Question

The rate of growth of some bacteria is dN/dt = 800 + 200e^t. If n(5) = 40,000 find N(t). Integrate to 5 get 5 = 800t + 200e^t + C? Is that right?

Answer 1

First integrate the function with respect to t:
\[\int\limits_{}^{}800+200e^{t} dt = 800t+200e^{t} +c\]
Now we have to figure out what the constant is by plugging in t = 5 and N(t) = 40000:
\[40000 = 800(5) + 200e^{5} +c \Rightarrow c = 36000-200e^{5}\]

Answer 2

c = 6317.37 ish
So the final answer would be:
\[N(t) = 800t+200e^{t} +6317.37\]