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
- A \(2(e-1)\)
- B \(2 e -1\)
- C \(2\)
- D \(e(e-1)\)
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…
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