Re: [積分] 請教一題積分問題

看板trans_math作者linch1 (tobias)時間15年前 (2009/03/02 09:29)推噓1(1推 0噓 0→)

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※ 引述《ADAH33 (逐漸消失的生命)》之銘言： : let f : R -> R be cont : f(x) = 1 + (0到1積分) kf(x-ky) dy + (0到x-k積分) f(y)dy : where k > 0 : find f(x) ∫(0->1) kf(x-ky) dy Let u=x-ky, du=-kdy, y=0 ==> u=x, y=1 ==> u=x-k =∫(x->x-k) -f(u)du =∫(x-k->x) f(u)du =∫(x-k->x) f(y)dy f(x) = ∫(0->1) kf(x-ky) dy + ∫(0->x-k) f(y)dy = 1 + ∫(x-k->x) f(y)dy+∫(0->x-k) f(y)dy = 1 + ∫(0->x) f(y)dy By the fundamental theorem of calculus, we have f'(x) = f(x). ==> f(x) = Ae^x. ( f(0)=A ) And f(x) = 1 + ∫(0->x) f(y)dy ==> f(0) = 1 = A f(x) = e^x -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 122.118.21.32

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ADAH33

03/02 11:16, , 1^F

03/02 11:16, 1^F

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