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Question Number 57992 by Tinkutara last updated on 15/Apr/19

Answered by MJS last updated on 16/Apr/19

$$\underset{1}{\stackrel{c}{\int }}(-x^5+8x^2)dx=\frac{16}{3}$$$$-\frac{1}{6}{c}^6+\frac{8}{3}{c}^3-\frac{5}{2}=\frac{16}{3}$$$$c^6-16c^3+47=0$$$$t=c^3$$$$t^2-16t+47=0$$$$t_1=8-\sqrt{17};t_2=8+\sqrt{17}$$$$c_1=\sqrt[3]{8-\sqrt{17}};c_2=\sqrt[3]{8+\sqrt{17}};\mathrm{all}\mathrm{other}\mathrm{solutions}\notin \mathbb{R}$$$$\mathrm{but}$$$$-x^5+8x^2=0\Rightarrow x_1=x_2=0;x_3=2;\mathrm{all}\mathrm{other}\mathrm{solutions}\notin \mathbb{R}$$$$\Rightarrow \mathrm{area}\mathrm{in}]-\mathrm{\infty };0[\mathrm{negative},\mathrm{in}]0;2[\mathrm{positive}$$$$\mathrm{and}\mathrm{in}]2;+\mathrm{\infty }[\mathrm{negative}$$$$c_2\approx 2.3>2\Rightarrow \mathrm{not}\mathrm{valid}$$$$\Rightarrow c=c_1=\sqrt[3]{8-\sqrt{17}}$$$$c+2=2+\sqrt[3]{8-\sqrt{17}}\approx 3.57095$$

Commented by MJS last updated on 16/Apr/19

$$\mathrm{we}\mathrm{should}\mathrm{start}\mathrm{with}$$$$\underset{1}{\stackrel{2}{\int }}(-x^5+8x^2)dx=\frac{49}{6}>\frac{16}{3}\Rightarrow 1<c<2$$

Commented by Tinkutara last updated on 18/Apr/19

Thanks Sir! Maybe the answer 1 in book is wrong.

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