The answer is not quite the same but here's the process:
7 + (2 + 6) 2 ÷ 4 ⋅ (1/2)^4 A: 23 B: 18 C: 9 D: 8
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1. The four subintervals are [0, 2], [2, 3], [3, 7], and [7, 8]. We construct trapezoids with "heights" equal to the lengths of each subinterval - 2, 1, 4, and 1, respectively - and the average of the corresponding "bases" equal to the average of the values of at the endpoints of each subinterval. The sum is then
which is measured in units of gallons, hence representing the amount of water that flows into the tank.
2. Since is differentiable, the mean value theorem holds on any subinterval of its domain. Then for any interval
, it guarantees the existence of some
such that
Computing the difference quotient over each subinterval above gives values of 0.275, 0.3, 0.3, and 0.26. But just because these values are non-zero doesn't guarantee that there is definitely no such for which
. I would chalk this up to not having enough information.
3. gives the rate of water flow, and
, so that the average rate of water flow over [0, 8] is the average value of
, given by the integral
If doing this by hand, you can integrate by parts, setting
For the remaining integral, consider the trigonometric substitution , so that
. Then
or approximately 3.0904, measured in gallons per hour (because this is the average value of ).
4. By the fundamental theorem of calculus,
and is increasing whenever
. This happens over the interval (-2, 3), since
on [-2, 0), and
on [0, 3).
5. First, by additivity of the definite integral,
Over the interval [-2, 0), we have , and over the interval [0, 6],
. So the integral above is