Q. 44 - Calculus

Question

In Exercises 39–44, write out the sigma notation for the Riemann sum described in such a way that the only letter which appears in the general term of the sum is k. Don’t calculate the value of the sum; just write it down in sigma notation.

$f\left(x\right)=\sqrt{1-x^2},\left[a,b\right]=\left[-1,1\right],midpoint sum,n=20$

Step-by-Step Solution

Answer

The midpoint sum is $\frac{1}{10}\sum _{k=1}^{20}\sqrt{1-{\left(\frac{4k-13}{20}\right)}^2}$

Step 1. Given Information

We have $f\left(x\right)=\sqrt{1-x^2},\left[a,b\right]=\left[-1,1\right],midpoint sum,n=20$

Step 2. Finding ∆ x and x k - 1

$∆x=\frac{b-a}{n}=\frac{1+1}{n}=\frac{2}{20}=\frac{1}{10}$

$$\begin{gathered} x_k=a+k∆x=-1+\frac{2k}{10} \\ {x}_{k-1}=a+\left(k-1\right)∆x=-1+\frac{2\left(k-1\right)}{10}=\frac{-1+2k-2}{10}=\frac{2k-3}{10} \end{gathered}$$

and $\frac{x_{k-1}+x_k}{2}=\frac{{\displaystyle \frac{2k-3}{10}}+{\displaystyle \frac{2k-10}{10}}}{2}=\frac{4k-13}{20}$

Step 3. Finding Reimann Sum

We have $f\left(x\right)=\sqrt{1-x^2}$

Riemann sum

$$\begin{gathered} =\sum _{k=1}^{20}f\left(\frac{x_{k-1}+x_k}{2}\right)∆x \\ =\sum _{k=1}^{20}f\left(\frac{4k-13}{20}\right)\frac{1}{10} \\ =\frac{1}{10}\sum _{k=1}^{20}f\left(\frac{4k-13}{20}\right) \\ =\frac{1}{10}\sum _{k=1}^{20}\sqrt{1-{\left(\frac{4k-13}{20}\right)}^2} \end{gathered}$$