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Question Number 41461 by rahul 19 last updated on 07/Aug/18

$$\mathrm{Find}\mathrm{area}\mathrm{of}\mathrm{square}\mathrm{inserted}\mathrm{in}\mathrm{curve}$$$$\mathrm{f}\left(x\right)=3x-x^3.$$

Commented by rahul 19 last updated on 07/Aug/18

Commented by rahul 19 last updated on 07/Aug/18

$$\mathrm{Ans}:{\left(216\right)}^{\frac{1}{3}}+{(-108)}^{\frac{1}{3}}.$$

Commented by MJS last updated on 07/Aug/18

$${216}^{\frac{1}{3}}=6$$$${(-108)}^{\frac{1}{3}}=-{108}^{\frac{1}{3}}=-{\left(2^2{3}^3\right)}^{\frac{1}{3}}=-3\sqrt[3]{4}$$$$\Rightarrow \mathrm{answer}=6-3\sqrt[3]{4}$$

Answered by MJS last updated on 07/Aug/18

$$\mathrm{idea}:\mathrm{intersect}y=d\mathrm{with}f\left(x\right)$$$$\mathrm{we}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{need}{x}_1<0$$$$\mathrm{solve}{x}_3-x_2=d\mathrm{for}d$$$$$$$$3x-x^3=d$$$$x^3-3x+d=0$$$$[f^{\prime }\left(x\right)=3-3x^2\Rightarrow \max \left(f\left(x\right)\right)=\left(\begin{array}{c}1\\ 2\end{array}\right)\Rightarrow$$$$\Rightarrow 0<d<2\iff x_1,x_2,x_3\in \mathbb{R}\Rightarrow$$$$\Rightarrow \mathrm{we}\mathrm{need}\mathrm{the}\mathrm{trigonometric}\mathrm{method}]$$$$x_1=-2\sin \left(\frac{1}{3}(\pi +\arcsin \frac{d}{2})\right)$$$$x_2=2\sin \left(\frac{1}{3}\arcsin \frac{d}{2}\right)$$$$x_3=2\cos \left(\frac{1}{3}(\frac{\pi }{2}+\arcsin \frac{d}{2})\right)$$$$x_3-x_2=d$$$$2\sqrt{3}\cos \left(\frac{1}{3}(\pi +\arcsin \frac{d}{2})\right)=d$$$$\frac{1}{3}(\pi +\arcsin \frac{d}{2})=\arccos \frac{d\sqrt{3}}{6}$$$$\pi +\arcsin \frac{d}{2}=3\arccos \frac{d\sqrt{3}}{6}$$$$\tan (\pi +\arcsin \frac{d}{2})=\tan \left(3\arccos \frac{d\sqrt{3}}{6}\right)$$$$\frac{d}{\sqrt{4-d^2}}=\frac{(d^2-3)\sqrt{12-d^2}}{d(d^2-9)}$$$$\mathrm{this}\mathrm{leads}\mathrm{to}$$$$d^6-18d^4+108d^2-108=0$$$$d=\sqrt{u}$$$$u^3-18u^2+108u-108=0$$$$u=v+6$$$$v^3+108=0$$$$v=-3\sqrt[3]{4}$$$$u=6-3\sqrt[3]{4}$$$$d=\sqrt{6-3\sqrt[3]{4}}\approx 1.11256$$$$\mathrm{area}\mathrm{of}\mathrm{square}=d^2=6-3\sqrt[3]{4}\approx 1.23780$$

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