# Thread: prob

1. ## prob

I don't know what to do with this. Help please.

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

t=0, r(t)=2
t=1, r(t)=8
t=2, r(t)=26
t=3, r(t)=32
t=4, r(t)=46
t=5, r(t)=52
t=6, r(t)=60

(a) Give upper and lower estimates for the total quantity Q(6) of erupted materials after 6 seconds.
Q(6) = ? tonnes (lower estimate)
Q(6) = ? tonnes (upper estimate)

(b) Use the Midpoint Rule to estimate Q(6).
Q(6) = ? tonnes

2. ## try

In this problem we are given several values of the function r(t). Q(t) is simply the integral of r(t) - that is, the area under the graph. When evaluating Q(6) using the midpoint rule means:
Q(6)~Q((6-0)/2)*6.
a better evaluation can be achieved by evaluation the area between t(n) and t(n+1) using the trapezoidal rule and then summing.

a lower estimate can be achieved in several ways, two of which:
(a) using the overall minimum (2) and calculation minimum*time
(b) a more realistic lower bound can be achieved by doing (a) for every segment of the form [n,n+1] (thus using 52 as the minimum in [5,6] ) and summing.

an upper estimate is achieved similarly.