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What is the minimum value of f(x) = 0 ∫ 3 e |x-t| dt where x ∈ [0,2] ? (0 is the lower limit and 3 is the upper limit)

What is the minimum value of f(x) = 03e|x-t|dt where x ∈ [0,2] ? (0 is the lower limit and 3 is the upper limit)

Grade:12

2 Answers

Neha
10 Points
6 years ago
e*3+2*e^2+2e-5
Shibashis Mallik
21 Points
6 years ago
please explain..

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