Question Number 94579 by pete last updated on 20/May/20
$$\mathrm{A}\:\mathrm{study}\:\mathrm{indicates}\:\mathrm{that}\:{x}\:\mathrm{months}\:\mathrm{from}\:\mathrm{now} \\ $$$$\mathrm{the}\:\mathrm{population}\:\mathrm{of}\:\mathrm{a}\:\mathrm{certain}\:\mathrm{town}\:\mathrm{will}\:\mathrm{be}\: \\ $$$$\mathrm{decreasing}\:\mathrm{at}\:\mathrm{the}\:\mathrm{rate}\:\mathrm{of}\:\mathrm{5}+\mathrm{3}{x}^{\frac{\mathrm{2}}{\mathrm{3}}} \:\mathrm{people} \\ $$$$\mathrm{per}\:\mathrm{month}.\:\mathrm{By}\:\mathrm{how}\:\mathrm{much}\:\mathrm{will}\:\mathrm{the}\:\mathrm{population} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{town}\:\mathrm{increase}\:\mathrm{per}\:\mathrm{the}\:\mathrm{next}\:\mathrm{8}\:\mathrm{months}. \\ $$$$\boldsymbol{{I}}\:\boldsymbol{{need}}\:\boldsymbol{{help}}\:\boldsymbol{{with}}\:\boldsymbol{{the}}\:\boldsymbol{{above}}\:\boldsymbol{{question}},\:\boldsymbol{{please}}. \\ $$
Answered by mr W last updated on 19/May/20
![assume you mean increasing rate. (dP/dx)=5+3x^(2/3) ∫_P_1 ^P_2 dP=∫_0 ^8 (5+3x^(2/3) )dx ΔP=[5x+(9/5)x^(5/3) ]_0 ^8 =(5×8+(9/5)×8^(5/3) )≈98](Q94590.png)
$${assume}\:{you}\:{mean}\:{increasing}\:{rate}. \\ $$$$\frac{{dP}}{{dx}}=\mathrm{5}+\mathrm{3}{x}^{\frac{\mathrm{2}}{\mathrm{3}}} \\ $$$$\int_{{P}_{\mathrm{1}} } ^{{P}_{\mathrm{2}} } {dP}=\int_{\mathrm{0}} ^{\mathrm{8}} \left(\mathrm{5}+\mathrm{3}{x}^{\frac{\mathrm{2}}{\mathrm{3}}} \right){dx} \\ $$$$\Delta{P}=\left[\mathrm{5}{x}+\frac{\mathrm{9}}{\mathrm{5}}{x}^{\frac{\mathrm{5}}{\mathrm{3}}} \right]_{\mathrm{0}} ^{\mathrm{8}} =\left(\mathrm{5}×\mathrm{8}+\frac{\mathrm{9}}{\mathrm{5}}×\mathrm{8}^{\frac{\mathrm{5}}{\mathrm{3}}} \right)\approx\mathrm{98} \\ $$
Commented by pete last updated on 20/May/20
$$\mathrm{thanks}\:\mathrm{very}\:\mathrm{much}\:\mathrm{sir} \\ $$