Question

The rate of growth of a microbe population is given by m’(t)=12te^(2t), where t is time in days. What is the total accumulated growth during the first 2 days?

Answer 1

To find the total accumulated growth during the first 2 days, we need to integrate the rate of growth function m'(t) from 0 to 2:

∫12te^(2t) dt

To solve this integral, we can use integration by parts. Let u = t and dv = 12e^(2t) dt. Then, du = dt and v = 6e^(2t).

Using the formula for integration by parts:

∫udv = uv - ∫vdu

we have:

12te^(2t) = 6te^(2t) - ∫6e^(2t) dt

Integrating the second term:

∫6e^(2t) dt = 3e^(2t)

Therefore, our integral becomes:

∫12te^(2t) dt = 6te^(2t) - 3e^(2t) + C

Now, we can evaluate the integral from 0 to 2:

[6(2)e^(2*2) - 3e^(2*2)] - [6(0)e^(2*0) - 3e^(2*0)]

= [12e^4 - 3e^4] - [0 - 3]

= 9e^4 - 3

So, the total accumulated growth during the first 2 days is 9e^4 - 3 ≈ 657.41 units.