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. t (h) 0 2 4 6 8 10
r(t) (L/h) 8.8 7.6 6.8 6.2 5.7 5.3

V=_ upper estimate
V= _lower estimate

The exact amount of oil that leaks out for 0 ≤ t ≤ 10 is given by the integral,

Then the upper and lower estimates of this integral correspond to the upper and lower Riemann/Darboux sums. Since r(t) is said to be decreasing, this means that the upper estimate corresponds to the left-endpoint Riemann sum, while the lower estimate would correspond to the right-endpoint sum.

So you have

• upper estimate:

(8.8 L/h) (2 h - 0 h) + (7.6 L/h) (4 h - 2h) + (6.8 L/h) (6 h - 4h) + (6.2 L/h) (8 h - 6h) + (5.7 L/h) (10 h - 8 h)

= (2 h) (8.8 + 7.6 + 6.8 + 6.2 + 5.7) L/h)

= 70.2 L

• lower estimate:

(7.6 L/h) (2 h - 0 h) + (6.8 L/h) (4 h - 2h) + (6.2 L/h) (6 h - 4h) + (5.7 L/h) (8 h - 6h) + (5.3 L/h) (10 h - 8 h)

= (2 h) (7.6 + 6.8 + 6.2 + 5.7 + 5.3) L/h)

= 63.2 L

$2416.6

step-by-step explanation:

per hour = $ 25

total work time = 35

commission = 15%

total she has to make = $ 4,500

let she sold x dollar products

so her commission will be 15 % of x

now

total = weekly basic pay + commission

4500= (25*35) + 15 % of x

4500=875 + * x

4500-875 =   * x

3625= * x

multiplying wwith 100 on both sides gives

36250 = 15 x

now

x =  

x= $2416.6

can you post the graph?