Who actually gets this problem?

Question

The rate of water usage for a business, in gallons per hour, is given by
W(t)=25te^(-t), where t is the number of hours since midnight. Find the average rate of water usage over the interval 0 is less than or equal to t and t is less than or equal to 6.

Answer 1

I get it ☺

Now *you* better get it together or you'll *never* pass that Calu-lost class.

The total water usage (call it G) at t hours is
integral from 0 to t of 25te^(-t) dt (Hint: Do it by parts ☺)
and the 'average' water usage is G/t.

Now get your little monkey αss moving and work the problem ☺


Doug

Answer 2

You should divide the total used water to the time which is 6 hours. To get the total used water, you have to calculate the integral of 25te^(-t) from 0 to 6.

∫25te^(-t) dt = -25[te^(-t) + e^(-t)]

so, ∫25te^(-t) dt from 0 to 6 is equall to [-25( 6e^(-6) + e^(-6) )] - [-25( 0e^(0) + e^(0) )] ≈ (-0.4338) - (-25) ≈ 24.5662

Finally you should divide this number which is the total used water to 6, the time.

24.5662 ÷ 6 ≈ 4.0943
its the avverage rate of water usage in this 6 hours.

Answer 3

Average value problem


1/(b-a) multiplied by the integral from a to b of W(t)

a=0, b=6 in this case

So compute the following:

(1/6) integral from 0 to 6 of 25t e^(-t) dt

Answer 4

W(6)=25*6e(-6)