Since the function is increasing, one has that, given any partition $P=\{t_0,\dots,t_n\}$
$M_i=\sup_{[t_{i-1},t_i]}f(x)=f(t_i)$ $m_i=\inf_{[t_{i-1},t_i]}f(x)=f(t_{i-1})$
Then one has that the upper and lower sums are
$U(f,P)=\sum_{i=1}^n M_i(t_{i}-t_{i-1})=\sum_{i=1}^n f(t_i)(t_{i}-t_{i-1})$ $L(f,P)=\sum_{i=1}^n m_i(t_{i}-t_{i-1})=\sum_{i=1}^n f(t_{i-1})(t_{i}-t_{i-1})$
We then have that $U(f,P)-L(f,P)=\sum_{i=1}^n (f(t_i)-f(t_{i-1}))(t_{i}-t_{i-1})$
Let $\epsilon >0$ be given and let $P$ now be such that $t_i-t_{i-1}<\delta$ for some $\delta$ we will determine shortly. Then
$U(f,P)-L(f,P)<\sum_{i=1}^n (f(t_i)-f(t_{i-1}))\delta$
$U(f,P)-L(f,P)<\delta \sum_{i=1}^n (f(t_i)-f(t_{i-1}))$
$U(f,P)-L(f,P)<\delta (f(t_n)-f(t_{0}))$
Thus it suffices to take $\delta <\dfrac {\epsilon}{f(t_n)-f(t_{0})}$ for any given $\epsilon$ and thus we have integrabililty.
On a side note, if $f$ is increasing over $[a,b]$ it is immediately bounded since $f(a)\leq f(x)\leq f(b)$ for every $x\in [a,b]$.