Show that $\int_a^b xdx = \frac{b^2-a^2}{2}$ using upper and lower Riemann sums

Here's my work so far:

Let $f(x) = x$ and suppose that $P = \{x_0, x_1, x_2,...,x_n\}$ is the partition for interval $[a,b]$, creating $n$ number of subintervals with equal lengths.

$L_f(P)$ denotes the lower Riemann sum associated with partition P and $U_f(P)$ denotes the upper Riemann sum associated with partition P. So from what I know, $U_f(P)$ will always be greater than or equal to the true definite integral. Likewise, $L_f(P)$ will always be less than or equal to the true definite integral. Like so...

$$L_f(P) \leq \int_a^b xdx \leq U_f(P)$$

To calculate $L_f(P)$, for each integer $i$ in $[1,n]$, let $m_i = \textrm{min}\{f(x) \,|\, x \ \textrm{in} \ [x_{i-1}, x_i]\}$. Then...

\begin{align} L_f(P) &= \sum_{i=1}^n m_i(x_i - x_{i-1}) = (\frac{b-a}{n})(a+ ...+m_{n}) \end{align}

how can I continue? My goal is to show the upper and lower sums are equal.

1 Answer

You are close to being done. Note that $x_i = a + \frac{i}{n}(b-a)$, so\begin{align*} \sum_{i=1}^{n}m_i &= \sum_{i=0}^{n-1}(a+ \frac{i}{n}(b-a)) \\ &= na + \frac{b-a}{n}\sum_{i=0}^{n-1} i \\ &= na + \frac{b-a}{n}\frac{n(n-1)}{2} \end{align*}The lower Riemann sum then becomes\begin{align*} L_f(P)=\frac{b-a}{n}\cdot\sum_{i=1}^nm_i &= a(b-a) + \frac{n-1}{n} \frac{(b-a)^2}{2} \end{align*}A very similar computation gives us that$$U_f(P) = a(b-a)+\frac{n+1}{n} \frac{(b-a)^2}{2}.$$Letting $n$ approach $\infty$ we can see that we get the same limit namely\begin{align*} a(b-a) + \frac{(b-a)^2}{2} &= \frac{2ab - 2a^2 + b^2 + a^2 - 2ab}{2} = \frac{b^2 - a^2}{2} \end{align*}