Find Interest earned on 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 on the interval [a,b] is given by

 

1/(b-a) * int(f,a,b)

 

Thus, the average balance in the account in November is

 

1/(30-0) int(7.5t^2-300t+5000) = 1/30*82500 = 2750.

 

If the account earns 0.2% of the average balance, then the interest earned will be 

 

.0002(2750) = $5.50

Answer 2

Assuming we are dealing with a conventional 30 day month, we compute the average daily balance as

(1/30)∫₀³⁰(7.5t² -300t +5000 )dt=(1/30)(2.5t³-150t²+5000t) evaluated at t=30 (it's 0 at t=0)

=(2.5×30³-150×30²+5000×30)/30  The interest is just .002 times this number.

=.0002×(2.5×30³-150×30²+5000×30)/30=5.50