# When to use indefinite & definite integration to solve a problem

In a certain country , the population is projected to grow at a rate of

$$P'(t) = 400(1+\frac{2t}{\sqrt{25+t^2}} )$$

People per year $t$ years from now. The current population is $60,000$ . What will be the population $5$ years from now ?

To solve this ,

Why can't I do definite integration ? From 0 to 5

Why have I do indefinite integration to find the Constant $C$ first , before being able to sub 5 years into $t$ to find the population $5$ years from now ?

• Because it is a continuous process. Then, integration is required. Is it clear for you ? – Claude Leibovici Aug 18 '17 at 7:42

Method 1:

Separating variables, and integrating both sides, we obtain: $$\int dP=\int 400\left(1+\frac{2t}{\sqrt{25+t^2}}\right)~dt$$ $$P=400(2\sqrt{t^2+25}+t)+C$$ Using the initial condition $P(0)=60000$ to solve for $C$, we obtain: $$60000=400(2\sqrt{25}+0)+C \implies C=56000$$ Therefore, we obtain the population at $t=5$ as: $$P(5)=400(2\sqrt{5^2+25}+5)+56000=2000(2\sqrt{2}+1)+56000 \approx 63656.8542$$

Method 2:

Setting the appropriate bounds and separating variables: $$\int_{60000}^{P(5)} dP=\int_0^5 400\left(1+\frac{2t}{\sqrt{25+t^2}}\right)~dt$$ Evaluating the integrals, we obtain: $$P(5)-60000=2000(2\sqrt{2}-1)$$ Solving for $P(5)$ gives us the population at $t=5$ years: \begin{align}P(5)&=2000(2\sqrt{2}-1)+60000\\&=2000(2\sqrt{2}+1)+56000 \end{align} The answer is the same as in Method 1.

$P(5)=P(0)+\int_0^5 P'(t) dt$ and $P(0)=60.000$.

The two methods are equivalent and both possible.

Let $P(t)$ be some antiderivative of $P'(t)$ (to a constant), and let $P_0,P_5$ be the populations now and in five years.

1) $P(t) = \int P'(t)\,dt+C$, then $P_0=P(0)+C$, $P_5=P(5)+C$ and by elimination of $C$, $P_5=P_0+P(5)-P(0)$.

2) $P_5=P_0+\int_0^5 P'(t)\,dt=P_0+\left.P(t)\right|_0^5$. (In the final expression, there is no need to introduce an integration constant, as it cancels out.)