Approximation for Left and Right hand sums How do you find the difference of area for $L_n$-$R_n$?
I know the difference is: $L_n$-$R_n$ = $\frac{b-a}{n}(f(a)-f(b))$, BUT I'm still unable to understand how that is. Please help, thanks. $$\\$$
$L_n$ stands for left-hand sum
$R_n$ stands for right-hand sum
 A: Let the interval $[a,b]$ be divided into $n$ intervals of length $h = \frac{b-a}{n}$ using $n-1$ points $x_1, x_2, \ldots, x_{n-1}$ where $x_i = a + ih$. Let $x_0 = a$ and $x_n = b$. Then,
$$L_n = \sum_{i=1}^{n} \frac{b-a}{n}f(x_{i-1})$$
$$R_n = \sum_{i=1}^{n} \frac{b-a}{n}f(x_{i})$$
$$\implies L_n-R_n = \sum_{i=1}^{n} \frac{b-a}{n}(f(x_{i-1})-f(x_{i})) = \frac{b-a}{n} \sum_{i=1}^{n}(f(x_{i-1})-f(x_{i})) = \frac{b-a}{n}(f(x_0)-f(x_n)) = \frac{b-a}{n}(f(a)-f(b))$$
A: $$L_n = \sum_{j=0}^{n-1} \frac{b-a}{n}f(x_{j})$$
which can be written as: $L_n$= $\frac{b-a}{n}$[$f(x_0)$+$f(x_1)$+$f(x_2)$+$f(x_3)$+...+$f(x_{n-1})$]. $$\\$$
$$R_n = \sum_{j=1}^{n} \frac{b-a}{n}f(x_{j})$$
which can be written as: $R_n$= $\frac{b-a}{n}$[$f(x_1)$+$f(x_2)$+$f(x_3)$+...+$f(x_{n-1})$+$f(x_n)$]. $$\\$$
When you subtract $R_n$ from $L_n$ you can cancel out all the terms expect for the first term in $L_n$, the last term in $R_n$, and $\frac{b-a}{n}$. 
That will leave you with: $\frac{b-a}{n}$$(f(x_0)-f(x_n))$. Where $x_0$=$a$ and $x_n$=$b$.
Thus giving you: $\frac{b-a}{n}$$(f(a)-f(b))$
