Answer
$\displaystyle \frac{125-13\sqrt{13}}{48} \approx1.628$
Work Step by Step
To find the the line integral, we simply compute $\displaystyle \int_{a}^bf(\vec{r}(t))||\vec{r}'(t)||dt$
where $\quad\displaystyle f(x,y,) = \frac{\sqrt{y}}{x}\quad$ , $\quad \vec{r}(t) = \langle t^3, t^4 \rangle\quad$ , $\quad a = \displaystyle \frac{1}{2}\quad$ , and $\quad b = 1$
$\vec{r}'(t) = \langle 3t^2, 4t^3 \rangle$
$||\vec{r}'(t)|| = \sqrt{(3t^2)^2 + (4t^3)^2} = \sqrt{9t^4+16t^6} = t^2\sqrt{9+16t^2}$
$\displaystyle \int_{1/2}^1\frac{\sqrt{t^4}}{t^3}(t^2\sqrt{9+16t^2})dt = \int_{1/2}^1t\sqrt{9+16t^2}dt$
We can solve this integral using a calculator or $u$-substitution. By $u$-substitution let
$u = 9 + 16t^2\quad$ and therefore $\quad du = 32tdt$
$\displaystyle \frac{1}{32}\int u^{1/2}dt = \frac{1}{32}[(\frac{2}{3})(9+16t^2)^{3/2}]^{t = 1}_{t = 1/2}$
$\displaystyle\frac{1}{48}[(9+16(1)^2)^{3/2} - (9+16(\frac{1}{2})^2)^{3/2}] = \frac{1}{48}((25)^{3/2}-(13)^{3/2})$
$\displaystyle \frac{1}{48}(5^3-13\sqrt{13}) = \frac{125-13\sqrt{13}}{48} \approx1.628$
