Q. 21 - Calculus

Question

Express each of the types of Riemann sums that follow in general sigma notation and also as an expanded sum. You may assume that f is a function defined on $[a,b],$n is a positive integer, $∆x=\frac{b-a}{n}, \mathrm{and} {\mathrm{k}}_{\mathrm{x}}=\mathrm{a}+\mathrm{k}∆\mathrm{x}.$

midpoint sum.

Step-by-Step Solution

Answer

The general sigma notation of midpoint sum and its expanded sum is following.
$\sum _{k=1}^{n}f\left(\frac{x_{k-1}-x_k}{2}\right)∆x=\sum _{k=1}^{n}f\left[\frac{\left(a+\left(k-1\right)\left(\frac{b-a}{n}\right)\right)-\left(a+k\left(\frac{b-a}{n}\right)\right)}{2}\right]\left(\frac{b-a}{n}\right)$

Step 1. Given information.

The given type of Riemann sum is the midpoint sum.

Step 2. the midpoint sum

Consider the function defined on the interval $\left[a,b\right]$and n is a positive integer.

Then The n-rectangle midpoint sum for f on $\left[a,b\right]$is $\sum _{k=1}^{n}f\left(\frac{{\stackrel{}{x}}_{k-2}-x_k}{2}\right)∆x.$

Where $∆\mathrm{x}=\frac{\mathrm{b}-\mathrm{a}}{2}, {\stackrel{*}{\mathrm{x}}}_{\mathrm{k}}=\mathrm{a}+\mathrm{k}∆\mathrm{x}, \& {\stackrel{*}{\mathrm{x}}}_{\mathrm{k}}\in [x_{\mathrm{k}-1},x_{\mathrm{k}}]$

then expanded sum will be the following.

$$\begin{gathered} \sum _{k=1}^{n}f\left(\frac{x_{k-1}-x_k}{2}\right)∆x=\sum _{k=1}^{n}f\left[\frac{\left(a+\left(k-1\right)∆x\right)-\left(a+k∆x\right)}{2}\right]∆x \\ \sum _{k=1}^{n}f\left(\frac{x_{k-1}-x_k}{2}\right)∆x=\sum _{k=1}^{n}f\left[\frac{\left(a+\left(k-1\right)\left(\frac{b-a}{n}\right)\right)-\left(a+k\left(\frac{b-a}{n}\right)\right)}{2}\right]\left(\frac{b-a}{n}\right) \end{gathered}$$