The definitions I am using are
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a manifold with boundary is something locally homeomorphic to $(0,1] \times \mathbb{R}^n$ or $\mathbb{R}^n$.
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an oriented manifold is one where the transition functions between any two charts have positive Jacobian.
Its clear that $[0,1]$ is a manifold with boundary. Is it oriented? I seem to be getting the conclusion that it is not, because around $0$, the local chart is $x \to 1-x$, and around $1$ the local chart is $x \to x$, and these have opposite orientations. But Stokes theorem surely should apply to this setting and reduce to the fundamental theorem of calculus. Recall that Stokes theorem says that if $M$ is a compact oriented $n$-manifold with boundary $\partial M$ with the induced orientation, and $\omega$ is an $n-1$ form, then $$\int_{\partial M} \omega = \int_M d\omega.$$
Taking $\omega=f$ a zero form, i.e., function and $M=[0,1]$, I expect to recover $$f(1)-f(0) = \int_0^1 f'(x) dx ,$$ so $[0,1]$ I suspect is an oriented manifold with boundary, but I'm not seeing exactly why.
In a related vein, I think I can see intuitively that the closed unit disk is an oriented manifold with boundary, and then Stokes gives Green's theorem.
@Bill, in a comment below you wrote "the chart containing 1 orients the interval 0→1." Well, let me be more precise: the chart $U_1=(0,1] \to (0,1]$ given by $x \to x$ (for $x \in (0,1]$) orients $(0,1]$. I cannot extend this particular chart to include $0$ because then $[0,1]$ would not be a homeomorphic to $(0,1]$ which is in my definition of manifold with closed boundary. To get a chart including the zero, I need to include another chart, e.g $U_2=[0,1)$ and then the map $x \to 1-x$ would be a homeomorphism $U_2 \to (0,1]$ as required in the definition I am using for manifold with boundary. But now, these charts have opposite orientations. I don't know how to come up with two (or more) charts that do not give opposite orientations. The definition of manifold with boundary I am using is the one on p.25 of Voisin's Hodge Theory and Complex Analytic Geometry 1, and its easily seen to be equivalent to the more standard one homeomorphic to an open subset of the closed upper half plane. I agree with your second commment.
Best Answer
Ok, this is a bit late and you might have solved it already by yourself. But I stumbled upon this question on Google because I too had this problem, I'm hoping people with the same problem can find this solution. There is a quirk that few books discuss which only happens in dimension 1. With your definition (and most) the interval $[0,1]$ does not have an oriented atlas. This is easy to see because, as you said, the local charts at the endpoints have opposite orientations and any other collection of charts is going to flip monotonicity at some point and at this point the orientations will be incompatible.
To fix this you define a manifold with boundary distinguishing two cases (sorry to change your definition, but it's only a superficial change): in dimension $n >1$ a (topological) manifold with boundary is a second countable Hausdorff topological space locally homeomorphic to $\mathbb{H}^n = \{x \in \mathbb{R}^n\,:\, x^n \geq 0\}$ (this is just the usual definition). If $n = 1$ then you define a $1$-dimensional topological manifold with boundary as a second countable Hausdorff topological space where at each point $p$ there is a neighborhood $U$ of $p$ and a map $\varphi : U \to \mathbb{R}$ where $\varphi(U)$ is open in either $\mathbb{H}^1$ or $\mathbb{H}^1_{-} = \{x \in \mathbb{R} : x \leq 0\}$ with the relative topology and $\varphi_{|U}$ is a homeomorphism.
Now you define everything else (smooth manifold, orientation, etc.) as usual. With this definition you can give $[0,1]$ the structure of a smooth manifold with boundary using the charts: $(U = [0,1)$, $\varphi(x) = x)$ and $(V = (0,1]$, $\psi(x) = x-1)$, which are clearly compatible orientation-wise.
This is the approach given in Loring Tu: An introduction to manifolds, he specially discusses this difficulty in page 254; example 22.9.