回復 6# subway 的帖子
#6
k/(k+3)² = (1/k)* 1/ [1+3/k]²
所求=∫ {0 to 1} 1/(1+3x)² dx
(令u=3x+1 , du=3dx , x=0時,u=1 ;x=1時,u=4)
= (1/3) ∫ {1 to 4} 1/u² du
=(-1/3)* u^(-1) | {1 to 4}
=(-1/3)* (1/4 -1/1)
=1/4
#9
疊合~
2y+ycosx=cox+2sinx
2y=(1-y)cosx+2sinx
2y=[(1-y)²+4]^0.5* cos(x+a)
|2y|/[(1-y)²+4]^0.5<=1
解得-5/3<=y<=1
#12
歐拉公式~
#14
唬人的~題目雖是空間敘述
但可用柯西不等式~
# 15
所求=∫ {0 to 1} [cos(πx/4)]² dx
=(1/2)∫ {0 to 1} [cos(πx/2)+1] dx (兩倍角公式)
=(1/2) [ (2/π)*sin(πx/2) +x ] | {0 to 1}
=(1/2)[ 2/π +1]
=1/π +1/2
