Given a triangle with vertices $0,1,i$ I'm asked to evaluate the integral $$\int_{\partial \triangle} \! z \ dz$$ by the Cauchy-Goursat theorem I know that the integral is equal to zero because $f(z)=z$ is an entire function; however when I calulte the integral without the theorem I end up with this result that makes no sense: $$\int_{\partial \triangle} \! z \ dz = \int_0^1 \! t \ dt + \int_1^i \! (i-1) \left(it+1-t \right) \ dt + \int_i^0 \! -i\ i(1-t)\ dt = -i$$.
I used the parametrization of the paths joining the vertices of the triangle, so I know that there is an error but I don't see where.
Second integral should go from $0$ to $i$, as $y=t$. Also I think you messed up on the third integral. These are the parametrizations I used:
For path $0\to1$: $$ \left\{ \begin{aligned} &x=t\\ &y=0 \end{aligned} \right\}\implies z(t)=t\ \text{ for } t\in(0,1)\implies \text dz=\text dt $$
For path $1\to i$: $$ \left\{ \begin{aligned} &x=t\\ &y=1-t \end{aligned} \right\}\implies z(t)=t+i(1-t)=(1-i)t+i\ \text{ for } t\in(1,0)\implies \text dz=(1-i)dt $$ For path $i\to0$: $$ \left\{ \begin{aligned} &x=0\\ &y=t \end{aligned} \right\}\implies z(t)=it\ \text{ for } t\in(1,0)\implies \text dz=i\text dt $$ Finally, we can now compute the integral $$\begin{aligned} \int_{\partial\triangle}zdz&=\int_0^1 t\text dt+\int_1^0[(1-i)t+i](1-i)\text dt+\int_1^0iti\text dt\\ &=\int_0^1 t\text dt-\int_1^0t\text dt+\int_1^0[(1-i)^2t+(1+i)]\text dt\\ &=2\left.\frac{t^2}{2}\right|_0^1+\left[(1-i)^2\frac{t^2}{2}+(1+i)t\right|_1^0\\ &=1-(1-i)^2\frac{1}{2}-(1+i)=-\frac{1}{2}(1-2i-1)-i=i-i=0\ \square, \end{aligned} $$ which concurs with your initial reasoning.