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maria [59]
3 years ago
14

A population of insects increases at the rate of 200 10t 13t2 what is the change in the population of insects between day 0 and

day 3
Mathematics
1 answer:
RUDIKE [14]3 years ago
4 0

Answer:

762 days

Step-by-step explanation:

Given

Rate = 200 + 10t + 13t^2

Let the rate be R.

So the rate change with time is represented as:

\frac{dR}{dt} = 200 +10t + 13t^2

So:

dR = (200 + 10t + 13t^2)\ dt

To get the number of insects between day 0 and day 3, we need to integrate dR and set the bounds to 0 and 3

i.e.

dR = (200 + 10t + 13t^2)\ dt becomes

\int\limits^3_0 {dR} \, =\int\limits^3_0 (200 + 10t + 13t^2)\ dt

R =\int\limits^3_0 (200 + 10t + 13t^2)\ dt

Integrate

R = 200t + \frac{10t^2}{2} + \frac{13t^3}{3} [3,0]

Solve for R by substituting 0 and 3 for t

R = (200*3 + \frac{10*3^2}{2} + \frac{13*3^3}{3}) - ( 200*0 + \frac{10*0^2}{2} + \frac{13*0^3}{3})

R = (200*3 + \frac{10*9}{2} + \frac{13*27}{3}) - (0 + \frac{10}{2} + \frac{0}{3})

R = (200*3 + 5*9 + 13*9) - (0)

R = 600 + 45 + 117 - 0

R =762

<em>The population of insect between the required interval is 762</em>

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