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Question Number 142461 by bramlexs22 last updated on 01/Jun/21

Commented by qaz last updated on 01/Jun/21

$${\mathrm{S}}_1={\int }_0^{{\mathrm{a}}_1}(\mathrm{c}-8\mathrm{x}+{27\mathrm{x}}^3)\mathrm{dx}={\mathrm{a}}_1\mathrm{c}-{4\mathrm{a}}_1^2+\frac{27}{4}{\mathrm{a}}_1^4$$$${\mathrm{S}}_2={\int }_{{\mathrm{a}}_1}^{{\mathrm{a}}_2}(8\mathrm{x}-{27\mathrm{x}}^3-\mathrm{c})\mathrm{dx}={\int }_{{\mathrm{a}}_2}^{{\mathrm{a}}_1}(\mathrm{c}-8\mathrm{x}+{27\mathrm{x}}^3)\mathrm{dx}$$$${\mathrm{S}}_1={\mathrm{S}}_2\Rightarrow -({\mathrm{ca}}_2-{4\mathrm{a}}_2^2+\frac{27}{4}{\mathrm{a}}_2^4)=0$$$${\mathrm{a}}_2\ne 0\Rightarrow \mathrm{c}-{4\mathrm{a}}_2+\frac{27}{4}{\mathrm{a}}_2^3=0$$$$\mathrm{c}={8\mathrm{a}}_2-{27\mathrm{a}}_2^3$$$$\Rightarrow {\mathrm{a}}_2=\frac{4}{9}.....\mathrm{c}=\frac{32}{27}$$

Answered by benjo_mathlover last updated on 01/Jun/21

Commented by EDWIN88 last updated on 01/Jun/21

$$\mathrm{good}\mathrm{explanation}$$

Answered by mr W last updated on 01/Jun/21

$$y=8x-27x^3=c$$$$say8a-27a^3=c$$$${\int }_0^a(c-8x+27x^3)dx=0$$$$ca-4a^2+\frac{27}{4}{a}^4=0$$$$c-4a+\frac{27}{4}{a}^3=0$$$$c-4a+\frac{1}{4}(8a-c)=0$$$$\Rightarrow a=\frac{3c}{8}$$$$\frac{3c}{8}(8-27\times \frac{9c^2}{64})=c$$$$c^2=\frac{16\times 64}{81\times 9}$$$$\Rightarrow c=\frac{32}{27}$$

Commented by mr W last updated on 01/Jun/21

Commented by mr W last updated on 01/Jun/21

$${\int }_0^a={\int }_0^b+{\int }_b^a=A+(-A)=0$$

Commented by benjo_mathlover last updated on 01/Jun/21

$$oouhgreat...$$$$$$

Commented by benjo_mathlover last updated on 01/Jun/21

$$whyareaI=\underset{0}{\stackrel{a}{\int }}(c-8x+27x^3)dx=0?$$

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