Question

To calculate the volume of the log using the midpoint rule.

Short answer

The volume of the \(\log \) is \(5.80\;{{\rm{m}}^3}\).

Step-by-step solution

Given data

The log is cut at \(1\;{\rm{m}}\) intervals.

The log is \(10\;{\rm{m}}\) long.

The end points and cross sectional areas of the logs are as follows:

\(x(m)\)	\(0\)	\(1\)	\(2\)	\(3\)	\(4\)	5	\(6\)	\(7\)	\(8\)
\(A\)	\(0.68\)	\(0.65\)	\(0.64\)	\(0.61\)	\(0.58\)	\(0.53\)	\(0.55\)	\(0.52\)	\(0.50\)

The number of intervals as \(n = 5\).

The midpoints are \(1\;{\rm{m}},3\;{\rm{m}},5\;{\rm{m}},7\;{\rm{m}}\), and \(9\;{\rm{m}}\).

Concept used of Midpoint Rule

Midpoint Rule

\(\int_a^b f (x)dx \approx \sum\limits_{i = 1}^n f \left( {{{\bar x}_i}} \right)\Delta x = \Delta x\left( {f\left( {{{\bar x}_1}} \right) + \ldots \ldots + f\left( {{{\bar x}_n}} \right)} \right)\)

Simplify the function

The expression to find the volume as shown below.

\(V = \int_a^b A (x)dx\) …..(1)

The expression to find the volume of the log uses midpoint rule as shown below.

\(\int_a^b A (x)dx = \frac{{b - a}}{n}(A(1) + A(3) + A(5) + A(7) + A(9))\) ….(2)

Substitute 0 for \(a,10\;{\rm{m}}\) for b, 5 for \(n,0.65\;{{\rm{m}}^2}\) for \(A(1),0.61\;{{\rm{m}}^2}\) for \(A(3),0.59\;{{\rm{m}}^2}\) for \(A(5),0.55\;{{\rm{m}}^2}\) for \(A\) (7), and \(0.50\;{{\rm{m}}^2}\) for \(A(9)\) in Equation (2).

\(\begin{aligned}{c}\int_a^b A (x)dx = \frac{{10 - 0}}{5}(0.65 + 0.61 + 0.59 + 0.55 + 0.50)\\ = 2 \times 2.90\\ = 5.80\;{{\rm{m}}^3}\end{aligned}\)

Therefore, the volume of the \(\log \) is \(5.80\;{{\rm{m}}^3}\).