Réponses

2014-02-19T23:27:53+01:00
Bonsoir,

L=\int_0^1\sqrt{1+(f'(x))^2}dx=\int_0^1\sqrt{1+(2x)^2}dx\\\\L=\int_0^1\sqrt{1+4x^2}dx

Calculons la primitive par changement de variable.

2x=sh(t)\Longrightarrow x=\dfrac{1}{2}sh(t)\Longrightarrow dx=\dfrac{1}{2}ch(t)\ dt\\\\\sqrt{1+4x^2}=\sqrt{1+(2x)^2}=\sqrt{1+sh^2(t)}=\sqrt{ch^2(t)}=ch(t)\\\\\int\sqrt{1+4x^2}dx=\int [ch(t)\times\dfrac{1}{2}ch(t)]\ dt\\\\\int\sqrt{1+4x^2}dx=\dfrac{1}{2}\int ch^2(t)\ dt\\\\\int\sqrt{1+4x^2}dx=\dfrac{1}{2}\int \dfrac{ch(2t)+1}{2}\ dt\\\\\int\sqrt{1+4x^2}dx=\dfrac{1}{4}\int [ch(2t)+1]\ dt

\int\sqrt{1+4x^2}dx=\dfrac{1}{4}[\dfrac{sh(2t)}{2}+t]\\\\\int\sqrt{1+4x^2}dx=\dfrac{1}{4}[\dfrac{2sh(t)ch(t)}{2}+t]\\\\\int\sqrt{1+4x^2}dx=\dfrac{1}{4}[sh(t)ch(t)+t]


x = 0 ==> sh(t) = 0  et  t= 0
x = 1 ==> sh(t) = 2 et ch(t) = √(1 + sh²(t)) = √(1 + 4) = √5.

Par conséquent, 

L=\int_0^1\sqrt{1+4x^2}dx=\dfrac{1}{4}[(2\sqrt{5}+sh^{-1}(2)) - (0 + 0)]\\\\L\approx\dfrac{1}{4}[(2\sqrt{5}+1,443635475]\\\\L\approx1,478942858