evaluate the $\int_{0}^1 x\,dx$ using the definition

Question

From the definition -"$f$" is integrable on [a,b] if there exists a number $A$ so that for any $\epsilon > 0$ there exists a $\delta > 0$ such that if the sequence ${X_i}$ from $i=0,n$ is a partition then the corresponding Riemann Sum "S" satisfies absolute value $|S - A | < \epsilon$

First I let $f(x) = x$

Then I let $a = 0 < x_1 < x_2 <......< x_{n-1} < x_n = b$ be the partition of $[0,1]$.

From there I found the Reimman Sum "S" to be equal to $x(x_n - x_0)$

I got this integral equals "$x$" which is wrong.

Can someone please give me some insight into what I am missing/ guidance>?

Answer 1

The left-sided Riemann sum $S$ for $x$ on $[0,1]$ is expressed as

$$S=\lim_{N\to \infty}\sum_{n=0}^{N-1} x_n(x_{n+1}-x_n) \tag 1$$

where $x_0=0$ and $x_N=1$. Using Summation by Parts in $(1)$, we find

$$S=\lim_{N\to \infty}\left(x_N^2-x_0^2-\sum_{n=0}^{N-1}x_{n+1}(x_{n+1}-x_n)\right) \tag 2$$

It is straightforward to show that the sum on the right-hand side of $(2)$, which is the right-handed Riemann sum, approaches $S$ as $N\to \infty$. Therefore, we have

$$\begin{align} S&=\lim_{N\to \infty}\left(x_N^2-x_0^2\right)-S\\\\ 2S&=1\\\\ S&=\frac12 \end{align}$$

as expected!

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NOTE:

Here we show that the left-sided and right-sided Riemann sums are equal. To see that $S=\lim_{N\to \infty }\sum_{n=0}^{N-1}x_{n+1}(x_{n+1}-x_n)$ we form the difference

$$\begin{align} \sum_{n=0}^{N-1}x_{n+1}(x_{n+1}-x_n)-\sum_{n=0}^{N-1}x_{n}(x_{n+1}-x_n)&=\sum_{n=0}^{N-1}(x_{n+1}-x_n)^2\\\\ &\le \max_{0\le n\le N-1}\left(x_{n+1}-x_n\right)\sum_{n=0}^{N-1}(x_{n+1}-x_n)\\\\ &=\max_{0\le n\le N-1}\left(x_{n+1}-x_n\right)(x_N-x_0)\\\\ &\to 0\,\,\text{as}\,\,N\to \infty \end{align}$$

as was to be shown!

Answer 2

lets make the partition equally spaced.

$x_i = \frac in$ if you really wanted to you could say $x_i = \frac {(b-a)i}n+a$ but since b = 1 and a = 0, it really isn't necessary.

$\int_0^1 x \,dx = \lim_\limits{n\to\infty} \frac 1n\sum_\limits{i=1}^n x_i = \lim_\limits{n\to\infty} \frac {\frac12 (n)(n+1)}{n^2} $

Now, it is usually sufficient to say at this time.

$\lim_\limits{n\to\infty} \frac {\frac12 (n)(n+1)}{n^2} = \frac 12$

But since you raised the question...if you wanted to take that back to its definition.

$\forall \epsilon>0,\exists N>0$ such that $n>N \implies |\frac{\frac12 (n)(n+1)}{n^2} - \frac 12| < \epsilon$

$|\frac{\frac12 (n)(n+1)}{n^2} - \frac 12| = |\frac{1}{2n}|$

Let $N = \frac1{2\epsilon}, n>N\implies |\frac{\frac12 (n)(n+1)}{n^2} - \frac 12| < \epsilon$