Answer
$\mathop \smallint \limits_C^{} xy{\rm{d}}s = \frac{1}{6}\sqrt 5 $
Work Step by Step
We have $f\left( {x,y} \right) = xy$.
Using the parametrization ${\bf{r}}\left( t \right) = \left( {t,2t - 1} \right)$ for $0 \le t \le 1$, we obtain
$f\left( {{\bf{r}}\left( t \right)} \right) = t\left( {2t - 1} \right) = 2{t^2} - t$
$ds = ||{\bf{r}}'\left( t \right)||dt = \sqrt {\left( {1,2} \right)\cdot\left( {1,2} \right)} dt = \sqrt 5 dt$
By Eq. (4) in Section 17.2:
$\mathop \smallint \limits_C^{} f\left( {x,y} \right){\rm{d}}s = \mathop \smallint \limits_a^b f\left( {{\bf{r}}\left( t \right)} \right)||{\bf{r}}'\left( t \right)||{\rm{d}}t$
$\mathop \smallint \limits_C^{} xy{\rm{d}}s = \sqrt 5 \mathop \smallint \limits_0^1 \left( {2{t^2} - t} \right){\rm{d}}t$
$\mathop \smallint \limits_C^{} xy{\rm{d}}s = \sqrt 5 \left( {\frac{2}{3}{t^3} - \frac{1}{2}{t^2}} \right)|_0^1 = \sqrt 5 \left( {\frac{2}{3} - \frac{1}{2}} \right) = \frac{1}{6}\sqrt 5 $
So, $\mathop \smallint \limits_C^{} xy{\rm{d}}s = \frac{1}{6}\sqrt 5 $.
