Solution
$ \mathbf{F}(x,y)=-x\hat{i}+y\hat{j} $
$$
\int_C \mathbf{F}\cdot d\mathbf{r}
$$
$$
=\int_C(-x\,dx+y\,dy)
$$
Parameterization:
$$
x=\cos t,\qquad
y=\sin t
$$
$$
dx=-\sin t\,dt,\qquad
dy=\cos t\,dt
$$
$$
\int_{0}^{\pi/4}
\left[
(-\cos t)(-\sin t)
+
(\sin t)(\cos t)
\right]dt
$$
$$
=
\int_{0}^{\pi/4}
2\sin t\cos t\,dt
$$
where
$$
0\le t\le\frac{\pi}{4}
$$
$$
2\sin t\cos t=\sin2t
$$
$$
=\int_{0}^{\pi/4}\sin2t\,dt
$$
$$
=\left[-\frac12\cos2t\right]_{0}^{\pi/4}
$$
$$
=-\frac12\cos\frac{\pi}{2}
+\frac12\cos0
$$
$$
=0+\frac12
=\frac12
$$
$$
\boxed{\int_C \mathbf{F}\cdot d\mathbf{r}=\frac12}
$$
Option A