Calculation of Path integral If C represents the straight line segment from $(1,1)$ to $(1,0)$, compute $\int_{C}(x+y^2) ds$.
C can be parameterized  as $\mathbf c(t)=(1,t)$ where t goes from $1$ to $0$. Then $\int_{C}(x+y^2) ds = \int\limits_{1}^{0}(1+t^2)\left\lVert \mathbf c'(t)  \right\rVert dt = - \frac{4}{3}.$
C can be parameterized  as $\mathbf c(t)=(1,1-t)$ where t goes from $0$ to $1$. Then $\int_{C}(x+y^2) ds = \int\limits_{0}^{1}(2-2t+t^2)\left\lVert \mathbf c'(t)  \right\rVert dt = \frac{4}{3}.$
Since the value of path integral/scalar line integral does not depend on choice of parameterisation, why am I getting different answers?
 A: There is a fundamental misunderstanding here concerning the meaning of “the value of path integral/scalar line integral does not depend on choice of parameterisation”.
If $U\subset\mathbb R^n$, if $f\colon U\longrightarrow\Bbb R$ is continuous map, and if $\gamma\colon[a,b]\longrightarrow U$ (with $a,b\in\Bbb R$ and $a<b$) is a continuously differentiable curve, then we define the line integral of $f$ along the curve $\gamma$ as$$\int_\gamma f=\int_a^bf\bigl(\gamma(t)\bigr)\lVert\gamma'(t)\rVert\,\mathrm dt.$$When we say that this line integral is independent of the choice of parameterisation, what we mean is this: if $\beta\colon[c,d]\longrightarrow[a,b]$ (with $c,d\in\Bbb R$ and $c<d$) is a bijection such that:


*

*$\beta(c)=a$;

*$\beta(d)=b$;

*both $\beta$ and $\beta^{-1}$ are differentiable


and if $\gamma^\star=\gamma\circ\beta$, then $\int_{\gamma^\star}f=\int_\gamma f$. And a curve $\gamma^\star$ obtained from $\gamma$ by this process is what we call a reparametrization of $\gamma$.
When we see a curve as a line in space obtained as a point moves through that space, a (differentiable) parametrization of that curve is a differentiable map $\gamma\colon[a,b]\longrightarrow U$ (with, again, $a,b\in\Bbb R$ and $a<b$) such that, as $t$ growths from $a$ to $b$, $\gamma(t)$ is the point mentioned above describing the curve. For instance, if your curve is the right half of the unit circle in $\Bbb R^2$ obtained as a point goes from $(0,-1)$ to $(0,1)$ along that half-circle always advancing and never going back, then one such parametrization will be$$\begin{array}{rccc}\gamma\colon&\left[-\frac\pi2,\frac\pi2\right]&\longrightarrow&\Bbb R^2\\&t&\mapsto&\bigl(\cos(t),\sin(t)\bigr)\end{array}$$and another one will be$$\begin{array}{rccc}\gamma^\star\colon&[-1,1]&\longrightarrow&\Bbb R^2\\&t&\mapsto&\left(\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}\right).\end{array}$$
Note that if you define $\beta\colon[-1,1]\longrightarrow\left[-\frac\pi2,\frac\pi2\right]$ by $\beta(t)=2\arctan(t)$, then $\gamma^\star=\gamma\circ\beta$.
The problem with what you did is that when you talk about $c(t)$ “where $t$ goes from $1$ to $0$”, that doesn't describe a curve from an interval $[a,b]$ (with $a<b$) into $\Bbb R^2$. And, since you are going from $1$ to $0$, it is actually natural that you get the wrong sign.
A: I don't know what's happening but I'll wager my phone that the correct value is the second one. This is because the function is defined over a line in the first quadrant -- hence $x,\,y$ are always positive. Of course the differential is positive as well. Thus the integral cannot be negative.
