Question

A population of honeybees increased at a rate of r(t) bees per week, where the graph of \({\rm{r}}\)is as shown. Use the Midpoint Rule with six subintervals to estimate the increase in the bee population during the first \({\rm{24}}\)weeks

Short answer

The resulted value by using midpoint formula is \({\rm{64000}}{\rm{.}}\)

Step-by-step solution

Graph of the function

The graph of the function \({\rm{f(x)}}\) can be found in the image below. This is the graph's estimated area, as determined by the rectangles.

Width of the Subintervals

With six rectangles on the range of\({\rm{x}}\)on \({\rm{(0,24),}}\) the width of the subintervalsis given by the formula:

\({\rm{\Delta x = }}\frac{{{\rm{b - a}}}}{{\rm{n}}}{\rm{ = }}\frac{{{\rm{24 - 0}}}}{{\rm{6}}}{\rm{ = 4}}\)

The following are the endpoints:

\(\begin{array}{*{20}{l}}{{x_0} = 0,}&{{x_1} = 4,}&{{x_2} = 8,}&{{x_3} = 12,}&{{x_4} = 16,}&{{x_5} = 20,}&{{x_6} = 24}&{}\end{array}\)

The midpoints are:

\(\begin{array}{*{20}{l}}{}&{x_1^* = \frac{{{x_0} + {x_1}}}{2} = 2}\\{}&{x_2^* = \frac{{{x_1} + {x_2}}}{2} = 6}\\{}&{x_3^* = \frac{{{x_2} + {x_3}}}{2} = 10}\\{}&{x_4^* = \frac{{{x_3} + {x_4}}}{2} = 14}\\{}&{x_5^* = \frac{{{x_4} + {x_5}}}{2} = 18}\\{}&{x_6^* = \frac{{{x_5} + {x_6}}}{2} = 22}\end{array}\)

Sum of the rectangles

The sum of the rectangles yields to

\(\begin{aligned}{c}\sum\limits_{i = 1}^6 f \left( {x_i^*} \right)\Delta x &\approx f\left( {x_1^*} \right)\Delta x + f\left( {x_2^*} \right)\Delta x + f\left( {x_3^*} \right)\Delta x + f\left( {x_4^*} \right)\Delta x + f\left( {x_5^*} \right)\Delta x + f\left( {x_6^*} \right)\Delta x\\ &= f(2)\Delta x + f(6)\Delta x + f(10)\Delta x + f(14)\Delta x + f(18)\Delta x + f(22)\Delta x\\ &= (f(2) + f(6) + f(10)) \times \Delta x + (f(14) + f(18) + f(22)) \times \Delta x\\ &= (0 + 1000 + 7000) \times 4 + (7000 + 1000 + 0) \times 4\\ &= 64000\end{aligned}\)

Therefore, the increase in bee population during first \({\rm{24}}\)weeks is \({\rm{64000}}{\rm{.}}\)