TITLE: Definite integral over $(0,1)$ rather than $[0,1]$
QUESTION [8 upvotes]: I'm wanting to calculate a simple definite integral (but haven't done so since high-school), and am seeking some clarification.
$$f(x)=\begin{cases}x^{2}(x-1) & 0<x<1\\ 0 & \text{otherwise}\end{cases}$$
I am interested in the area under the curve that is between $0$ and $1$. However, I can't just calculate $$\int_{0}^{1}x^{2}(x-1)\;dx$$ as the function isn't $x^{2}(x-1)$ for $0\leq x\leq1$, correct? Is there a method that I've completely forgotten about that will allow me to get what I'm after?
Thanks for your patience in answering what I'm sure is a very basic question.

REPLY [4 votes]: Changing the value of an integrable function at a finite number of points will have no effect on the ultimate value of the definite integral, and below I will explain why.
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 $\displaystyle U(f, \mathcal{P}) = \sum_i (x_{i+1} - x_i)\sup \Big( \{ f(x) \ | \ x \in [x_i, x_{i+1}] \} \Big)$
And so with in mind, it's possible to show that changing the value of an integrable function at a single point has no effect on the value of the definite integral.  Basically, the idea is that you take your original partitions and refine them to enclose the points in question within intervals of arbitrarily small width, limiting the relevant terms to zero in the $L(f, \mathcal{P})$ and $U(f, \mathcal{P})$ summations.
For further discussion, see Chapter 6 of Walter Rudin's Principles of Mathematical Analysis (PDFs are freely available online).