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Question Number 123591 by ajfour last updated on 26/Nov/20

Commented by ajfour last updated on 26/Nov/20

$$y=(x-p)(x^3-x-c)$$$$Findpifthedifferentlyshaded$$$$areasareequal.$$

Commented by MJS_new last updated on 26/Nov/20

$$\mathrm{my}\mathrm{try}:$$$$\mathrm{choose}v,w>0\mathrm{and}\mathrm{let}$$$$\alpha =\frac{v+w}{\sqrt{v^2+vw+w^2}}\wedge \beta =-\frac{v}{\sqrt{v^2+vw+w^2}}\wedge \gamma =-\frac{w}{\sqrt{v^2+vw+w^2}}$$$$\Rightarrow$$$$(x-\alpha )(x-\beta )(x-\gamma )=$$$$=x^3-x-c\wedge c=\frac{v(v+w)w}{{(v^2+vw+w^2)}^{3/2}}$$$$\Rightarrow$$$$\underset{0}{\stackrel{p}{\int }}(x-p)(x^3-x-c)dx+\underset{p}{\stackrel{\alpha }{\int }}(x-p)(x^3-x-c)dx=0$$$$\Rightarrow$$$$p=\frac{2(15c-2\alpha (3{\alpha }^2-5))\alpha }{15(4c-\alpha ({\alpha }^2-2))}=$$$$=\frac{2(v+w)(4v^2+13vw+4w^2)}{(v^2+4vw+w^2)\sqrt{v^2+vw+w^2}}$$$$...\mathrm{did}\mathrm{the}v-w-\mathrm{thing}\mathrm{to}\mathrm{make}\mathrm{sure}\mathrm{it}\mathrm{looks}$$$$\mathrm{like}\mathrm{your}\mathrm{graphic}...$$

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