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Net change in population via definite integral

A population of bacteria in a culture is growing at a rate of N′(t)=500(1+t)2N'(t) = \dfrac{500}{(1+t)^2} cells per hour, where tt is measured in hours after the culture was started.

Use the Net Change Theorem to find the total change in the bacteria population over the time interval [0,8][0,8]. That is, evaluate ∫08500(1+t)2 dt.\int_0^8 \frac{500}{(1+t)^2}\,dt.

Hint

Use uu-substitution with u=1+tu = 1+t, so du=dtdu = dt. Rewrite the integral in terms of uu, adjust the bounds, and then apply the Net Change Theorem (FTC Part 2) to evaluate.

Step-by-step solution

Let u=1+tu = 1+t, so du=dtdu = dt. When t=0t=0, u=1u=1; when t=8t=8, u=9u=9. The integral becomes ∫19500u2 du=500∫19u−2 du.\int_1^9 \frac{500}{u^2}\,du = 500\int_1^9 u^{-2}\,du.

An antiderivative of u−2u^{-2} is −u−1-u^{-1}, so 500∫19u−2 du=500[−1u]19=500(−19+1)=500⋅89=40009500\int_1^9 u^{-2}\,du = 500\left[-\frac{1}{u}\right]_1^9 = 500\left(-\frac{1}{9} + 1\right) = 500\cdot\frac{8}{9} = \frac{4000}{9}. By the Net Change Theorem, this represents the total change in population, N(8)−N(0)=40009≈444.4N(8)-N(0) = \frac{4000}{9} \approx 444.4 cells.

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