is $\frac{x^{2}-4}{x-2}$ Riemann integrable in [1,3]?

Question

\begin{align*} f(x)=\frac{x^{2}-4}{x-2} \end{align*} As this function is not defined at 2. what would we say about the Riemann integrability in [1,3] interval? As per the conditions, Riemann integration is only for bounded functions. So Do we need to redefine the function to call it as Riemann integrable? By assigning the limit value or by some fixed number. or Can we say this is Riemann integrable without redefining it, means by break-ing the given interval into [1,2) and (2,3]?

Answer 1

The concept of Riemann integrable function from an interval $[a,b]$ into $\Bbb R$ is only defined for bounded functions from $[a,b]$ into $\Bbb R$. Since the domain of your function is $[1,3]\setminus\{2\}$, then the definition does not apply; your function neither is nor isn't Riemann integrable.

You can look at is is an improper integral: compute $\lim_{x\to2^-}\int_1^xf(t)\,\mathrm dt$ and $\lim_{x\to2^+}\int_x^3f(t)\,\mathrm dt$. It happens that both limits exist; then $\int_1^3f(t)\,\mathrm dt$ is their sum.

Answer 2

Yes for Reimann Integrability you have to assign a value for $X=2$ and you can assign any real number at that point. Other wise you can use concept of improper integrals..

Answer 3

$x^2 - 4 = (x-2)(x+2)$

So:

$$f(x) = \frac{x^2-4}{x-2} = x+2,$$ everywhere $f$ is defined.

You technically can't say that $f$ is Riemann integrable since as you said, Riemann integrability is only defined for bounded, everywhere defined functions on closed intervals. But since $f(x) = x+2$ everywhere except a point (at which $f$ is not even defined) and integration is unchanged if you change a function by a point, there isn't any harm in thinking $\int_1^3 f(x) dx = \int_1^3 x+2 \ dx$.

On the other hand you could also go about defining it using the improper integrals, i.e. $\int_1^3 f(x) dx := \lim_{a \to 2} \int_1^a f(x) dx + \lim_{b \to 2} \int_b^3 f(x) dx$.

This will give you the same answer.

Technical annoyances like this is why we define generalizations of the Riemann integral, like the Lebesgue integral.

Answer 4

It depends, the function is undefined at $x=2$, although we can ignore it if we want.

Clearly the function is continuous everywhere else, and therefore it is integrable if we don't pay attention to that fact.