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Question Number 155382 by MathsFan last updated on 29/Sep/21

Answered by puissant last updated on 29/Sep/21

$$Q={\int }_0^{10}\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+...}}}}dx$$$$u=\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+....}}}}\to u^2=x+\sqrt{x+\sqrt{x+\sqrt{x+....}}}$$$$\Rightarrow u^2-u=x\Rightarrow u-u+\frac{1}{4}=x+\frac{1}{4}$$$$\Rightarrow {(u-\frac{1}{2})}^2=x+\frac{1}{4}\Rightarrow u=\frac{1}{2}+\sqrt{\frac{4x+1}{4}}$$$$Q={\int }_0^{10}(\frac{1}{2}+\frac{1}{2}\sqrt{4x+1})dx=\frac{1}{2}{\int }_0^{10}(1+\sqrt{4x+1})dx$$$$=\frac{1}{2}{[x+\frac{2}{3}\times \frac{1}{4}\sqrt{{(4x+1)}^3}]}_0^{10}=\frac{1}{2}{[x+\frac{1}{6}\sqrt{{(4x+1)}^3}]}_0^{10}$$$$=\frac{1}{2}[(10+\frac{\sqrt{68921}}{6})-(0+\frac{1}{6})]$$$$$$$$\therefore \because Q=\frac{1}{12}(59+2\sqrt{68921})..$$$$$$$$..............\mathcal{L}epuissant...............$$$$$$

Commented by SANOGO last updated on 30/Sep/21

$$ledurgar$$

Commented by MathsFan last updated on 30/Sep/21

$$correctsir$$

Commented by peter frank last updated on 01/Oct/21

$$\mathrm{good}$$

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