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Question Number 70074 by cat2315 last updated on 30/Sep/19

∫_1 ^2 [3+(1/t^2 )]dt=

∫12[3+1t2]dt=

Commented by mathmax by abdo last updated on 30/Sep/19

if [..]means the integr part we get ∫_1 ^2 [3+(1/t^2 )]dt =∫_1 ^2 (3+[(1/t^2 )])dt =3 +∫_1 ^2 [(1/t^2 )]dt for 1≤t≤2 ⇒(1/2)≤(1/t)≤1 ⇒(1/4) ≤(1/t^2 ) ≤1 ⇒[(1/t^2 )]=0 ⇒ ∫_1 ^2 [3+(1/t^2 )]dt =3

if[..]meanstheintegrpartweget∫12[3+1t2]dt=∫12(3+[1t2])dt=3+∫12[1t2]dtfor1⩽t⩽2⇒12⩽1t⩽1⇒14⩽1t2⩽1⇒[1t2]=0⇒∫12[3+1t2]dt=3

Answered by Rio Michael last updated on 30/Sep/19

∫_1 ^2 [3+ t^(−2) ]dt = [3t − t^(−1) ]_1 ^2 = [3t −(1/t)]_1 ^2 = [(6 −(1/2)) − 2] = (7/2)

∫12[3+t−2]dt=[3t−t−1]12=[3t−1t]12=[(6−12)−2]=72

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