Oil leaked from a tank at a rate of r(t) liters per hour. the rate decreased as time passed, and values of the rate at two hour time intervals are shown in the table. find lower and upper estimates for the total amount of oil that leaked out. v ? l (lower estimate) v ? l (upper estimate) t (h) 0 2 4 6 8 10 r(t) (l/h) 8.6 7.9 6.8 6.4 5.7 5.3

The given data is
t, h: 0 2 4 6 8 10
r(t), L/h: 8.6 7.9 6.8 6.4 5.7 5.3

The lower and upper estimates for the total amount that leaked may be computed as the Left and Right Riemann sums.

The shape of the graph of r versus will determine which of the two sums yields an upper or lower sum.
The plot of the graph is shown below.

The Left Riemann sum is
Sl = 2*(8.6+7.9+6.8+6.4+5.7) = 70.8 L

The Right Riemann sum is
Sr = 2*(7.9+6.8+6.4+5.7+5.3) = 64.2 L

Answer:
The lower estimate for oil leakage is 64.2 L
The upper estimate for oil leakage is 70.8 L

tifanihayyu tifanihayyu · Answer 2 · 2020-02-02T20:37:07+01:00

Tower and upper estimates for the total amount of oil that leaked out is 63.4 L and 80 .6 L

Explanation:

Oil leaked from a tank at a rate of [tex]r(t)[/tex] liters per hour. the rate decreased as time passed, and values of the rate at two hour time intervals are shown in the table. find lower and upper estimates for the total amount of oil that leaked out!

The volumetric flow rate is the fluid volume which passes per unit time. It is usually represented by the symbol Q. It is a quantity that represents the variation of fluid volume of concerning time is known as the volume flow rate.

Data :

The oil flow rate at t=0 is r1 = 8.6 L/h
The oil flow rate at t=2 is r2 = 7.9 L/h
The oil flow rate at t=4 is r3 = 6.7 L/h
The oil flow rate at t=6 is r4 = 6.4 L/h
The oil flow rate at t=8 is r5= 5.6 L/h
The oil flow rate at t=10 is r6 = 5.1 L/h
The width of interval N = 2 h

The lower estimate of oil : [tex]V_l = N (r_2 + r_3 + r_4+r_5+r_6)[/tex]

[tex]V_l = 2h * (7.9+6.7+6.4+5.6+5.1) L/h = 63.4 L[/tex]

Whereas the upper estimate of oil: [tex]V_u = N (r_1 + r_2 + r_3 + r_4+r_5+r_6)[/tex]

[tex]V_l = 2h * (8.6+7.9+6.7+6.4+5.6+5.1) L/h = 80.6 L[/tex]

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