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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?
 
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4 Answers

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

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Answer updated to account for new information.

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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
 
Answer: $6.34.
Dear Bella,
 
July has 31 days, so, if I'm reading the problem correctly
 
B(t) = (5.5)t2 - 300t + 5000
 
Bella, I'm afraid there might be something wrong with this equation.  As written, it predicts that our poor account holder will be losing money.  For 0 > t ≤ 54.5455, 300*t will be greater than (5.5)*t2 and the $5000 in the account will be diminished.  Not until day 55 arrives will the (5.5)t2 term be larger than the (300)*t term.
 
Something doesn't flush her, Bella.  Have a look at that original equation again and see if you entered it correctly.

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