PROBLEM Integrate $\int_{i}^{i+1}{zdz}$ along a straight line parallel to the $x$ axis.
The definition of a complex line integral states let $f(z)$ be a continuous complex-valued function of a complex variable, and let $\mathbb{C}$ be a smooth curve in the complex plane parametrized by $$z(t)=x(t)+iy(t)\text{ for } a\leq t \leq b$$ Then the complex line integral of $f$ over $\mathbb{C}$ is given by $$\int_{\mathbb{C}}f(z)dz=\int_{a}^{b}{f(z(t))z'(t)dt}$$ The smooth condition guarantees the $z'(t)$ is continuous and, hence that the integral exists.
SOLUTION
Along a straight line parallel to the $x$ axis $z(t)$ can be parametrized by $$z(t)=t+i$$ It follows from the definition of a complex line integral \begin{align*} \int_{i}^{i+1}{zdz} &= \int_{i}^{i+1}{(t+i)(1)dt}\\ &=\int_{i}^{i+1}{tdt}+\int_{i}^{i+1}{idt}\\ &=\frac{t^{2}}{2}\Bigg|^{i+1}_{i}+it\Bigg|^{i+1}_{i}\\ &=\frac{(i+1)^{2}-(i)^{2}}{2}+i((i+1)-(i))\\ &=\frac{-1+2i+1+1}{2}+i=\frac{1+2i}{2}+i\\ &=\frac{1}{2}+2i \end{align*}
The solutions of selected answers in Mathematical Methods in the Physical Sciences by Boas states that the answer is $\frac{1}{2}+i$. What am I doing wrong?
When you plug in the parametrization of $z(t)$, then you need to change the integration to $\int_0^1$ instead of $\int_i^{i+1}$.