ExamBro
ExamBro
JEE Mains · Maths · STD 12 - 7.2 definite integral

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

  1. A \(2(e-1)\)
  2. B \(2 e -1\)
  3. C \(2\)
  4. D \(e(e-1)\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(2(e-1)\)

Step-by-step Solution

Detailed explanation

For \(x \leq 0\) \(f(x)=\int \limits_0^2 e^{t-x} d t=e^{-x}\left(e^2-1\right)\) For \(0 < x < 2\) \(f(x)=\int \limits_0^x e^{x-t} d t+\int \limits_x^2 e^{t-x} d t=e^x+e^{2-x}-2\) For \(x \geq 2\) \(f(x)=\int \limits_0^2 e^{x-t} d t=e^{x-2}\left(e^2-1\right)\) For…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app