Riemann Integral of $|f| \le N(b-a)$

Question

I need help proving the following:

Let $f$ be Riemann integrable on $[a,b]$, and suppose $|f(x)| \le N$ for all $x \in [a,b]$. Prove, from the definition of a Riemann integral, that

$\int_a^b |f| \le N(b-a) $.

Here is my attempt at the proof, I feel like I am close, but can't see where to go from here (if I went it the correct direction to start with that is):

Suppose $f$ is Riemann integrable on $[a,b]$ and that $|f(x)|\le N, \forall x \in [a,b]$.

Then $-N \le f(x) \le N, \forall x \in [a,b]$.

So $-N(b-a) \le L(f,P)$ and $U(f,P) \le N(b-a)$.

Now since $f$ is Riemann integrable, by definition, $\int_a^b f = L(f,P)=U(f,P)$.

So $-N(b-a) \le \int_a^b f \le N(b-a)$

$\implies |\int_a^b f| \le N(b-a)$.

I can't seem to get any further from here.

Edit: The definition I'm using is that $f$ is Riemann integrable on $[a,b]$ if $U(f,P)=L(f,P)$ where $$U(f,P)= \sum_{i=0}^n=M_i(x_i-x_{i-1})$$ and $$L(f,P)= \sum_{i=0}^n=m_i(x_i-x_{i-1})$$

Thanks.

Answer 1

For any partition $\;P=\{x_0:=a\,,\,x_1\,...\,x_n=b\;$ of $\;[a,b]\;$ , you have that for any $\;c_i\in[x_i,\,x_{i+1}]\;$

$$\sum_{k=0}^{n-1}|f(c_i)|(x_{i+1}-x_i)\le N\sum_{k=0}^{n-1}(x_{i+1}-x_i)=N(b-a)$$