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?

## Find Interest earned on interest-bearing account

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# 2 Answers

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

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

(1/30)∫

_{0}^{30}(7.5t^{2}-300t +5000 )dt=(1/30)(2.5t^{3}-150t^{2}+5000t) evaluated at t=30 (it's 0 at t=0)=(2.5×30

^{3}-150×30^{2}+5000×30)/30 The interest is just .002 times this number.=.0002×(2.5×30

^{3}-150×30^{2}+5000×30)/30=5.50
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