Replace x and y with the corresponding components of r(t), where
[tex]\mathbf r(t)=\langle x(t),y(t)\rangle=\langlet+3,4-2t\rangle[/tex]
We have
[tex]\displaystyle\int_Cf(x,y)\,\mathrm ds=\int_1^2f(x(t),y(t))\sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2}\,\mathrm dt[/tex]
[tex]=\displaystyle\int_1^2(2(t+3)+4(4-2t))\sqrt{1^2+(-2)^2}\,\mathrm dt[/tex]
[tex]=\displaystyle\sqrt5\int_1^2(22-6t)\,\mathrm dt[/tex]
[tex]=\sqrt5(22t-3t^2)\bigg|_1^2=\boxed{13\sqrt5}[/tex]
I'm tempted to say the answer is B, but it doesn't seem to match up exactly. It's possible that choice contains a typo.
