Interest-bearing account

Question

Every month, a particular interest-bearing account earns 0.2 percent interest on the average balance for that month. The function B(t) =7.5t^2 -300t +5000 represents one investor's balance in this account during the month of November. B(t) gives the number of dollars in the account on day t, with t=0 being the beginning of the first day. How much interest will the account earn this month?

Answer 1

The average value of a function f(t) on an interval [a,b] is given by
∫_a^b f(t) dt /(b-a).
In your case, the average of B(t)=7.5t²-300t+500 on [0,30] is given by
∫₀³⁰ B(t) dt /30 = 2750.
0.2% interest on this balance is 2750*0.002 = $5.5.