Evaluate $$I=\int_\gamma\left(\sqrt{x^2-y}+\frac{x^2}{\sqrt{x^2-y}}\right)dx-\frac{x}{2\sqrt{x^2-y}} \ dy,$$
where $\gamma$ is a part of the curve $x=y^2$ from $(1,-1)$ to $(4,-2).$
Parameterizing by
$$\left\{ \begin{array}{rcr} x & = & t \\ y & = & -t^2 \\ \end{array} \right.\implies \left\{ \begin{array}{rcr} dx & = & dt \\ dy & = & -2t \ dt \\ \end{array} \right., \quad t\in[1,2].$$
So,
$$ \begin{align} I &=\int_\gamma \sqrt{t^2+t^2} + \frac{t^2}{\sqrt{t^2+t^2}}+\frac{t}{\sqrt{t^2+t^2}} \ dt\\ &=\int_\gamma |t|\sqrt{2}+\frac{t(t^2+1)}{|t|\sqrt{2}} \ dt \quad (|t|=t, \ t\geq0.)\\ &=\int_1^2 t\sqrt{2}+\frac{1}{\sqrt{2}}(t^2+1) \ dt = \frac{19\sqrt{2}}{6}. \end{align} $$
Correct anser is $11\sqrt{2}.$ What have I missed?
Your parametrization is wrong. You want $x=y^2$, but $(x,y)=(t,-t^2)$ does not satisfy that (and it's also unclear why you have $t$ going from $1$ to $2$). You could instead parametrize it as $(x,y)=(t^2,-t)$ for $t$ going from $1$ to $2$.