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Question Number 21503 by mondodotto@gmail.com last updated on 25/Sep/17

Answered by $@ty@m last updated on 26/Sep/17

$$Let4x^2=t$$$$\Rightarrow 8xdx=dt$$$$Changeoflimits:$$$$Asx\to 4,t\to 64$$$$Asx\to y,t\to 4y^2$$$$I=\underset{64}{\stackrel{4y^2}{\int }}\sqrt{1-t}\frac{dt}{8}$$$$ATQ,0=\frac{1}{8}\underset{64}{\stackrel{4y^2}{\int }}\sqrt{1-t}dt$$$$0={\left[\frac{2{(1-t)}^{\frac{3}{2}}}{3}\right]}_{64}^{4y^2}$$$${(1-4y^2)}^{\frac{3}{2}}-{(-63)}^{\frac{3}{2}}=0$$$${(1-4y^2)}^{\frac{3}{2}}={(-63)}^{\frac{3}{2}}$$$$1-4y^2=-63$$$$4y^2=64$$$$y=\pm 4$$

Commented by mrW1 last updated on 26/Sep/17

$$1-{4\mathrm{x}}^2⩾0$$$$\Rightarrow {\mathrm{x}}^2⩽\frac{1}{4}$$$$\Rightarrow -\frac{1}{2}⩽\mathrm{x}⩽\frac{1}{2}$$$$\mathrm{i}.\mathrm{e}.\mathrm{function}\mathrm{x}\sqrt{1-{4\mathrm{x}}^2}\mathrm{is}\mathrm{defined}\mathrm{only}\mathrm{for}$$$$\mathrm{x}\in [-\frac{1}{2},\frac{1}{2}].$$

Commented by $@ty@m last updated on 26/Sep/17

$$Doesitmeanthequestioniswrong?$$

Commented by mrW1 last updated on 26/Sep/17

$$\mathrm{yes}$$

Commented by abwayh last updated on 26/Sep/17

$$I\mathrm{think}\mathrm{the}\mathrm{solution}\mathrm{is}\mathrm{correct},$$$$\mathrm{check}\mathrm{the}\mathrm{integral}\mathrm{at}\mathrm{y}=\stackrel{-}{+}4\mathrm{you}\mathrm{will}\mathrm{find}\mathrm{the}\mathrm{result}$$$$\mathrm{equal}\mathrm{to}\mathrm{zero}$$

Commented by mrW1 last updated on 26/Sep/17

$$\mathrm{as}\mathrm{for}\mathrm{the}\mathrm{definition}\mathrm{of}\mathrm{definite}$$$$\mathrm{integral}{\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right),\mathrm{the}\mathrm{function}\mathrm{f}\left(\mathrm{x}\right)\mathrm{must}$$$$\mathrm{be}\mathrm{continuous}\mathrm{on}\mathrm{the}\mathrm{interval}[\mathrm{a},\mathrm{b}].$$$$$$$$\mathrm{but}\mathrm{x}\sqrt{1-{4\mathrm{x}}^2}\mathrm{is}\mathrm{not}\mathrm{defined}\mathrm{on}$$$$[-4,-\frac{1}{2}]\mathrm{and}[\frac{1}{2},4].\mathrm{so}$$$${\int }_{-4}^{-\frac{1}{2}}\mathrm{x}\sqrt{1-{4\mathrm{x}}^2}\mathrm{dx}\mathrm{and}{\int }_{\frac{1}{2}}^4\mathrm{x}\sqrt{1-{4\mathrm{x}}^2}\mathrm{dx}$$$$\mathrm{are}\mathrm{also}\mathrm{not}\mathrm{defined}.$$$$$$$${\int }_{-4}^4\mathrm{x}\sqrt{1-{4\mathrm{x}}^2}\mathrm{dx}\mathrm{is}\mathrm{also}\mathrm{not}\mathrm{defined},\mathrm{since}$$$${\int }_{-4}^4\mathrm{x}\sqrt{1-{4\mathrm{x}}^2}\mathrm{dx}={\int }_{-4}^{-\frac{1}{2}}\mathrm{x}\sqrt{1-{4\mathrm{x}}^2}\mathrm{dx}+{\int }_{-\frac{1}{2}}^{\frac{1}{2}}\mathrm{x}\sqrt{1-{4\mathrm{x}}^2}\mathrm{dx}+{\int }_{\frac{1}{2}}^4\mathrm{x}\sqrt{1-{4\mathrm{x}}^2}\mathrm{dx}$$

Commented by abwayh last updated on 26/Sep/17

$$\mathrm{thank}\mathrm{you}\mathrm{sir};{\mathrm{I}}^{{}^{\prime }}\mathrm{m}\mathrm{understand}$$

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