A question on multivariate integral Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a given function. Suppose $\boldsymbol{f}:\mathbb{R}^{N} \rightarrow \mathbb{R}^{N}$ is the vector version of $f$, e.g., $\boldsymbol{f}(\boldsymbol{t})=[f(t_1),f(t_2),\cdot\cdot\cdot,f(t_N)]^{\top}$, for $t=[t_1,t_2,\cdot\cdot\cdot,t_N]^{\top} \in \mathbb{R}^N$. Let $\mathcal{C}$ be a smooth curve in $\mathbb{R}^N$, from $\boldsymbol{0}$ to $\boldsymbol{x}$, where $\boldsymbol{x}=[x_1,x_2,\cdot\cdot\cdot,x_N]^{\top} \in \mathbb{R}^N$ is a fixed column vector. 
The question is when the integral 
\begin{equation}
\int_{\mathcal{C}} \boldsymbol{f}(\boldsymbol{t}) \cdot d\boldsymbol{t}
\end{equation}
is equal to
\begin{equation}
\sum_{i=1}^N \int_0^{x_i} f(t) dt.
\end{equation}
Thanks a lot.
 A: I think they are always identical due to your definition for $\mathbf{f}$. Consider parametrizing your curve $\mathcal{C}$ as $\mathbf t(s)$ for $s\in[0,1]$ then the first integral becomes
$$\int_0^1 \mathbf f(\mathbf t(s)) \cdot \left.\frac{d\mathbf{t}}{ds}\right|_s ds$$
where $d\mathbf t/ds$ is the tangent vector to the curve. This gives
$$\int_0^1 \sum_{i=1}^n \mathbf f(\mathbf t(s))_i \left(\left.\frac{d\mathbf{t}}{ds}\right|_s\right)_i ds$$
Because $\mathbf f(\mathbf t) =  [f(t_1),\ldots f(t_n)]^T$ the $i$-th component of $\mathbf f(\mathbf t(s))$ is just $f(t_i(s))$ and similarly the $i$-th component of the tangent is just the gradient $t_i$ wrt $s$. This means that the integral becomes
$$\sum_{i=1}^n \int_0^1 f(t_i(s))\, \left.\frac{dt_i}{ds}\right|_s ds$$
which we can see is equal to
$$\sum_{i=1}^n \int_0^{x_i}f(t) dt$$
In general this does not hold because we cannot "pull the  index inside" the argument of $\mathbf{f}(\mathbf t)$ for an arbitrary vector field. You might be interested to read about conservative fields, which are vector fields where the path integral is the same whatever path you take between two points.
Scalar potential:
A vector field is conservative if there is a scalar field $\phi:\mathbb{R}^n\mapsto\mathbb{R}$ such that $\mathbf{f}(\mathbf t) = \nabla \phi(\mathbf t)$. This implies that $\frac{\partial \phi}{\partial t_i}(x_i)=f(x_i) \Rightarrow \nabla\phi(\mathbf x)=\mathbf{f}(\mathbf x)$. We can see that for a scalar field the value at a point $\mathbf x$ is given by
$$\phi(\mathbf x) = \int_0^\mathbf{x} \nabla \phi (\mathbf t) \cdot d\mathbf{t}$$
or using the equality above
$$\phi(\mathbf x) = \int_0^\mathbf{x} \mathbf f (\mathbf t) \cdot d\mathbf{t}$$
which is exactly the integral we were interested in. Thereby confirming our intuition. Here is a picture of this idea where $f(t)=\sin(t)$ in two dimenions, so $\phi(x,y)=-\cos(x)-\cos(y)$

