Answer
$5.80\, m^{3}$
Work Step by Step
Step 1: State the midpoint rule.
For an interval [a,b], the midpoint rule can be used to determine the definite integral of a function where $width = \frac{b-a}{n}$.
Step 2: Solve for the midpoints.
In the given problem, the width is:
$width = \frac{10-0}{5} = 2$.
Dividing the interval into 5 subintervals, we have: [0,2],[2,4],[4,6],[6,8], and [8,10].
Solving for the midpoints at each subinterval:
Midpoint of [0,2] = $\frac{0+2}{2} = 1\,m$
Midpoint of [2,5] = $\frac{2+4}{2} = 3\,m$
Midpoint of [4,6] = $\frac{4+6}{2} = 5\,m$
Midpoint of [6,8] = $\frac{6+8}{2} = 7\,m$
Midpoint of [8,10] = $\frac{8+10}{2} = 9\,m$
Step 3: Use the data to determine the areas from each midpoint.
At x=1m, $A=0.65\,m^{2}$
At x=3m, $A=0.61\,m^{2}$
At x=5m, $A=0.59\,m^{2}$
At x=7m, $A=0.55\,m^{2}$
At x=9m, $A=0.50\,m^{2}$
Step 4: Estimate the volume using the values of the area above.
Volume = width(Summation of area)
Volume = $2m(0.65+0.61+0.59+0.55+0.50)m^{2}$
Volume = $5.80\, m^{3}$
