Question

A population of insects increases at a rate 230 + 8t + 0.9t2 insects per day (t in days). Find the insect population after 6 days, assuming that there are 50 insects at t = 0. (Round your answer to the nearest insect.)

Answer 1

Answer:

The insect population after 6 days is of 1639 insects.

Step-by-step explanation:

A population of insects increases at a rate 230 + 8t + 0.9t2 insects per day

This means that $r(t) = 230 + 8t + 0.9t^2$

The population of insects after x days is given by:

$P(t) = \int_{0}^{x}r(t)dt$

So

$P(x) = \int_{0}^{x} (230 + 8t + 0.9t^2)$

$P(x) = 230t + 4t^2 + 0.3t^3 + K|_{0}^{x}$

$P(x) = 230x + 4x^2 + 0.3x^3 + K$

In which K is the initital population(which is 50). So

$P(x) = 230x + 4x^2 + 0.3x^3 + 50$

After 6 days:

$P(6) = 230*6 + 4*6^2 + 0.3*6^3 + 50 = 1638.8$

Rounding to the nearest insect, 1639

The insect population after 6 days is of 1639 insects.