Integrating on open vs. closed intervals 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).
