How do I prove that this function is Riemann Integrable?

Question

$f(x)=1/(x+3)$ over the interval [0,1]

Here the function is a monotone function. So it is Riemann integrable. But is there a way I could prove that the given function is Riemann Integrable. I tried solving the problem by using partitions but I could not prove it.

Answer 1

Hint:

Claim: A continuous function $f: [a,b] \to \mathbb R$ is Riemann integrable.

Proof: Given $\varepsilon > 0$ since $f$ is continuous in a compact, it is also uniformly continuous at $[a,b]$, then there exists $\delta > 0$ such that $x,y \in [a,b]$, $$|y - x| < \delta \implies |f(y) - f(x)| < \frac{\varepsilon}{b-a}$$

Take a partition $P$ of $[a,b]$ on which every interval ahas length $< \delta$. For every interval $[t_{i-1}, t_i]$ there exists $x_i,y_i$ such that $m_i = f(x_i)$ and $M= f(y_i)$, then $\omega = f(y_i) - f(x_i) < \frac{\varepsilon}{b-a}$. It follows that $$\sum \omega_i (t_i - t_{i-1}) < \varepsilon$$

Answer 2

Here is a way of generalizing/extending what Aaron said, basically rewording it, since redundancy can help: You want to have upper- and lower sums so that , for any $\epsilon>0$, $U-L <\epsilon$.

As pointed out by Aaron, a continuous function in a(n) (compact) interval $[x_k, x_{k+1}]$ $f(x)$ takes maximum and minimum values $M,m$ on the interval. Now, you want to use these $M,m$ for your upper- and lower- sums. For this you need to make $M-m$ as small as you want. Can you see that making the intervals $[x_k, x_{k+1}]$ smaller, using uniform continuity can help you do that?

Answer 3

Even easier: the integrand function is a Lipschitz function, hence for every partition of $[0,1]$ the difference between the upper sum and the lower sum is bounded by the Lipschitz constant $L$ times the width of the largest interval of the partition. Riemann integrability readily follows.