Re: [積分] 微積分基本定理
※ 引述《betray911015 (回頭太難)》之銘言：
: 1.
: d π
: ----∫ sinxy dy
: dx 0
: 2. x
: when [f(t)]^2 =36 +∫ {[f(t)]^2 +[f'(t)]^2} dt, it can be shown that
: 0
: f(x) = af'(x), then a = ?
: 麻煩會的人，可以詳寫過程嘛，謝謝
2. 題目應該是像這樣子吧
x
(f(x))^2 = 36 + ∫ ((f(t))^2 + (f'(t))^2) dt
0
等號兩邊對x微分得
(2)(f(x))(f'(x)) = (f(x))^2 + (f'(x))^2
(f(x))^2 - (2)(f(x))(f'(x)) + (f'(x))^2 = 0
(f(x) - f'(x))^2 = 0
f(x) - f'(x) = 0
f(x) = f'(x) => a = 1
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