Answer
$\dfrac{73 \sqrt {73}-125}{48}$
Work Step by Step
Consider $I=\int_{C}y ds=\int_{1}^{2}(t^3/t^4) \times \sqrt {{(\dfrac{dx}{dt})^{2}}+{(\dfrac{dy}{dt})^{2}}}ds=\int_{1}^{2}(t^3/t^4) \times (t^2) \sqrt {9+16t^2}dt$
Plug $9+16t^2=a \implies da=32tdt$
$=(1/32)\int_{25}^{73} \sqrt a da=\dfrac{1}{32}|(2/3)(a^{3/2}|_{25}^{73}=\dfrac{1}{48}(73\sqrt{73}-25 \sqrt{25})=\dfrac{73 \sqrt {73}-125}{48}$
