Answer
$$2\sqrt2 - 1$$
Work Step by Step
We set up the integral to find the mass:
$$M=\int_C \delta (t) |\mathbf{v}(t)|dt$$
where C is given by the curve $$\mathbf{r}(t)=t^2\mathbf{i}+2t\mathbf{j}$$ and the $t$ range is $$0 \leq t \leq 1$$
Thus, we have:
$$M=\int_{0}^{1}\frac{3t}{2}\sqrt{(2t-0)^2+(2)^2}dt=3\int_{0}^{1}t\sqrt{t^2+1}dt=\frac{3}{2}\int_{1}^{2}u^{1/2}=\frac{3}{2}\frac{2}{3}(2\sqrt{2}-1)=2\sqrt{2}-1$$
