An animal population is increasing at a rate of 13 51t13 51t per year (where t is measured in years). By how much does the animal population increase between the fourth and tenth years.

ΔP = 567

Step-by-step explanation:

The increasing rate of the population is  13,51*t.

That rate by definition is:

dP/dt   where P is the population therefore

dP/dt = 13,51*t

dt  =  13,51*t*dt

Integrating on both sides of the equation  we get:

∫dp = ∫ 13,51*t*dt

P =  13,51*t²/2 + K         ( K is population for t = 0 )

Now the population in 10 years    P(₁₀)

P(₁₀) =  13,51* (10)² /2  + K

P(₁₀) =  675,5  + K      (1)

And   P(₄)     is    

P(₄)  = 13,51*(4)²/2  * K

P(₄)  = 108,08 + K        (2)

Then substracting

P(₁₀) - P(₄)  =  ( 675,5  + K ) -  ( 108,08 + K )

ΔP =  567,42

But we don´t have fraction of animal, then

ΔP = 567

overpriced

step-by-step explanation:

no solution

step-by-step explanation:

there is not enough information in the problem to solve.