Define
$f(x)= \begin{cases}
3 & 0 \le x \le 2 \\
2 & 2<x<3 \\
4 & 3 \le x \le 6
\end{cases}
$
over $[0,6]. $ Is this function Riemann integrable over $[0,6]$?
I believe that it is, but I'm not sure if my solution is correct.
Basically, I considered the partition of the interval: $\{0, 2 – \delta, 2 +\delta, 3 -\delta, 3 + \delta, 6 \}$ and showed that over this partition, the lower sum is $20 – 3 \delta$ and the upper sum is $20 + 3 \delta$.
Since the difference is only $6 \delta$, we can make delta arbitrarily small, meaning the function is integrable.
Is this correct?
Best Answer
Yes, your answer is correct by the Cauchy criterion of Riemann Integrability. Your function $f$ is bounded over $[0,6]$, and you can take $\delta = \dfrac{\epsilon}{6}> 0$. The partition you used works !