Answer
$\int_Cxds=\frac{5^{3/2}-1}{12}$
Work Step by Step
We can parameterize the parabola$y=x^{2}$
$C:x=t,y=t^{2}$
As we move from $(0,0)$ to $(1,1)$ , $t$ increases from $0$ to $1$.
Therefore, we will integrate from 0 to 1.
$ds=\sqrt {(\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}}dt$
$ds=\sqrt {{1}^{2}+({2t})^{2}}dt$
$ds=\sqrt {4t^{2}+1}dt$
Now, $\int_Cxds=\int _{0}^{1}t\sqrt {4t^{2}+1}dt$
$=\frac{1}{8}\int _{0}^{1}8t\sqrt {4t^{2}+1}dt$
Substitute $4t^{2}+1=u$ and $8tdt=du$
Limits of integration will change from $\int _{0}^{1}$ to$\int _{1}^{5}$.
$=\frac{1}{8}\int _{1}^{5}\sqrt {u}du$
$=\frac{1}{8}[\frac{2u^{3/2}}{3}] _{1}^{5}$
$=\frac{5^{3/2}-1}{12}$
Hence, $\int_Cxds=\frac{5^{3/2}-1}{12}$
