: 93年
: 3.
x
lim {∫e^(t^2) dt} / e^(x^2)
x→∞ 0
e^x^2 1
l i m ---------- l i m ---------
= x→∞ 2xe^x^2 = x→∞ 2x = 0
: 4.
: lim √n㏑㏑(1/x)
: x→0+
題目應該打錯了
lim √x㏑㏑(1/x)
x→0+
l i m
= x→0+ ㏑㏑(1/x) / (1/√x) = 0
: 94年
: 5.
Let f[(x-1)/(x+1)] = x Find f'(0)
x - 1
令t = ------
x + 1
1 + t 2
f(t) = -------- => f'(t) = --------- = > f'(0) = 2
1 - t (1 - t)^2 ^^^
: (記得這題之前好像有看過有人解過?)
: 6.
: ∫∫ xy dxdy, where B = {(x,y) | x^2 + y^2 ≦1, x≧0, y≧0}
: B
令x=rcosΘ,y=rsinΘ , dxdy=|J|dΘdr=rdΘdr
π/2 1 3
∫∫ xy dxdy = ∫ ∫ r sinΘcosΘdrdΘ
B 0 0
1 1 1
= ---β(1,1) * --- = ---
2 4 8
^^^^^^
: 95年
: 7.
: 1
: ∫ f(1-x) / [f(x) + f(1-x)] dx, where f is a continuous function
: 0
1
Let I = ∫ f(1-x) / [f(x) + f(1-x)] dx
0
1 1
I + I = ∫ f(1-x) / [f(x) + f(1-x)]dx + ∫ f(x)/[f(x) + f(1-x)] dx
0 0
1 1
2I = 1 => I = ∫ f(1-x) / [f(x) + f(1-x)] dx = ---
0 2
^^^^^^^
有錯請指教~3Q
--
○ "○ ○" ○"
(|\ (|\ )) ))
/`○rz /`○r27\" ○r27\"
○╭○rz ○rz ○rz-st○
￣′ ○=^
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 61.229.153.95
※ 編輯: GayerDior 來自: 61.229.165.205 (03/22 20:41)