Answer
$\displaystyle \frac{80\sqrt{10}-13\sqrt{13}}{27} \approx7.634$
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 $\quad\displaystyle f(x,y) = \frac{x^2}{y^{4/3}}\quad$ , $\quad \vec{r}(t) = \langle t^2, t^3 \rangle\quad$ , $\quad a = \displaystyle 1\quad$ , and $\quad b = 2$
$\vec{r}'(t) = \langle 2t, 3t^2 \rangle$
$||\vec{r}'(t)|| = \sqrt{(2t)^2 + (3t^2)^2} = \sqrt{4t^2+9t^4} = t\sqrt{4+9t^2}$
$f(\vec{r}(t)) = \displaystyle \frac{(t^2)^2}{(t^3)^{4/3}} = \frac{t^4}{t^4} = 1$
$\displaystyle \int_{1}^21\cdot (t\sqrt{4+9t^2})dt$
We can solve this integral using a calculator or $u$-substitution. By $u$-substitution let
$u = 4+9t^2\quad$ and therefore $\quad du = 18tdt$
$\displaystyle \frac{1}{18}\int u^{1/2}dt = \frac{1}{18}[(\frac{2}{3})(4+9t^2)^{3/2}]^{t = 2}_{t = 1}$
$\displaystyle\frac{1}{27}[(4+9(2)^2)^{3/2} - (4+9(1)^2)^{3/2}] = \frac{1}{27}((40)^{3/2}-(13)^{3/2})$
$\displaystyle \frac{1}{27}((2\sqrt{10})^3-13\sqrt{13}) = \frac{80\sqrt{10}-13\sqrt{13}}{27} \approx7.634$
