Find-0-5-1-2-x-2-2x-1-dx-

Question Number 208670 by hardmath last updated on 20/Jun/24

$$\mathrm{Find}:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{5}} \:\sqrt{\frac{\mathrm{1}}{\mathrm{2}}\:\left(\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{2x}\:+\:\mathrm{1}\right)}\:\mathrm{dx}\:\:=\:\:? \\ $$

Commented by essaad last updated on 20/Jun/24

Answered by Berbere last updated on 20/Jun/24

$${x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}=\left({x}−\mathrm{1}\right)^{\mathrm{2}} … \\ $$$$\sqrt{{x}^{\mathrm{2}} }=\begin{cases}{{x};{x}\geqslant\mathrm{0}}\\{−{x}\leqslant\mathrm{0}}\end{cases} \\ $$

Answered by mathzup last updated on 21/Jun/24

$${I}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\mathrm{5}} \sqrt{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{dx}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\mathrm{5}} \mid{x}−\mathrm{1}\mid{dx} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}\right){dx}\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int_{\mathrm{1}} ^{\mathrm{5}} \left({x}−\mathrm{1}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left[{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{1}} +\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left[\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−{x}\right]_{\mathrm{1}} ^{\mathrm{5}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}×\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left\{\frac{\mathrm{25}}{\mathrm{2}}−\mathrm{5}−\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\frac{\mathrm{15}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}+\frac{\mathrm{8}}{\:\sqrt{\mathrm{2}}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\mathrm{8}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{17}}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$

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