Calculus anti derivative problem?

Question

The projected rate of increase in enrollment at a new college is estimated by: 

dE/dt=7,000(t+16)^(-3/2) 

where E(t) is the projected enrollment in t years. If the enrollment is 1,000 now (t=0), find the projected enrollment 9 years from now.

Answer 1

The projected rate of increase in enrollment at a new college is : dE / dt = 7000(t + 16)^{(- 3/2)}.

⇒ dE = 7000(t + 16)^{(- 3/2)} * dt

Let, u = t + 16

du = dt.

⇒ dE = 7000(u)^{(- 3/2)} * du

ʃ dE = ʃ [7000(u)^{(- 3/2)} * du]

E = 7000ʃ [(u)^{(- 3/2)} du]

Apply formula : ʃ xⁿ dx = x^{n + 1} /(n + 1) + C.

E = 7000[u^{(- 3/2 + 1)}] / (- 3/2 + 1)] + C

E = 7000[u^{(- 1/2)}] / (- 1/2)] + C

E = - 2 * 7000 * u^{(- 1/2)} + C

E = - 14000 * u^{(- 1/2)} + C

Put, u = t + 16.

E = - 14000(t + 16)^{(- 1/2)} + C.

At, t = 0, E = 1000.

- 14000(0 + 16)^{(- 1/2)} + C = 1000

C = 1000 + 14000(16)^{(- 1/2)}

C = 1000 + 14000(4)^{(- 1)}

C = 1000 + (14000/4)

C = 1000 + 3500

C = 4500.

Therefore, E = - 14000(t + 16)^{(- 1/2)} + 4500.

At, t =  9,

E = - 14000(9 + 16)^{(- 1/2)} + 4500

E = - 14000(25)^{(- 1/2)} + 4500

E = - 14000(5)^{(- 1)} + 4500

E = (- 14000/5) + 4500

E = - 2800 + 4500

E = 1700.

Therefore, the projected enrollment 9 years from now is 1700.