Science:Math Exam Resources/Courses/MATH103/April 2015/Question 04/Solution 1

The most difficult part of this problem is setting up the integral.

Consider a region formed between a circle of radius $2$ and a circle of radius $2-dr$ where $dr$ is infinitesimally thin.

The population of ants inside this small region is given by

$b(r)\times Area=b(r)2\pi rdr$

We can get the total population by adding up the contributions of an infinite number of these small regions in our circle of radius $2$, or in other words, we can get the total population by integrating the above equation from $0$ to $2$.

Population = $\int _{0}^{2}{\frac {2\pi rdr}{r^{2}+1}}=2\pi \int _{0}^{2}{\frac {rdr}{r^{2}+1}}$

Use substitution:

$u=r^{2}+1$

$du=2rdr$

${\frac {du}{2}}=rdr$

$2\pi \int _{0}^{2}{\frac {rdr}{r^{2}+1}}={\frac {2}{2}}\pi \int _{0^{2}+1}^{2^{2}+1}{\frac {du}{u}}=\pi \int _{1}^{5}{\frac {du}{u}}$

$=\pi (ln(5)-ln(1))=\pi ln(5)$

Conclusion: There are $\color {blue}1000\times \pi ln(5)$ ants