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 Let g (x) = ∫0x f (t) dt, where f is such that 1/2 ≤ f(t) ≤ 1 for t [0,1] and

0 ≤ f(t) ≤1/2 for t [1,2]. Then g (2) satisfies the inequality:

7 years ago


Answers : (1)


g(2) = ∫02 f (t) dt = ∫01 f (t) dt   +  ∫12 f (t) dt

1/2 ≤ f(t) ≤ 1 for t ∈ [0,1] and 0 ≤ f(t) ≤1/2 for t ∈[1,2].

 011/2.dt ≤ ∫01 f (t).dt ≤ ∫01 1.dt   and 12 0.dt ≤ ∫12 f (t) dt ≤∫12 1/2.dt

1/2 ≤ ∫01 f (t).dt ≤  1   and     0 ≤ ∫12 f (t) dt ≤ 1/2

on adding above gives,

1/2 ≤ ∫01 f (t).dt + ∫12 f (t) dt ≤ 3/2

1/2 ≤ g(2) ≤ 3/2


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