Search 75,762 tutors
FIND TUTORS
Ask a question
0 0

Interest-bearing account

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?

Comments

The average value of a function f(t) on an interval [a,b] is given by
ab 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.

Comment

Tutors, please sign in to answer this question.

1 Answer


The average value of a function f(t) on an interval [a,b] is given by
ab 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.