A bacteria population is 6000 at time t = 0 and its rate of growth is 1000 · 9t bacteria per hour after t hours. what is the population after one hour? (round your answer to the nearest whole number.)

 dP/dt = 1000 * 9^t is the bacteria population's growth rate.where:P(t) is the population at time t, andP(0) = 6000. 
The population after one hour would be: 
P(1) = P(0) + <Integral of dP/dt from 0 to 1> 
P(1) = P(0) + [ 1000 * 9^1 / ln(9) - 1000 * 9^0 / ln(9) ] 
P(1) = 6000 + 3641 = 9461 would be the answer