boxed{2(e^2 - 1)} \).