use the midpoint of each subinterval for the value of each ck to find the riemann sum s5 for the following…

Explanation:

Step1: Calculate the width of sub - intervals

The formula for the width $\Delta x$ of sub - intervals is $\Delta x=\frac{b - a}{n}$. Here, $a=-8$, $b = 2$, and $n = 5$. So, $\Delta x=\frac{2-(-8)}{5}=\frac{10}{5}=2$.

Step2: Determine the sub - intervals and mid - points

The sub - intervals are $[-8,-6],[-6,-4],[-4,-2],[-2,0],[0,2]$. The mid - points $c_k$ of these sub - intervals are: For $[-8,-6]$, $c_1=\frac{-8+( - 6)}{2}=-7$; For $[-6,-4]$, $c_2=\frac{-6+( - 4)}{2}=-5$; For $[-4,-2]$, $c_3=\frac{-4+( - 2)}{2}=-3$; For $[-2,0]$, $c_4=\frac{-2 + 0}{2}=-1$; For $[0,2]$, $c_5=\frac{0 + 2}{2}=1$.

Step3: Calculate the Riemann sum

The Riemann sum $S_5=\sum_{k = 1}^{5}f(c_k)\Delta x$. $f(c_1)=\frac{1}{-7 + 9}=\frac{1}{2}$, $f(c_2)=\frac{1}{-5 + 9}=\frac{1}{4}$, $f(c_3)=\frac{1}{-3+9}=\frac{1}{6}$, $f(c_4)=\frac{1}{-1 + 9}=\frac{1}{8}$, $f(c_5)=\frac{1}{1+9}=\frac{1}{10}$. $S_5=\Delta x\left(f(c_1)+f(c_2)+f(c_3)+f(c_4)+f(c_5)\right)$ $S_5 = 2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}\right)$ $S_5=2\left(\frac{60 + 30+20 + 15+12}{120}\right)$ $S_5=2\times\frac{137}{120}=\frac{137}{60}\approx2.28$.

Answer:

$2.28$