Respuesta :

LammettHash
[tex]y=-0.1480x^3+1.353x^2-4.068x+5.651[/tex]
[tex]\implies y'=-0.444x^2+2.706x-4.068[/tex]

The arc length is given by the integral

[tex]\displaystyle\int_{0.2}^{5.25}\sqrt{1+(y')^2}\,\mathrm dx[/tex]

which has a value of about [tex]7.6683[/tex].