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

To find the maximum value of \( f(x) = \int_0^1 e^{|t-x|} \, dt \) for \( 0 \leq x \leq 1 \), we need to evaluate the integral by considering the definition of the absolute value function. The expression \( |t-x| \) can be split into cases depending on the value of \( t \) relative to \( x \).

Consider the following cases: 

1. Case 1: \( 0 \leq t \leq x \)
   In this interval, \( |t-x| = x-t \). Therefore, the integral over this range is: \(\int_0^x e^{x-t} \, dt\).
2. Case 2: \( x < t \leq 1 \)
   In this interval, \( |t-x| = t-x \). Therefore, the integral over this range is: \(\int_x^1 e^{t-x} \, dt\).

Now, let's evaluate both integrals and combine them:

• For the first case:

\(\int_0^x e^{x-t} \, dt = e^x \left[ -e^{-t} \right]_0^x = e^x (1 - e^{-x})\)

• For the second case:

\(\int_x^1 e^{t-x} \, dt = e^{-x} \left[ e^t \right]_x^1 = e^{-x} (e - e^x)\)

Combining both results, we find:

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

The expression \(f(x) = e - 1\) is constant and does not depend on \( x \) within the interval \([0,1]\).

Therefore, the maximum value of \( f(x) \) is:

Answer: \( e - 1 \).