<p>The minimum value of $$f(x) = \int_0^4 e^{|x-t|} \, dt$$ where $$x \in [0, 3]$$ is:</p>
Step-by-Step Solution
Key Concept: To minimize the integral of absolute value, we must split it into regions where the expression inside changes sign, then find the critical point using calculus. The key insight is that the minimum occurs where the derivative equals zero, which corresponds to balancing the contributions from both sides of the split.
<p><strong>Step 1: Split the integral at t = x</strong></p><p>Since |x - t| = (x - t) when t ≤ x and |x - t| = (t - x) when t > x, we write:</p><p>$$f(x) = \int_0^x e^{x-t} \, dt + \int_x^4 e^{t-x} \, dt$$</p><p><strong>Step 2: Evaluate the first integral</strong></p><p>$$\int_0^x e^{x-t} \, dt = e^x \int_0^x e^{-t} \, dt = e^x[-e^{-t}]_0^x = e^x(1 - e^{-x}) = e^x - 1$$</p><p><strong>Step 3: Evaluate the second integral</strong></p><p>$$\int_x^4 e^{t-x} \, dt = e^{-x} \int_x^4 e^t \, dt = e^{-x}[e^t]_x^4 = e^{-x}(e^4 - e^x) = e^{4-x} - 1$$</p><p><strong>Step 4: Combine results</strong></p><p>$$f(x) = e^x - 1 + e^{4-x} - 1 = e^x + e^{4-x} - 2$$</p><p><strong>Step 5: Find critical points using differentiation</strong></p><p>$$f'(x) = e^x - e^{4-x}$$</p><p>Setting $f'(x) = 0$: $e^x = e^{4-x}$</p><p>This gives: $x = 4 - x$, so $x = 2$</p><p><strong>Step 6: Verify x = 2 is in [0, 3] and check it's a minimum</strong></p><p>Since $f''(x) = e^x + e^{4-x} > 0$ for all x, this is indeed a minimum.</p><p><strong>Step 7: Calculate the minimum value</strong></p><p>$$f(2) = e^2 + e^{4-2} - 2 = e^2 + e^2 - 2 = 2e^2 - 2 = 2(e^2 - 1)$$</p><p><strong>∴ Answer: c</strong></p>
Correct Answer: c