Question

The minimum value of the function $f(x)=\int\limits_0^2 e^{|x-t|} d t$ is :

• A. 2
• B. $2(e-1)$
• C. $2 e-1$
• D. $e(e-1)$

Answer 1

The Correct Option is B

For x≤0 
f(x)=0∫2​et−xdt=e−x(e²−1) 
For 0<x<2 
f(x)=0∫x​ex−tdt+∫x²​et−xdt=ex+e²−x−2 
For x≥2 
f(x)=0∫2​ex−tdt=ex−2(e²−1) 
For x≤0,f(x) is ↓ and x≥2,f(x) is ↑ 
∴ Minimum value of f(x) lies in x∈(0,2) 
Applying A.M≥G.M, 
minimum value of f(x) is 2(e−1)

Answer 2

1. Split the integral at \(t = k\), where the absolute value changes: \[ f(x) = \int_{0}^{k} e^{k-t} dt + \int_{k}^{2} e^{t-k} dt. \] 2. Evaluate the first part of the integral: \[ \int_{0}^{k} e^{k-t} dt = e^k \int_{0}^{k} e^{-t} dt. \] Simplify: \[ \int_{0}^{k} e^{-t} dt = \left[-e^{-t}\right]_{0}^{k} = 1 - e^{-k}. \] Thus: \[ \int_{0}^{k} e^{k-t} dt = e^k (1 - e^{-k}) = e^k - 1. \] 3. Evaluate the second part of the integral: \[ \int_{k}^{2} e^{t-k} dt = e^{-k} \int_{k}^{2} e^{t} dt. \] Simplify: \[ \int_{k}^{2} e^{t} dt = \left[e^t\right]_{k}^{2} = e^2 - e^k. \] Thus: \[ \int_{k}^{2} e^{t-k} dt = e^{-k}(e^2 - e^k) = e^2 e^{-k} - 1. \] 4. Combine the results: \[ f(x) = (e^k - 1) + (e^2 e^{-k} - 1). \] 5. Simplify: \[ f(x) = e^k + e^2 e^{-k} - 2. \] 6. To minimize \(f(x)\), let \(z = e^k\), so: \[ f(x) = z + \frac{e^2}{z} - 2. \] Differentiate \(f(x)\) with respect to \(z\): \[ \frac{df}{dz} = 1 - \frac{e^2}{z^2}. \] Set \(\frac{df}{dz} = 0\) to find the critical point: \[ 1 = \frac{e^2}{z^2} \implies z^2 = e^2 \implies z = e. \] 7. Substitute \(z = e\) back into \(f(x)\): \[ f(x) = e + \frac{e^2}{e} - 2 = e + e - 2 = 2(e - 1). \] Thus, the minimum value of \(f(x)\) is \(2(e - 1)\). The integral involves an absolute value, so it must be split into two parts based on the point \(t = k\). Simplify each part and minimize the resulting function by treating it as a function of \(z = e^k\).