### Re: [數學]f(x)=x^2的弧長如何算?

Anonymous 寫到:f(x)=x^2的弧長如何算?#ed_op#BR#ed_cl#如果想算x=0~4的弧長#ed_op#BR#ed_cl#該如何做#ed_op#BR#ed_cl#面積是用微績分可算#ed_op#BR#ed_cl#這我知道#ed_op#BR#ed_cl#但弧長也是用微績分吧???#ed_op#BR#ed_cl#請幫忙...thx
#ed_op#DIV#ed_cl#我們要求弧長的公式:#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#在二維空間,y=f(x)的圖像中,#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#ds#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#=dx#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#+dy#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#ds#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#=dx#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#+[f'(x)]#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#dx#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#ds#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#=dx#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#{1+[f'(x)]#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#}#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#兩邊開方,#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#得,#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#ds=√{1+[f'(x)]#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#}dx#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#兩邊積分.#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#s=∫√{1+[f'(x)]#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#}dx#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#把x#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#代入#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#就能求到#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#註:我全部自己求,沒有抄襲或侵犯版權#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#

dλ(t) 就構成L的微量弧
L弧長=∫|dλ(t)|= ∫|λ’(t)|dt=∫|λ’(t)|dt
=∫√(x'(t)^2+ y'(t)^2)dt=∫√(1+ y'(x)^2)dx   for x'(t)>0

f(x)=x^2的弧長如何算?