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 width of each sub - interval $\Delta x=\frac{b - a}{n}$. Here, $a=-5$, $b = 0$, and $n = 5$. So, $\Delta x=\frac{0-(-5)}{5}=1$.

Step2: Determine the sub - intervals and mid - points

The sub - intervals are $[-5,-4],[-4,-3],[-3,-2],[-2,-1],[-1,0]$. The mid - points $c_k$ are: $c_1=-4.5$, $c_2=-3.5$, $c_3=-2.5$, $c_4=-1.5$, $c_5=-0.5$.

Step3: Calculate $f(c_k)$ for each $k$

For $k = 1$: $f(c_1)=\sqrt{5-(-4.5)}=\sqrt{9.5}\approx3.082$. For $k = 2$: $f(c_2)=\sqrt{5-(-3.5)}=\sqrt{8.5}\approx2.915$. For $k = 3$: $f(c_3)=\sqrt{5-(-2.5)}=\sqrt{7.5}\approx2.739$. For $k = 4$: $f(c_4)=\sqrt{5-(-1.5)}=\sqrt{6.5}\approx2.550$. For $k = 5$: $f(c_5)=\sqrt{5-(-0.5)}=\sqrt{5.5}\approx2.345$.

Step4: Calculate the Riemann sum $S_5$

The Riemann sum $S_5=\sum_{k = 1}^{5}f(c_k)\Delta x$. Since $\Delta x = 1$, $S_5=f(c_1)+f(c_2)+f(c_3)+f(c_4)+f(c_5)$. $S_5\approx3.082 + 2.915+2.739+2.550+2.345=13.63$.

Answer:

$13.63$