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

answer: d. $20

step-by-step explanation:

just took the test

answer: c, the green line

step-by-step explanation:

there is no black line, eliminate b.

simple interest implies periodic payments, but the red line shows no payments, eliminate d.

simple interest has equal payments. if paid say every day, the graph is about a straight line (actually a stairstep).

the yellow line is not straight, eliminate a.

check: the green line is straight, answer is c.

simple interest with positive interest rate gives a straight line sloping up to the right. green line:

g(t) = $100 + ($200-$100) ÷ (20 years) × (t years)

= $100 + (4%/year) × ($100)(t years)

incidentally, the yellow line curving upward represents continuously compounded interest

y = $100 × 2^(t years / (14 assuming the yellow line hits $200 at 14 years. 2^0 is 1, so y(0) = $100; 2^(14/14) is 2, so y(14) = $200.