Question:
Find $\int\overline{z}$, when the contour is a parabola. Interval is from 0 to 1.
My Attempt:
$z = x + iy \Rightarrow \overline{z} = x - iy$
$f(z) = x - iy$
Since the contour is a parabola, $\gamma(t) = t + it^2$ and $y = x^2$.
$Re(\gamma) = t$, $Im(\gamma) = t^2$
$\gamma\prime(t) = 1 + 2ti$
$f(\gamma(t)) = (t - t^2)$
So, $\int_{c}\overline z = \int_{0}^{1}(t-it^2)(1+2t)$
$= \int_{0}^{1}(t - 2t^3) + i(2t^2 - t^2)$
$= (\frac{t^2 - t^4}{2}) + i(\frac{t^3}{3})$ this must be evaluated from 0 to 1
$= \frac{1}{3}i$
Is my answer right?
If $f(x,y)=x-iy$ and $\gamma(t):=x(t)+iy(t)=t+it^2$ with $t\in [0,1]$ then
$$f(\gamma(t)):=f(x(t),y(t))=x(t)-iy(t)=t-it^2$$
As $\gamma'(t)=1+2it$, we arrive at
$$\int_{\gamma} f:=\int_{0}^{1}f(\gamma(t))\gamma'(t)dt=\int_{0}^{1}(t-it^2)(1+2it)dt= \int_{0}^{1}(t+2t^3)dt+i\int_{0}^{1}t^2dt= \\ \frac{1}{2}+\frac{1}{2}+\frac{1}{3}i= 1+\frac{1}{3}i.$$