Expert: Socrates - 4/21/2008

Question
The rate at which water is sprayed on a field of vegetables is given by R(t)=2 times the square root of 1+5t cubed, where t is in minutes and R(t) is in gallons per minute. During the time interval 0 is less than or equal to t is less than or equal to 4, what is the average rate of water flow, in gallons per minute?

Answer
The average value of the function is the integral divided by the length of the interval.

We want (2)(1/4) S (1+5t)^3/2 dt = (1/2) S (1+5t)^3/2 dt

Change variables, let x = 1 + 5t , dx/dt = 5 , dt = (1/5)dx

(1/2) S (1+5t)^3/2 dt = (1/2) S x^3/2 (1/5)dx = (1/10) S x^3/2 dx

An anti derivative for x^3/2 is (2/5)x^5/2 . Since x = 1 + 5t , letting t = 0 and t = 4 ,  the integral in x will be from 1 to 21.

We then want the difference of (1/25)x^5/2 at 21 and 1 .

This gives (1/25)(21^5/2 - 1) as the average rate of flow , or approximately 80.8 gallons per minute