Q. 24 - 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, $∆x=\frac{b-a}{n}, \mathrm{and} {\mathrm{k}}_{\mathrm{x}}=\mathrm{a}+\mathrm{k}∆\mathrm{x}.$

trapezoid sum.

Step-by-Step Solution

Answer

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

Step 1. Given information.

The given type of Riemann sum is the trapezoid sum.

Step 2. trapezoid sum

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

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

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.

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