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## Find the interest earned on an 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?

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.
Hi Bella,

Can I assume that you have calculus for this? In order to calculate an average, I would integrate B(t) with respect to t, then divide it by the total time (31 days for July  30 days for November).

Int(5.5t2-300t+5000)dt = (5.5/3)t3-150t2+5000t |310 = \$65,466.83

Int(7.5t2-300t+5000)dt = (7.5/3)t3-150t2+5000t |300 = \$82,500

Average daily balance = \$82,500/30 = \$2,750.

The interest is 0.2% on the average balance, or

(0.002)(\$2,750) = \$5.50

Answer updated to account for new information.
The interest will be 0.003*<B(t)>, where <> means average.

<B(t)>=1/31*∫031B(t)dt=1/31*[55/30*313-150*312+5000*31]≈2111.83

Interest compounded will be 0.003*2111.83≈6.34