I need help with this problem:
Compute the line integral$\int_{C^+}(x^2-y)dx+(y^2+x)dy$ where $C^+$ is the parabollic arc $y=x^2+1$, $0\leq x\leq 1$ oriented from $(0,1)$ to $(1,2)$.
First I parametrized the arc $C^+$ by setting $x=t$, then $y=t^2+1$, so the parametrization would be $\alpha(t)=(t,t^2+1), t\in[0,1]$. then I tried to solve this by using the formula $\int_Cfds=\int_\alpha fds=\int_{0}^1f(\alpha(t))\Vert\alpha'(t)\Vert dt+\int_1^2 f(\alpha(t))\Vert\alpha'(t)\Vert dt$. But I don't know how to compute that, since I have an integral with respect to $x$ and one with respect to $y$.
With $x=t$ and $y=t^2+1$ giving $dx=dt$ and $dy=2t \, dt$, and with $t$ going from $0$ to $1$, the integral becomes $$ \int_{C^+}(x^2-y)dx+(y^2+x)dy = \int_0^1 (t^2-(t^2+1)) \, dt + ((t^2+1)^2+t) \, 2t \, dt \\ = \int_0^1 (2t^5+4t^3+2t^2+2t-1) \, dt = \left[ \frac13 t^6 + t^4 + \frac23 t^3 + t^2 - t \right]_0^1 \\ = \frac13 + 1+ \frac23 + 1 - 1 = 2. $$