| (1) 関数 f(x) = x^2/4 - log(x)/2 の導関数は f'(x) = 1/2(x - 1/x) なので、弧長は
L = ∫[1,2] √(1 + f'(x)^2) dx = ∫[1,2] √(1 + 1/4(x^2 - 2 + 1/x^2)) dx = ∫[1,2] √(1/4(x^2 + 2 + 1/x^2)) dx = ∫[1,2] √(1/4(x + 1/x)^2) dx = 1/2 ∫[1,2] (x + 1/x) dx = 1/2 ( [ 1/2 x^2 ]_[1,2] + [ log(x) ]_[1,2] ) = 1/2 ( 2 - 1/2 + log(2) - log(1) ) = 3/4 + 1/2 log(2)
となります。
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