Why are the Left and Right Riemann sums for $\int_a^b f(x)\ dx$ equal?

Question

• For finding the area under $\operatorname{f}\left(x\right) = x$ from $a$ to $b$,
  • if a right Riemann sum is $$ \lim_{n \to \infty} \frac{b - a}{n} \sum_{i = 1}^{n} \operatorname{f}\left(a + i\,\frac{b - a}{n}\right) \quad\mbox{and} $$
  • if a left Riemann sum is $$ \lim_{n \to \infty}\frac{b - a}{n} \sum_{i = 0}^{n - 1} \operatorname{f}\left(a + i\,\frac{b - a}{n}\right), $$

then why would they be equal $?$.

Is it because the limits as $n\to \infty$ make it so the summation's indexing doesn't matter when it moves down by one $?$.

Answer 1

Note that the difference between the two sums is $$(b-a)(f(b)-f(a))/n$$ which approaches zero as $n$ goes to infinity.

Thus the two limits are the same provided that they exist.

Answer 2

The difference between the two terms is

$$d_n=\frac{b-a}{n}(f(b)-f(a)).$$

As a function Riemann integrable is bounded, let say $\vert f(x) \vert \le M$ for $x \in [a,b]$, we get

$$\vert d_n \vert \le \frac{2M(b-a)}{n}$$ and

$$\lim\limits_{n \to \infty} d_n=0.$$

Indeed the two sums are converging to the Riemann integral $\int_a^b f$.