Answer:
Step-by-step explanation:
[tex]length~of~a~curve~l =\int\limits^2_0 \sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2} )dt\\\frac{dx}{dt}=6t\\\frac{dy}{dt}=2*3t^2=6t^2\\l=\int\limits^2_0 {\sqrt{(6t)^2+(6t^2)^2} } \, dt\\=\int\limits^2_0 {6t\sqrt{1+t^2} } \, dt\\=3\int\limits^2_0 {(1+t^2)^{2}(2t) } \, dt[/tex]
=[tex]3\frac{(1+t^2)^3}{3}| t \rightarrow 0~to~2 \\\\=[(1+t^2)^3 ~|t \rightarrow 0 ~to~2 \\=(1+2^2)^3-(1+0^2)^3\\=125-1\\=124[/tex]
