函數的遞減與積分的相關性

f(x)遞減, 故 f(k+1) <= f(x) <= f(k), for x in [k, k+1]
對x積分, x=k~k+1, 得
∫[k~k+1] f(k+1) dx <=∫[k~k+1] f(x) dx <= ∫[k~k+1] f(k) dx
f(k+1)* [k+1-k] <= ∫[k~k+1] f(x) dx <= f(k)*[k+1-k]
即 f(k+1) <= ∫[k~k+1] f(x) dx <= f(k)

幾何意義:設 A(k, 0), B(k+1,0), D(k, f(k)), C(k+1, f(k+1))
E(k, f(k+1)), F(k+1, f(k))
f(k+1)= 矩形ABCE面積 <= 不規則形 ABCD面積 <= 矩形ABFE面積=f(k)


圖片參考：http://imgcld.yimg.com/8/n/AE03435620/o/151104280335813872104850.jpg

煩惱即是菩提 ( 知識長 )
我一開始覺得奇怪

但看著看著
就發現
原來是

f(k+1)* [k+1-k] <= ∫[k~k+1] f(x) dx <= f(k)*[k+1-k]
變成
f(k+1) *[1] <= ∫[k~k+1] f(x) dx <= f(k)*[1]
變成
f(k+1) <= ∫[k~k+1] f(x) dx <= f(k)


或是

2011-04-28 17:47:15 補充：
或是


f(k+1)* [k+1-k] <= ∫[k~k+1] f(x) dx <= f(k)*[k+1-k]
變成
f(k+1) <=( ∫[k~k+1] f(x) dx )/ [k+1-k] <= f(k)
變成
f(k+1) <=( ∫[k~k+1] f(x) dx )/ 1 <= f(k)
變成
f(k+1) <= ∫[k~k+1] f(x) dx <= f(k)


這都可以

這太*了!!

2011-04-28 17:48:55 補充：
f(k+1) <= ∫[k~k+1] f(x) dx <= f(k)

這個定理 好妙!!!

可用積分均值定理..