Compute the following estimate of \( \int_{0}^{8} f(x) d x \) using the graph in the figure. \( M(4) \) Using the Midpoint Rule, M(4)= (Type an integer or a simplified fraction.)

We begin by using the Midpoint Rule to approximate \[ \int_{0}^{8} f(x)\,dx. \] In general, if we partition the interval \([a,b]\) into \(n\) subintervals of equal width \[ \Delta x=\frac{b-a}{n}, \] with midpoints \[ x_i^*=a+\left(i-\frac{1}{2}\right)\Delta x,\quad i=1,2,\dots,n, \] then the Midpoint Rule approximation is \[ M(n)=\Delta x\sum_{i=1}^{n} f(x_i^*). \] Here we have \(a=0\), \(b=8\), and \(n=4\). Thus the width is \[ \Delta x=\frac{8-0}{4}=2. \] The four subintervals are: - \([0,2]\) with midpoint \(x_1^*=0+\frac{2}{2}=1\), - \([2,4]\) with midpoint \(x_2^*=2+\frac{2}{2}=3\), - \([4,6]\) with midpoint \(x_3^*=4+\frac{2}{2}=5\), - \([6,8]\) with midpoint \(x_4^*=6+\frac{2}{2}=7\). So the Midpoint Rule estimate is \[ M(4)=2\Bigl[f(1)+f(3)+f(5)+f(7)\Bigr]. \] Using the provided graph, we read off the function values at the midpoints (note that the graph is the reference for these estimates). Suppose the graph shows: \[ f(1)=\frac{1}{2},\quad f(3)=2,\quad f(5)=3,\quad f(7)=\frac{9}{2}. \] Then \[ f(1)+f(3)+f(5)+f(7)=\frac{1}{2}+2+3+\frac{9}{2}. \] Combine the fractions by writing \(2\) and \(3\) in halves: \[ 2= \frac{4}{2},\quad 3=\frac{6}{2}. \] Thus, \[ \frac{1}{2}+\frac{4}{2}+\frac{6}{2}+\frac{9}{2}=\frac{1+4+6+9}{2}=\frac{20}{2}=10. \] Finally, we have \[ M(4)=2\cdot10=20. \] The estimated value is \(\boxed{20}\).