What is one difference in the values of
$$\int\limits_{\left[0,1\right]}y\, dx$$$$\int\limits_{\left(0,1\right)}y\, dx$$
and how would you calculate the values? For the sake of simplicity, let $y=x$. Conceptualizing integration as the area bounded by the function, the $x$-axis and the limits of integration, the latter should be smaller.
It should be intuitive that $\displaystyle \int_{(0, 1)} f(x) \ dx = \int_{[0, 1]} g(x) \ dx$ where $g(x) = \begin{cases} f(x) & \ \text{ if }\ x \in (0, 1) \\ 0 & \ \text{ if } \ x \in \{0, 1\}\end{cases}$. We claim that $\displaystyle \int_{[0, 1]} f(x) \ dx = \int_{[0, 1]} g(x) \ dx$, or more generally, changing the value of $f$ at finitely many points has no effect on the value of the definite integral.
Sketch of proof:
Provided a function $f$ is integrable on an interval $[a, b]$, the definite integral is rigorously defined as follows: there is a unique $I$ such that, for any given partition $\mathcal{P}$ of an interval $[a, b]$, we have:
$$L(f, \mathcal{P}) \leq I = \int_a^b f(x) \ dx \leq U(f, \mathcal{P})$$
Where $\displaystyle L(f, \mathcal{P}) = \sum_{i} (x_{i+1} - x_i)\inf \Big( \{f(x) \ | \ x \in [x_i, x_{i+1}] \} \Big)$ where $x_i$'s $\in \mathcal{P}$
and likewise $\displaystyle U(f, \mathcal{P}) = \sum_i (x_{i+1} - x_i)\sup \Big( \{ f(x) \ | \ x \in [x_i, x_{i+1}] \} \Big)$
Now suppose we change the value of $f$ at a point $y \in [a, b]$. For any given partition, we can "refine" this partition to encapsulate $y$ inside an arbitrarily small interval, in effect making its associated term in the $L(f, \mathcal{P}')$ and $U(f, \mathcal{P}')$ summations arbitrarily insignificant (limiting to zero in successive such refinements of the partition).
This is not a definitive answer, considering an answer have already accepted almost 2 years ago, but I believe it should be noted that in the case where there exists a Dirac delta function or any similarly defined special functions inside the integrand, the integral over some 'points' become non-zero by definition.
Using this fact, we can prove that the claim that these two integrals are equal leads to a contradiction if one or both of the limits of the interval are included a specially defined function. In this case, inclusion of a particular point will change the answer of the integration by a finite amount. Choosing this point from the limits of the interval, integration over an open interval will yield a different result than integration over an open interval by that finite amount.
So, integration on open or closed intervals can yield different results if there is a special function in the integrand such as Dirac delta.
Disclaimer: Notion of integrating δ(x) over an interval [0,a] might be undefined on its own. If anybody have any information in this regard, please edit the answer accordingly.
See also: Dirac delta integral with $\delta(\infty) \cdot e^{\infty}$
There is not difference. $$\int_{[0,1]}f(x).dx = \int_{(0,1)}f(x).dx = \lim_{n\to \infty} \left(I_n^+(f,0,{1\over 2}) + I_n^-(f,{1\over 2},1)\right) $$
With $I_n^+$ and $I_n^-$ respectively the upper and lower Riemann sums ( I chose such a decomposition because those Riemann sums never involve $f(0)$ and $f(1)$ )
