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Question Number 94579 by pete last updated on 20/May/20

$$\mathrm{A}\mathrm{study}\mathrm{indicates}\mathrm{that}x\mathrm{months}\mathrm{from}\mathrm{now}$$$$\mathrm{the}\mathrm{population}\mathrm{of}\mathrm{a}\mathrm{certain}\mathrm{town}\mathrm{will}\mathrm{be}$$$$\mathrm{decreasing}\mathrm{at}\mathrm{the}\mathrm{rate}\mathrm{of}5+3x^{\frac{2}{3}}\mathrm{people}$$$$\mathrm{per}\mathrm{month}.\mathrm{By}\mathrm{how}\mathrm{much}\mathrm{will}\mathrm{the}\mathrm{population}$$$$\mathrm{of}\mathrm{the}\mathrm{town}\mathrm{increase}\mathrm{per}\mathrm{the}\mathrm{next}8\mathrm{months}.$$$$\mathit{I}\mathit{n}\mathit{e}\mathit{e}\mathit{d}\mathit{h}\mathit{e}\mathit{l}\mathit{p}\mathit{w}\mathit{i}\mathit{t}\mathit{h}\mathit{t}\mathit{h}\mathit{e}\mathit{a}\mathit{b}\mathit{o}\mathit{v}\mathit{e}\mathit{q}\mathit{u}\mathit{e}\mathit{s}\mathit{t}\mathit{i}\mathit{o}\mathit{n},\mathit{p}\mathit{l}\mathit{e}\mathit{a}\mathit{s}\mathit{e}.$$

Answered by mr W last updated on 19/May/20

$$assumeyoumeanincreasingrate.$$$$\frac{dP}{dx}=5+3x^{\frac{2}{3}}$$$${\int }_{P_1}^{P_2}dP={\int }_0^8(5+3x^{\frac{2}{3}})dx$$$$\mathrm{\Delta }P={[5x+\frac{9}{5}{x}^{\frac{5}{3}}]}_0^8=(5\times 8+\frac{9}{5}\times 8^{\frac{5}{3}})\approx 98$$

Commented by pete last updated on 20/May/20

$$\mathrm{thanks}\mathrm{very}\mathrm{much}\mathrm{sir}$$

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