Area approximations with sigma notation

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Remember the Riemann sum, written as: $f(x_1^*) \Delta x + f(x_2^*) \Delta x + \ldots + f(x_n^*) \Delta x$

The only change from one term to the next, is the subscript of the sample point. That subscript runs from $1$ to $n$ . That means we can write this Riemann sum in sigma notation as: $\sum _{k=1}^n f(x_k^*) \Delta x$

We found formulas for the sample points in certain cases.

The left Riemann sum is $\sum _{k=1}^n f(a + (k-1)\Delta x) \Delta x$

The right Riemann sum is $\sum _{k=1}^n f(a + k \Delta x) \Delta x$

The midpoint Riemann sum is $\sum _{k=1}^n f\left (a + \left (k-\frac {1}{2}\right )\Delta x\right ) \Delta x$

Approximate the area under the graph of $f(x) = x^2+1$ on the interval $[2,4]$ using $100$ rectangles and right-hand endpoints.

Let’s start by finding $\Delta x$ .

$\Delta x = \frac {b-a}{n} = \frac {1}{50}$ .

For right-hand endpoints, $x_k^* = a + k \Delta x$ . That is, $x_k^* = \answer {2} + k \answer {1/50}$ .

The Riemann sum is:

$\begin{align*} \sum_{k=1}^{100} f\left( x_k^* \right) \Delta x &=\sum_{k=1}^{100} f\left( 2 + \frac{k}{50} \right) \frac{1}{50}\\ &=\sum_{k=1}^{100}\left( \left( 2 + \frac{k}{50} \right)^2+1 \right)\frac{1}{50}\\ &= \sum_{k=1}^{100} \left( \frac{k^2}{50^2} + \frac{2}{25}k+5\right) \frac{1}{50}\\ &= \frac{1}{50}\sum_{k=1}^{100} \left( \frac{k^2}{50^2} + \frac{2}{25}k+5\right)\\ &= \frac{1}{50}\left(\frac{1}{50^2}\sum_{k=1}^{100} k^2 + \frac{2}{25}\sum_{k=1}^{100}k+ \sum_{k=1}^{100}5\right) \end{align*}$

We have formulas for calculating the individual sums here!

$\begin{align*} \sum_{k=1}^{100} k^2 &= \frac{(100)(100+1)(2\cdot100+1)}{6} \\ \\ &= \answer{338350}\\ \sum_{k=1}^{100} k &= \frac{(100)(100+1)}{2} \\ \\ &= \answer{5050}\\ \sum_{k=1}^{100} 5 &= 5 \cdot 100 \\ &= \answer{500} \end{align*}$

Then

$\begin{align*} \sum_{k=1}^{100} f\left( x_k^* \right) \Delta x &=\frac{1}{50}\left(\frac{1}{50^2}\sum_{k=1}^{100} k^2 + \frac{2}{25}\sum_{k=1}^{100}k+ \sum_{k=1}^{100}5\right) \\ &= \frac{1}{50}\left(\frac{1}{50^2}\answer{338350} + \frac{2}{25}\answer{5050}+ \answer{500}\right) \\ &= \answer{62/3} \end{align*}$

The process used in this example is the same for almost all of these approximations. Plug $\Delta x$ and $x_k^*$ into the Riemann sum formula and simplify, then use our summation formulas.