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Question Number 130158 by benjo_mathlover last updated on 23/Jan/21

$$\mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{segment}\mathrm{of}\mathrm{the}\mathrm{curve}$$$${\mathrm{y}}^2={\mathrm{x}}^3-{\mathrm{x}}^2\mathrm{if}\mathrm{the}\mathrm{line}\mathrm{x}=2\mathrm{is}\mathrm{the}\mathrm{chord}$$$$\mathrm{determining}\mathrm{the}\mathrm{segment}$$

Answered by liberty last updated on 23/Jan/21

$$\mathrm{Area}={\int }_1^2[\mathrm{x}\sqrt{\mathrm{x}-1}-(-\mathrm{x}\sqrt{\mathrm{x}-1})]\mathrm{dx}$$$$=2{\int }_1^2\mathrm{x}\sqrt{\mathrm{x}-1}\mathrm{dx}$$$$=4{\int }_0^1({\mu }^2+1){\mu }^2\mathrm{d}\mu ;\mathrm{where}\mu =\sqrt{\mathrm{x}-1}$$$$=4{[\frac{{\mu }^5}{5}+\frac{{\mu }^3}{3}]}_0^1=\frac{32}{15}$$

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