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?

## Interest-bearing account

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# 1 Answer

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)=7.5t

^{2}-300t+500 on [0,30] is given by

∫

_{0}

^{30}B(t) dt /30 = 2750.

0.2% interest on this balance is 2750*0.002 = $5.5.

## Comments

∫

_{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

∫

_{0}^{30}B(t) dt /30 = 2750.0.2% interest on this balance is 2750*0.002 = $5.5.

Comment