In my math course I was told the following
Let $f$ be a function defined on $[a;b] \subset \mathbb{R}$
We partition this interval in $n$ smaller intervals :
$$[x_0,x_1],[x_1,x_2]...[x_i,x_{i+1}],[x_{n-1},x_n]$$
we define $h$ the length of these intervals : $$h=\frac{b-a}{n}$$
then $$x_n=a+nh$$
It is also clear that when $n$ tends to $+ \infty$, $h$ tends to 0
$hf(x_i)$
is the area of a rectangle under a curve drawn above the text
We define $S_n=\sum_{i=0}^{n-1} h f(x_i)$$
then we say that $f$ is integrable if
$$\exists ! l=\lim_{n\to \infty} S_n \in \mathbb{R}$$
and we represent this as $$l=\int_a^b f(x) dx $$
We admitted that
1. all continuous functions are integrable
2. $f\geq g \implies \int_a^b f(x) dx \geq \int_a^b g(x) dx \, f,g $ are integrable functions on this interval
3. $\int_a^b f(x) dx + \int_b^c f(x) dx = \int_a^c f(x) dx $
Can anyone help me proving these properties ?
I'm stuck after the epsilon definition of the limit.
Thanks
Tom
Let $f$ and $g {\in}R[a,b]$ .Let $p$ be a partition such that $P=[X0,X1],.......[Xn-1,Xn]$. Suppose $X_0=a$ and $X_n=b$.let $f$ and $g$ continuous and $f≥g$ then $g(x)-F(x)≤0$ so $g(x)-f(x) \in R[a, b]$ then ${\int_a^b(g(x)-f(x)}={\int_a^{X1}(g(x)-f(x))+\int_X1^{X2}(g(x)-f(x))...........\int_Xn-1^b(g(x)) -f(x))} ≤ 0(b-a)$ so $\int_{a}^{X_1}(g) + \int_{x_1}^{x_2}(g)+.......+ \int_{x_n-1}^{b}(g) - \int_{a}^{X_1}(f) \int_{x_1}^{x_2}(f)+........+ \int_{x_n-1}^{b}(f)) \le 0$ so ${\int _a^b(g) -\int_a^b(f) ≤0}$