My question is about finding the total distance traveled, between 0 and 9, given the velocity function defined by the table below, using the trapezoidal approximation with 3 sub-intervals.
There seems to be two different logical ways to approach this, that end up with different values.
I know this means we have to find the value of :
$\int_{0}^{9} |v(t)| dt$
and graphically all we know for sure is shown below:
But there seems to be 2 ways to interpret this, depending on what we assume happens between (4,5) and (6,-6). I will consider the two cases as $v_1(t)$ and $v_2(t)$, and their graphical representations are shown below:
(notice both function fit the discrete points given in the table)
Algebraically, in the $v_1(t)$ case, we find the trapezoid areas using the negatives, and then take the absolute value. $\int_{0}^{9} |v_1(t)| dt \approx \left| \frac{1+5}{2} \cdot 4 \right| + \left| \frac{5+-6}{2} \cdot 2 \right| + \left| \frac{-6+-4}{2} \cdot 3 \right| =12+1+15=28 $
And, in the $v_2(t)$ case, first we take the absolute value of the table values, and then find the trapezoid areas. $\int_{0}^{9} |v_2(t)| dt \approx \frac{|1|+|5|}{2} \cdot 4 + \frac{|5|+|-6|}{2} \cdot 2 + \frac{|-6|+|-4|}{2} \cdot 3 =12+11+15=38 $
Both ways seem to have merit. Are both answers correct?
The first variant $v_1$, while intuitively more exact, does not conform to the task description of using only the 3 given intervals.
And as you found yourself, you can not be sure how the intervals are filled, so both values are in the range of possible values of the "true" integral. But then the other intervals could also be filled with highly accelerated velocity profiles.