Integration word problem?

Question

Suppose that a volcano is erupting and readings of the rate r(t) at which solid materials are spwed into the atmosphere are given in table 1. The time measured in seconds and the units for r(t) are tonnes (metric tons) per second.

r(0) = 2
r(1) = 10
r(2) = 24
r(3) = 36
r(4) = 46
t(5) = 54
r(6) = 60

Give upper and lower estimates for the quantity Q(6) of erupted materials after 6 seconds. Justify in complete sentences how you know you have upper and lower estimates

And use midpoint rule to estimate Q(6)

This is what I did:

I know that ∫r(t)dt = Q(t)
that must mean Q(6) is the area under the curve of r(t) from 0 to 6. So how do I get the upper bound and lower bound?

For the midpoint rule this is what i did
Q(6) = ∫r(t)dt from 0 to 6 = [(2+10)+(10+24)+(24+36)+(36+46)+(46+54)+... = 201 tonnes

??????????

Answer 1

It looks like you've got most of it already.

Because the function is strictly increasing, the lower bound on the area will come from the left sum, the upper bound from the right sum.

Answer 2

First, one minor point: you didn't show the division by 2.

The idea behind the mid-point method is to use the average of the two endpoints. For example, for the interval between t = 0 and t = 1, which is one second, using R(0) would give 2 tons of stuff spewed out, using R(1) would give 10 tons of material spewed out, the mid-point method should give 6 tons of material ((2 + 10)/2).

Next, as the previous answer noted, your function is monotonically increasing - it goes up and never down. That means that using the starting value will give an under-estimate (for example the 2 for the interval [0,1]), using the ending value gives you an over-estimate (for example, 10 for that interval), while using the mid-point value gives you a more accurate estimate.

So I would use the starting values for your lower bound, the ending values for your upper bound and your midpoint estimate for your best guess.

Note that the middle values are the same for both so it is only the values at the end that change:

Is = r(0) + r(1) + ... + r(5)
Ie = r(1) + r(2) + ... + r(5) + r(6)
Im = (r(0)+r(1))/2 + (r(1)+r(2))/2 + ... + (r(5)+r(6))/2
= r(0)/2 + r(1)/2 + r(1)/2 + r(2)/2 + + r(5)/2 + r(5)/2 + r(6)/2
= r(0)/2 + r(1) + r(2) + ... + r(5) + r(6)/2