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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:
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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