Question

If \( f(x) = \int_0^1 e^{|t - x|} \, dt \) for \( 0 \leq x \leq 1 \), then the maximum value of \( f(x) \) is:

• A. \(e - 1\)
• B. \(2(e - 1)\)
• C. \((e - 1)\)
• D. \(2(\sqrt{e} - 1)\)

Hint:

When integrating \(e^{|t-x|}\), split the integral at \(t=x\) to remove the modulus.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Split the integral at \(t = x\) to handle the modulus.

Step 2: Detailed Explanation:
For \(0 \le x \le 1\): \[ f(x) = \int_0^x e^{x-t} dt + \int_x^1 e^{t-x} dt \] \[ = e^x \int_0^x e^{-t} dt + e^{-x} \int_x^1 e^{t} dt \] \[ = e^x (1 - e^{-x}) + e^{-x} (e - e^x) \] \[ = e^x - 1 + e^{1-x} - 1 \] \[ f(x) = e^x + e^{1-x} - 2 \] The maximum occurs at the endpoints because \(e^x + e^{1-x}\) is minimized at \(x = 0.5\) and maximized at endpoints. \(f(0) = e^0 + e^{1} - 2 = 1 + e - 2 = e - 1\) \(f(1) = e^1 + e^{0} - 2 = e + 1 - 2 = e - 1\)

Step 3: Final Answer:
Maximum value is \(e - 1\), which corresponds to option (A).