UnderBirdです。

x=aからx=bまでのy=f(x)の曲線の長さlは、
l=∫_\(a^{b}\) \(\sqrt{\quad}\){1+(y'\()^{2}\)}dx
で求めます。
ｙ＝（\(\frac{a}{2}\))(e^(\(\frac{x}{a}\))+e^(-\(\frac{x}{a}\)))のとき、
y'=（\(\frac{1}{2}\))(e^(\(\frac{x}{a}\))-e^(-\(\frac{x}{a}\)))
1+(y'\()^{2}\)=（\(\frac{1}{4}\))(e^(2\(\frac{x}{a}\))+2+e^(-2\(\frac{x}{a}\)))={(e^(\(\frac{x}{a}\))+e^(-\(\frac{x}{a}\)))/2}^2より
\(\sqrt{\quad}\){1+(y'\()^{2}\)}=(e^(\(\frac{x}{a}\))+e^(-\(\frac{x}{a}\)))/2
また、偶関数より
l=2∫_\(0^{x}\)1 (e^(\(\frac{x}{a}\))+e^(-\(\frac{x}{a}\)))/2 dx
=∫_\(0^{x}\)1 e^(\(\frac{x}{a}\))+e^(-\(\frac{x}{a}\)) dx
=a(e^(x\(\frac{1}{a}\))-e^(-x\(\frac{1}{a}\)))
となります

弧の長さL=∫\(\sqrt{\quad}\)(1+(y'\()^{2}\))dxとなる。
y'=(\(\frac{a}{2}\))(e^(\(\frac{x}{a}\))/a-e^(-\(\frac{x}{a}\))/a)=(\(\frac{1}{2}\))(e^(\(\frac{x}{a}\))-e^(-\(\frac{x}{a}\)))
(y'\()^{2}\)=(\(\frac{1}{4}\))(e^(2\(\frac{x}{a}\))-2+e^(-2\(\frac{x}{a}\)))
1+(y'\()^{2}\)=(\(\frac{1}{4}\))(e^(2\(\frac{x}{a}\))+2+e^(-2\(\frac{x}{a}\)))=[(\(\frac{1}{2}\))(e^(\(\frac{x}{a}\))+e^(-\(\frac{x}{a}\))]^2
L=∫(\(\frac{1}{2}\))(e^(\(\frac{x}{a}\))+e^(-\(\frac{x}{a}\)))dx=(\(\frac{1}{2}\))[ae^(\(\frac{x}{a}\))-ae^(-\(\frac{x}{a}\))]
=a[e^(x\(\frac{1}{a}\))-e^(-x\(\frac{1}{a}\))]