Define $g : [0,1.571] \to \mathbb{R}$ by
$$ { g }(x)=\begin{cases} \sin x \quad \text{if}\quad x \in [0,1/2] \\\cos x \quad \text{if}\quad x \in (1/2,1] \\ \tan x \quad \text {if} \quad x \in (1,1.571]\end{cases} $$
My question : Is $g$ Riemann integrable on $[0,1.571]?$
My attempt : I think Yes
$[0,1.571] = [0,1/2]\cup[1/2,1]\cup [1,1.571]$. Now by using the Darboux's concept , i,e if $g$ is integrable on $[a,b]$ iff there exists a partition of $[a,b]$ whose sum of areas under curves = $\sum_1^n M_j \varDelta x_j < \varepsilon$.
By Darboux's idea $g$ is integrable on $[0,1/2],[1/2,1]$ and $[1,1.571]$
This implies the existence of partitions $\mathcal P_1, \mathcal P_2$ and $\mathcal P_3$ of each interval such that
$$U(\mathcal P_1, g)-L (\mathcal P_1, g)< \epsilon$$
$$U(\mathcal P_2, g)-L (\mathcal P_2, g)< \epsilon$$
and $$U(\mathcal P_3, g)-L (\mathcal P_3, g)< \epsilon$$
Also, $\mathcal{P}=\mathcal P_1 \cup \mathcal P_2 \cup \mathcal P_3$
$$\implies U(\mathcal P, g)-L (\mathcal P, g)< \epsilon$$
Therefore $g$ is Riemann integrable on $[0,1.571]$
Is it true ?
Best Answer
$1.571>\frac\pi2$ and $\tan$ isn't Riemann (or Lebesgue) integrable in a any neighbourhood of $\frac\pi2$. Therefore $g$ isn't Riemann integrable.