Q 15. - Calculus

Question

Explain why the sum $\sum _{k=1}^{20}f\left(-3+\frac{k}{2}\right)(0.25)$ can't be a right sum for $f$ on $\left[a,b\right]=\left[-3,2\right]$

Step-by-Step Solution

Answer

The sum $\sum _{k=1}^{20}f\left(-3+\frac{k}{2}\right)(0.25)$ can't be a right sum as part of the sum $\sum _{k=1}^{20}f\left(-3+\frac{k}{2}\right)(0.25)$ indicates that $∆x=\frac{1}{2}$

Step 1. Given information

$\sum _{k=1}^{20}f\left(-3+\frac{k}{2}\right)(0.25)$ is the sum.

Step 2. Explanation

Consider the sum $\sum _{k=1}^{20}f\left(-3+\frac{k}{2}\right)(0.25)$

$\sum _{k=1}^{20}f\left(-3+\frac{k}{2}\right)(0.25)=f\left(-3+\frac{1}{2}\right)(0.25)+f\left(-3+\frac{2}{2}\right)(0.25)+f\left(-3+\frac{3}{2}\right)(0.25)+...$

Here, part of the sum denotes that $∆x=\frac{1}{2}$ and the other part indicates 0.25