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Question Number 108017 by mathdave last updated on 13/Aug/20

Answered by hgrocks last updated on 13/Aug/20

$$2+\sqrt{\mathrm{x}}={\mathrm{y}}^2\Rightarrow \sqrt{\mathrm{x}}={\mathrm{y}}^2-2$$$$\frac{1}{2\sqrt{\mathrm{x}}}.\frac{1}{2\sqrt{2+\sqrt{\mathrm{x}}}}\mathrm{dx}=\mathrm{dy}$$$$\frac{1}{\sqrt{2+\sqrt{\mathrm{x}}}}\mathrm{dx}=4({\mathrm{y}}^2-2)$$$$$$$$\mathrm{I}=12\underset{2}{\stackrel{3}{\int }}({\mathrm{y}}^2-2)\mathrm{dy}$$$$\mathrm{I}=12{[\frac{{\mathrm{y}}^3}{3}-2\mathrm{y}]}_2^3$$$$\mathrm{I}=12[\frac{19}{3}-2]$$$$=76-24$$$$\mathrm{I}=52$$$$$$$$★\mathbf{HG}★$$

Answered by mathmax by abdo last updated on 13/Aug/20

$$\mathrm{I}={\int }_4^{49}\frac{3}{\sqrt{2+\sqrt{\mathrm{x}}}}\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\sqrt{2+\sqrt{\mathrm{x}}}=\mathrm{t}\Rightarrow 2+\sqrt{\mathrm{x}}={\mathrm{t}}^2\Rightarrow$$$$\sqrt{\mathrm{x}}={\mathrm{t}}^2-2\Rightarrow \mathrm{x}={({\mathrm{t}}^2-2)}^2\Rightarrow \mathrm{I}=3{\int }_2^3\frac{2\left(2\mathrm{t}\right)({\mathrm{t}}^2-2)}{\mathrm{t}}\mathrm{dt}$$$$=12{\int }_2^3({\mathrm{t}}^2-2)\mathrm{dt}=12{[\frac{{\mathrm{t}}^3}{3}-2\mathrm{t}]}_2^3=12\{9-6-\frac{8}{3}+4\}$$$$=12\{7-\frac{8}{3}\}=(12.7)-32=84-32=52$$

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