The line integral of a scalar function $f(x,y)$ along a curve $\vec{r}(t)$ for $a \leq t \leq b$ is defined to be $$ \int\limits_{\vec{r}(t)} f(\vec{r}(t)) \, ds = \int_a^b f(\vec{r}(t)) \, ||\vec{r}\,'(t)|| \, dt $$ and my Calculus text says this definition is invariant of the parameterization of $\vec{r}(t)$.
Thus, since the two curves $$ \vec{r}_1(t) = \langle t,t^2 \rangle \; (0 \leq t \leq 1) \quad \text{and} \quad \vec{r}_2(t) =\langle t^2,t^4 \rangle \; (0 \leq t \leq 1) $$ determine exactly the same curve (when restricted to $t \in [0,1]$), the corresponding line integrals should be the same, but they aren't (let's just use $f(x,y) = 1$ for simplicity): $$ \int\limits_{\vec{r}_1(t)} f(\vec{r}_1(t)) \, ds = \int_0^1 \sqrt{1+4t^2} \, dt \approx 1.09 \\ \int\limits_{\vec{r}_2(t)} f(\vec{r}_2(t)) \, ds = \int_0^1 \sqrt{4t^2+16t^6} \, dt \approx 1.48. $$ Am I missing something?