Answer
$8868$ people
Work Step by Step
Given that,
$b(t)=2200e^{0.024t} $and $d(t)=1460e^{0.018t}$
For $0\leq t\leq 10,b(t)>d(t)$,
So, the area between the curves is given by
$\int_{0}^{10} [b(t)-d(t)] dt$
Now,
$=\int_{0}^{10}(2200e^{0.024t}-1460e^{0.018t})dt$
$=[\frac{2200}{0.024}e^{0.024t}-\frac{1460}{0.018}e^{0.018t}]_{0}^{10}$
$=(\frac{275,000}{3}e^{0.24}-\frac{730,000}{9}e^{0.18})-(\frac{275,000}{3}-\frac{730,000}{9})$
$\approx8868$ people