Number of subscription per month is given by
S0(t) = "1000( 1+ 0.3 t )^{3\/2}" for 0 <= t <= 60
= 0 for every other value
Number of subscribers at any time will be given by
"\\int 1000(1 + 0.3t)^{3\/2} dt" = "(\\frac{4000}{3})((1+0.3t)^{\\frac{5}{2}}) + c"
where c is some fix number depending on the number of subscribers in the starting.
Number of subscriber in five years = "\\int_{0}^{60} 1000(1 + 0.3t)^{3\/2} dt" = 2096750 (approx)
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