
Yefim S. answered 03/05/21
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x = t - t2; y = 4/3t3/2, 1 ≤ t ≤ 2;
Length of curve s = ∫12[(1 - 2t)2 + (2t1/2)2]1/2dt = ∫12(1 + 4t2)1/2dt
Cameron L.
asked 03/04/21Set up, but do not evaluate, the integral that represents the length of the curve given by
x = t - t2, , over the interval 1 ≤ t ≤ 2
Yefim S. answered 03/05/21
Math Tutor with Experience
x = t - t2; y = 4/3t3/2, 1 ≤ t ≤ 2;
Length of curve s = ∫12[(1 - 2t)2 + (2t1/2)2]1/2dt = ∫12(1 + 4t2)1/2dt
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