I'm working through chapter 4 of Alfohrs, and the first exercise is computing the integral $\int_\gamma x \: dz$ where $\gamma$ is the directed line segment from $0$ to $1+i$. What, exactly, does this mean?
I know that by definition, $\int_\gamma f(z)\:dz = \int_a^b f(z(t)) z'(t) \:dt$; however, I'm not quite sure what to do with this definition here. One thing that seems to be the case is $f$ is probably implied to be in the form $x + iy$ (is this a fair assumption?) If so, I could use the expansion $\int_\gamma f(z)\:dz = \int_\gamma (u dx - v dy) + i \int_\gamma (u dy + v dx)$, which would after substitution become $\int_\gamma x dx + i \int_\gamma x dy$.
If I can indeed have it in this form, I need to somehow parametrize $\gamma$, correct? My attempt here is saying that $\gamma(t) = t + it$ for $0 \leq t \leq 1$. Then I get that $\int_\gamma x dx + i \int_\gamma x dy = \int_0^1 x(t)x'(t)dt + i\int_0^1 x(t)y'(t)dt = \int_0^1 t dt + i\int_0^1 t dt = (t^2/2)\mid_0^1 + (t^2/2)\mid_0^1 = (t^2)\mid_0^1 = 1^2=0^2=1$.
Is any of this reasoning correct?
Edit
Fixed my last calculation in response to mrsamy's answer: it's now $\int_\gamma x dx + i \int_\gamma x dy = \int_0^1 x(t)x'(t)dt + i\int_0^1 x(t)y'(t)dt = \int_0^1 t dt + i\int_0^1 t dt = (t^2/2)\mid_0^1 + i(t^2/2)\mid_0^1 = 1/2 + 1/2 \cdot i$.
You are correct that the first step is parametrizing the curve $\gamma$.
Instead of $\gamma(t)$, you write $z(t)=t+it$, $0\le t\le 1$. So $z'(t)=1+i$ and thus $$ \int_\gamma f(z)dz=\int_0^1 f(z(t))z'(t)dt=\int_0^1t(1+i)dt=(1+i)\int_0^1tdt=\frac12(1+i). $$
You miss an $i$ in your calculation.