# 009B Sample Final 1, Problem 1

Suppose the speed of a bee is given in the table.

 Time (s) Speed (cm/s) ${\displaystyle 0.0}$ ${\displaystyle 125.0}$ ${\displaystyle 2.0}$ ${\displaystyle 118.0}$ ${\displaystyle 4.0}$ ${\displaystyle 116.0}$ ${\displaystyle 6.0}$ ${\displaystyle 112.0}$ ${\displaystyle 8.0}$ ${\displaystyle 120.0}$ ${\displaystyle 10.0}$ ${\displaystyle 113.0}$

(a) Using the given measurements, find the left-hand estimate for the distance the bee moved during this experiment.

(b) Using the given measurements, find the midpoint estimate for the distance the bee moved during this experiment.

Foundations:
1. The height of each rectangle in the left-hand Riemann sum is given by choosing
the left endpoints of each interval.
3. The height of each rectangle in the midpoint Riemann sum is given by
${\displaystyle {\frac {f(a)+f(b)}{2}}}$  where  ${\displaystyle a}$  is the left endpoint of the interval and  ${\displaystyle b}$  is the right endpoint of the interval.

Solution:

(a)

Step 1:
To estimate the distance the bee moved during this experiment,
we need to calculate the left-hand Riemann sum over the interval  ${\displaystyle [0,10].}$
Based on the information given in the table, we will have  ${\displaystyle 5}$  rectangles and
each rectangle will have width  ${\displaystyle 2.}$
Step 2:
Let  ${\displaystyle s(t)}$  be the speed of the bee during the experiment.
Then, the left-hand Riemann sum is

${\displaystyle {\begin{array}{rcl}\displaystyle {2(s(0)+s(2)+s(4)+s(6)+s(8))}&=&\displaystyle {2(125+118+116+112+120)}\\&&\\&=&\displaystyle {1182{\text{ cm}}.}\end{array}}}$

(b)

Step 1:
To estimate the distance the bee moved during this experiment,
we need to calculate the Riemann sum using the midpoint rule over the interval  ${\displaystyle [0,10].}$
Based on the information given in the table, we will have  ${\displaystyle 5}$  rectangles and
each rectangle will have width  ${\displaystyle 2.}$
Step 2:
Let  ${\displaystyle s(t)}$  be the speed of the bee during the experiment.
Then, the Riemann sum using the midpoint rule is

${\displaystyle {\begin{array}{rcl}\displaystyle {2{\bigg (}{\frac {s(0)+s(2)}{2}}+{\frac {s(2)+s(4)}{2}}+{\frac {s(4)+s(6)}{2}}+{\frac {s(6)+s(8)}{2}}+{\frac {s(8)+s(10)}{2}}{\bigg )}}&=&\displaystyle {1170{\text{ cm}}.}\end{array}}}$

(a)    ${\displaystyle 1182{\text{ cm}}}$
(b)    ${\displaystyle 1170{\text{ cm}}}$