Answer
Part a: $4\sqrt5$
Part b: $\frac{1}{12}(17\sqrt{17}-1)$
Work Step by Step
The question asks us to find the integral $\int_{C}xds$, where C is two different curves.
For part a, curve $C$ is a straight line segment given by $x=t$ and $y=\frac{t}{2}$ from $(0,0)$ to $(4,2)$.
Thus, the parametrization of $C$ is $r(t)=ti+\frac{t}{2}j$.
We then must find $|v(t)|$, which is the derivative of r above.
$|v(t)|=|1i+\frac{1}{2}|=\frac{\sqrt{5}}{2}$
Thus, combining everything:
$\int_{C}xds=\int_{0}^{4}{t \cdot \frac{\sqrt{5}}{2}}dt=4\sqrt5$
For part b, a similar approach is taken for the new curve given:
$(x,y)=(t,t^2)$ from $(0,0)$ to $(2,4)$
$f(x,y)=x$
$|v(t)|=\sqrt{(1)^2+(2t)^2}=\sqrt{4t^2+1}$
Combining all that together:
$\int_C x ds= \int_{0}^{2}t\sqrt{4t^2+1}dt=\frac{1}{12}(17\sqrt{17}-1)$
Where we used u-substitution ($u=4t^2+1$) to integrate.
