Can someone tell me whether this is correct thank you!
We know that if a function f is continuous on $[a,b]$, a closed finite interval, then f is uniformly continuous on that interval. This means that if we're given any $\varepsilon > 0$, there exists $\delta > 0$ such that if $x$ and $y$ are any two points in $[a,b]$ with $|x-y| < \delta$, then $|f(x) - f(y)| < \varepsilon$.
So let's say we're given an epsilon. To show that a continuous function f is integrable, we must find a delta such that: For all partitions $\Gamma= \{x_0< \ldots < x_n\}$ of $[a,b]$ with $|\Gamma|:=\max \{x_{i+1} - x_i\} < \delta$ we have $$\mathrm{S}_{\delta} - \mathrm{s}_{\delta} < \varepsilon,$$ where, $$\mathrm{S}_{\delta}:=\inf \Sigma\ M_i(x_{i+1} - x_i)\ \mathrm{and}\ \mathrm{s}_{\delta}:=\inf \Sigma\ m_i(x_{i+1} - x_i),$$
over all partitions $\Gamma$ that satisfies $|\Gamma|<\delta$, with $M_i:=\max f|_{[x_i, x_i+1]}$ and $m_i:=\min f|_{[x_i, x_i+1]}$.
So, to recap, for a given epsilon we must find a delta such that $\mathrm{S}_{\delta} - \mathrm{s}_{\delta} < \varepsilon$. Since $f$ is uniformly continuous on $[a,b]$, we can choose $\delta$ such that $$|x-y| < \delta \ \Rightarrow \ |f(x) - f(y)| < \varepsilon/(b-a).$$
Then, for this $\delta$, we have, for any partition $\Gamma$ with $|\Gamma| <\delta$, that $M_i - m_i < \epsilon/(b-a)$. Therefore,
$$\mathrm{S}_{\delta}-\mathrm{s}_{\delta} < \frac\varepsilon{(b-a)} (b-a) = \varepsilon.$$