Midpoint Riemann sum for water flow rate over 24 hours

[ricekrispie]

Homework Statement

t (hours) | R(t) (gallons per hour)
0 9,6
3 10,4
6 10,8
9 11,2
12 11,4
15 11,3
18 10,7
21 10,2
24 9,6​

The rate at which water flows out of a pipe, in gallons per hour, is given by a differentiable function R of time t. The table above shows the rate as measured every 3 hrs for a 24-hour period.

(a) Use a midpoint Riemann sum with 4 subdivisions of equal length to approximate $$\int_0^{24} R(t) dt$$. Using correct units, explain the meaning of your answer in terms of water flow.

(b) Is there some time t, 0 < t < 24, such that R´(t) = 0? Justify your answer.

(c) The rate of water flow R(t) can be approximated by Q(t) = (1/79)(768 + 23t - t^2).
Use Q(t) to approximate the average rate of water flow during the 24-hour time period. Indicate units of measure.


P.S.: THIS IS AN AP PROBLEM :(
P.S.2: SORRY, I DON´T HAVE A CLUE OF WHERE TO START

 

[NateTG]

For multi-character limits of integration, you need to use {}'s:
$$\int_0^{24}$$

The question is basically asking how much water has flowed out of the pipe.

Do you know what a Riemann sum is?

 

[ricekrispie]

no.. i have no idea

 

[NateTG]

no.. i have no idea

http://mathworld.wolfram.com/LowerSum.html
http://mathworld.wolfram.com/UpperSum.html

It's a method for approximating the area under a curve by rectangles.

You'll want to talk to your teacher/look at your notes before attempting part A then.

Part B doesn't depend no part A, so you can work on that in the meantime.

 

[NateTG]

no.. i have no idea

http://mathworld.wolfram.com/LowerSum.html
http://mathworld.wolfram.com/UpperSum.html

It's a method for approximating the area under a curve by rectangles.

You'll want to talk to your teacher/look at your notes before attempting part A then.

Part B doesn't depend on part A, so you can work on that in the meantime.