Question Number 92082 by mathmax by abdo last updated on 04/May/20
$${let}\:{f}\left(\alpha\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} {x}\sqrt{{x}^{\mathrm{2}} −{x}+\alpha}{dx}\:\:\:\:\:\:{with}\:\alpha>\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\left.\mathrm{1}\right)\:{explicit}\:\:{f}\left(\alpha\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{g}\left(\alpha\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{xdx}}{\:\sqrt{{x}^{\mathrm{2}} −{x}+\alpha}} \\ $$$$\left.\mathrm{3}\right)\:{find}\:{the}\:{value}\:{of}\:{intehrals}\:\int_{\mathrm{0}} ^{\mathrm{1}} {x}\sqrt{{x}^{\mathrm{2}} −{x}+\sqrt{\mathrm{2}}}{dx}\:{snd} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{xdx}}{\:\sqrt{{x}^{\mathrm{2}} −{x}+\sqrt{\mathrm{2}}}} \\ $$
Commented by mathmax by abdo last updated on 05/May/20
![1)we have x^2 −x+α =x^2 −2(x/2)+(1/4)+α−(1/4) =(x−(1/2))^2 +((4α−1)/4) we do the changement x−(1/2) =((√(4α−1))/2)sh(t) ⇒ f(α) =∫_(−argsh((1/( (√(4α−1)))))) ^(argsh((1/( (√(4α−1)))))) {(1/2) +((√(4α−1))/2)sh(t)}((√(4α−1))/2)ch(t)×((√(4α−1))/2)ch(t)dt =(1/8)(4α−1) ∫_(−ln((1/( (√(4α−1))))+(√(1+(1/(4α−1)))))) ^(ln((1/( (√(4α−1)) ))+(√(1+(1/(4α−1)))))) (1+(√(4α−1))sh(t))ch^2 t dt ⇒((8f(α))/(4α−1)) =∫_(x(α)) ^(y(α)) ch^2 t dt +(√(4α−1))∫_(x(α)) ^(y(α)) sh(t)ch^2 t dt =(1/2)∫_(x(α)) ^(y(α)) (1+ch(2t))dt +[(1/3)ch^3 t]_(x(α)) ^(y(α)) =(1/2)(y(α)−x(α))+(1/4)[sh(2t)]_(x(α)) ^(y(α)) +(1/3)[(((e^t +e^(−t) )/2))^3 ]_(x(α)) ^(y(α)) =((y(α)−x(α))/2) +(1/8)[ e^(2t) −e^(−2t) ]_(x(α)) ^(y(α)) +(1/(24))[ (e^t +e^(−t) )^3 ]_(x(α)) ^(y(α)) =ln((1/( (√(4α−1))))+((2(√α))/( (√(4α−1)))))+(1/8){(((1+2(√α))/( (√(4α−1)))))^2 −(((1+2(√α))/( (√(4α−1)))))^(−2) −( (((1+2(√α))/( (√(4α−1)))))^(−2) −(((1+2(√α))/( (√(4α−1)))))^2 } +(1/(24)){ ( (((1+2(√α))/( (√(4α−1)))))−(((1+2(√α))/( (√(4α−1)))))^(−1) )^3 −( (((1+2(√α))/( (√(4α−1)))))^(−1) −(((1+2(√α))/( (√(4α−1))))))^3 } rest to finish the calculus...](https://www.tinkutara.com/question/Q92123.png)
$$\left.\mathrm{1}\right){we}\:{have}\:{x}^{\mathrm{2}} −{x}+\alpha\:={x}^{\mathrm{2}} −\mathrm{2}\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\alpha−\frac{\mathrm{1}}{\mathrm{4}}\:=\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{4}\alpha−\mathrm{1}}{\mathrm{4}} \\ $$$${we}\:{do}\:{the}\:{changement}\:{x}−\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{\sqrt{\mathrm{4}\alpha−\mathrm{1}}}{\mathrm{2}}{sh}\left({t}\right)\:\Rightarrow \\ $$$${f}\left(\alpha\right)\:=\int_{−{argsh}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}\alpha−\mathrm{1}}}\right)} ^{{argsh}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}\alpha−\mathrm{1}}}\right)} \left\{\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\sqrt{\mathrm{4}\alpha−\mathrm{1}}}{\mathrm{2}}{sh}\left({t}\right)\right\}\frac{\sqrt{\mathrm{4}\alpha−\mathrm{1}}}{\mathrm{2}}{ch}\left({t}\right)×\frac{\sqrt{\mathrm{4}\alpha−\mathrm{1}}}{\mathrm{2}}{ch}\left({t}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{4}\alpha−\mathrm{1}\right)\:\int_{−{ln}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}\alpha−\mathrm{1}}}+\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}\alpha−\mathrm{1}}}\right)} ^{{ln}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}\alpha−\mathrm{1}}\:}+\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}\alpha−\mathrm{1}}}\right)} \left(\mathrm{1}+\sqrt{\mathrm{4}\alpha−\mathrm{1}}{sh}\left({t}\right)\right){ch}^{\mathrm{2}} {t}\:{dt} \\ $$$$\Rightarrow\frac{\mathrm{8}{f}\left(\alpha\right)}{\mathrm{4}\alpha−\mathrm{1}}\:=\int_{{x}\left(\alpha\right)} ^{{y}\left(\alpha\right)} \:{ch}^{\mathrm{2}} {t}\:{dt}\:\:+\sqrt{\mathrm{4}\alpha−\mathrm{1}}\int_{{x}\left(\alpha\right)} ^{{y}\left(\alpha\right)} \:{sh}\left({t}\right){ch}^{\mathrm{2}} {t}\:{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{{x}\left(\alpha\right)} ^{{y}\left(\alpha\right)} \left(\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)\right){dt}\:\:+\left[\frac{\mathrm{1}}{\mathrm{3}}{ch}^{\mathrm{3}} {t}\right]_{{x}\left(\alpha\right)} ^{{y}\left(\alpha\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({y}\left(\alpha\right)−{x}\left(\alpha\right)\right)+\frac{\mathrm{1}}{\mathrm{4}}\left[{sh}\left(\mathrm{2}{t}\right)\right]_{{x}\left(\alpha\right)} ^{{y}\left(\alpha\right)} \:+\frac{\mathrm{1}}{\mathrm{3}}\left[\left(\frac{{e}^{{t}} \:+{e}^{−{t}} }{\mathrm{2}}\right)^{\mathrm{3}} \right]_{{x}\left(\alpha\right)} ^{{y}\left(\alpha\right)} \\ $$$$=\frac{{y}\left(\alpha\right)−{x}\left(\alpha\right)}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{8}}\left[\:{e}^{\mathrm{2}{t}} −{e}^{−\mathrm{2}{t}} \right]_{{x}\left(\alpha\right)} ^{{y}\left(\alpha\right)} \:\:+\frac{\mathrm{1}}{\mathrm{24}}\left[\:\left({e}^{{t}} \:+{e}^{−{t}} \right)^{\mathrm{3}} \right]_{{x}\left(\alpha\right)} ^{{y}\left(\alpha\right)} \\ $$$$={ln}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}\alpha−\mathrm{1}}}+\frac{\mathrm{2}\sqrt{\alpha}}{\:\sqrt{\mathrm{4}\alpha−\mathrm{1}}}\right)+\frac{\mathrm{1}}{\mathrm{8}}\left\{\left(\frac{\mathrm{1}+\mathrm{2}\sqrt{\alpha}}{\:\sqrt{\mathrm{4}\alpha−\mathrm{1}}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{1}+\mathrm{2}\sqrt{\alpha}}{\:\sqrt{\mathrm{4}\alpha−\mathrm{1}}}\right)^{−\mathrm{2}} \right. \\ $$$$−\left(\:\left(\frac{\mathrm{1}+\mathrm{2}\sqrt{\alpha}}{\:\sqrt{\mathrm{4}\alpha−\mathrm{1}}}\right)^{−\mathrm{2}} −\left(\frac{\mathrm{1}+\mathrm{2}\sqrt{\alpha}}{\:\sqrt{\mathrm{4}\alpha−\mathrm{1}}}\right)^{\mathrm{2}} \right\}\:+\frac{\mathrm{1}}{\mathrm{24}}\left\{\:\left(\:\left(\frac{\mathrm{1}+\mathrm{2}\sqrt{\alpha}}{\:\sqrt{\mathrm{4}\alpha−\mathrm{1}}}\right)−\left(\frac{\mathrm{1}+\mathrm{2}\sqrt{\alpha}}{\:\sqrt{\mathrm{4}\alpha−\mathrm{1}}}\right)^{−\mathrm{1}} \right)^{\mathrm{3}} \right. \\ $$$$\left.−\left(\:\left(\frac{\mathrm{1}+\mathrm{2}\sqrt{\alpha}}{\:\sqrt{\mathrm{4}\alpha−\mathrm{1}}}\right)^{−\mathrm{1}} −\left(\frac{\mathrm{1}+\mathrm{2}\sqrt{\alpha}}{\:\sqrt{\mathrm{4}\alpha−\mathrm{1}}}\right)\right)^{\mathrm{3}} \right\}\:{rest}\:{to}\:{finish}\:{the}\:{calculus}… \\ $$
Commented by mathmax by abdo last updated on 05/May/20
$${another}\:{way}\:{let}\:{find}\:\int\:{x}\sqrt{{x}^{\mathrm{2}} \:+{a}}\:{dx}\:{with}\:{a}>\mathrm{0} \\ $$$$\int\:{x}\sqrt{{x}^{\mathrm{2}} \:+{a}}{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{2}{x}\right)\sqrt{{x}^{\mathrm{2}} \:+{a}}{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{2}{x}\right)\left({x}^{\mathrm{2}} \:+{a}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}}\left({x}^{\mathrm{2}} \:+{a}\right)^{\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}} \:+{c}\:=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}}{\mathrm{3}}\left({x}^{\mathrm{2}} \:+{a}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left({x}^{\mathrm{2}} \:+{a}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:+{c}\:\Rightarrow \\ $$$$\int\:{x}\sqrt{{x}^{\mathrm{2}} −{x}+\alpha}{dx}\:=\int\:{x}\sqrt{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\alpha−\frac{\mathrm{1}}{\mathrm{4}}}{dx} \\ $$$$=_{{x}−\frac{\mathrm{1}}{\mathrm{2}}={t}} \:\:\:\int\:\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right)\sqrt{{t}^{\mathrm{2}} \:+\alpha−\frac{\mathrm{1}}{\mathrm{4}}}{dt} \\ $$$$=\int\:{t}\sqrt{{t}^{\mathrm{2}} \:+\alpha−\frac{\mathrm{1}}{\mathrm{4}}}{dt}\:\:+\int\:\sqrt{{t}^{\mathrm{2}} \:+\alpha−\frac{\mathrm{1}}{\mathrm{4}}}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left({t}^{\mathrm{2}} \:+\alpha−\frac{\mathrm{1}}{\mathrm{4}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:+\:\int\:\sqrt{{t}^{\mathrm{2}} \:+\alpha−\frac{\mathrm{1}}{\mathrm{4}}}{dt}\:{also}\:{we}\:{must}\:{find}\: \\ $$$$\int\sqrt{{x}^{\mathrm{2}} \:+{a}}{dx}\:\:{ch}.{x}\:=\sqrt{{a}}{sh}\left({t}\right)\:{give} \\ $$$$\int\:\sqrt{{x}^{\mathrm{2}} \:+{a}}{dx}\:=\int\:\sqrt{{a}}{ch}\left({t}\right)\sqrt{{a}}{ch}\left({t}\right){dt}\:={a}\:\int\:\frac{\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)}{\mathrm{2}}{dt} \\ $$$$=\frac{{a}}{\mathrm{2}}{t}\:+\frac{\mathrm{1}}{\mathrm{4}}{sh}\left(\mathrm{2}{t}\right)\:+{c}\:=\frac{{a}}{\mathrm{2}}\:{argsh}\left(\frac{{x}}{\:\sqrt{{a}}}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}{sh}\left({t}\right){ch}\left({t}\right)\:+{c} \\ $$$$=\frac{{a}}{\mathrm{2}}{ln}\left(\frac{{x}}{\:\sqrt{{a}}}\:+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{a}}}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}×\frac{{x}}{\:\sqrt{{a}}}\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{a}}}+{c} \\ $$$$=\frac{{a}}{\mathrm{2}}{ln}\left(\frac{{x}+\sqrt{{x}^{\mathrm{2}} \:+{a}}}{\:\sqrt{{a}}}\right)\:+\frac{{x}}{\mathrm{2}{a}}\sqrt{{x}^{\mathrm{2}} \:+{a}}\:+{c} \\ $$$$=\frac{{a}}{\mathrm{2}}{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} \:+{a}}\right)+\frac{{x}}{\mathrm{2}{a}}\sqrt{{x}^{\mathrm{2}} \:+{a}}\:\:+{C}\:\Rightarrow \\ $$$$\int\:\sqrt{{t}^{\mathrm{2}} +\alpha−\frac{\mathrm{1}}{\mathrm{4}}}{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\alpha−\frac{\mathrm{1}}{\mathrm{4}}\right){ln}\left({x}+\sqrt{{t}^{\mathrm{2}} \:+\alpha−\frac{\mathrm{1}}{\mathrm{4}}}\right) \\ $$$$=\frac{{t}}{\mathrm{2}\left(\alpha−\frac{\mathrm{1}}{\mathrm{4}}\right)}\sqrt{{t}^{\mathrm{2}} \:+\alpha−\frac{\mathrm{1}}{\mathrm{4}}}\:+{c}\:\:\:\:{with}\:{t}={x}−\frac{\mathrm{1}}{\mathrm{2}}\:…. \\ $$$$ \\ $$
Answered by Kunal12588 last updated on 04/May/20
![1) ∫_0 ^1 x(√(x^2 −x+α)) dx =(1/2)∫_0 ^1 (2x−1)(√(x^2 −x+α)) dx+(1/2)∫_0 ^1 (√(x^2 −x+α)) dx =(1/2)∫_0 ^1 (√((x−(1/2))^2 +((4α−1)/4))) dx; ((4α−1)/4)>0 =(1/2)[(1/2)(x−(1/2))(√(x^2 −x+α))+((4α−1)/8) ln∣(x−(1/2))+(√(x^2 −x+α))∣]_0 ^1 =(1/2)[(((√α)/4)+((4α−1)/8)ln∣(1/2)+(√α)∣)−(−((√α)/4)+((4α−1)/8)ln∣−(1/2)+(√α)∣)] =(1/2)[((√α)/2)+((4α−1)/8)ln∣((2(√α)+1)/(2(√α)−1))∣] =((√α)/4)+((4α−1)/(16))ln∣((4α+4(√α)+1)/(4α−1))∣ 2) ∫_0 ^1 ((xdx)/( (√(x^2 −x+α)))) =(1/2)∫_0 ^1 (((2x−1)dx)/( (√(x^2 −x+α))))+(1/2)∫_0 ^1 (dx/( (√(x^2 −x+α)))) =(1/2)∫_0 ^1 (dx/( (√((x−(1/2))^2 +((√((4α−1)/4)))^2 )))) =(1/2)[ln∣(x−(1/2))+(√(x^2 −x+α))∣]_0 ^1 =(1/2)[ln∣(1/2)+(√α)∣−ln∣(√α)−(1/2)∣] =(1/2)[ln∣(((√α)+(1/2))/( (√α)−(1/2)))∣]=(1/2)ln∣((2(√α)+1)/(2(√α)−1))∣=(1/2)ln∣((4α+4(√α)+1)/(4α−1))∣](https://www.tinkutara.com/question/Q92092.png)
$$\left.\mathrm{1}\right)\:\int_{\mathrm{0}} ^{\mathrm{1}} {x}\sqrt{{x}^{\mathrm{2}} −{x}+\alpha}\:{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{2}{x}−\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} −{x}+\alpha}\:{dx}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{{x}^{\mathrm{2}} −{x}+\alpha}\:{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{4}\alpha−\mathrm{1}}{\mathrm{4}}}\:{dx};\:\frac{\mathrm{4}\alpha−\mathrm{1}}{\mathrm{4}}>\mathrm{0} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{2}}\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)\sqrt{{x}^{\mathrm{2}} −{x}+\alpha}+\frac{\mathrm{4}\alpha−\mathrm{1}}{\mathrm{8}}\:{ln}\mid\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\sqrt{{x}^{\mathrm{2}} −{x}+\alpha}\mid\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\left(\frac{\sqrt{\alpha}}{\mathrm{4}}+\frac{\mathrm{4}\alpha−\mathrm{1}}{\mathrm{8}}{ln}\mid\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\alpha}\mid\right)−\left(−\frac{\sqrt{\alpha}}{\mathrm{4}}+\frac{\mathrm{4}\alpha−\mathrm{1}}{\mathrm{8}}{ln}\mid−\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\alpha}\mid\right)\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\sqrt{\alpha}}{\mathrm{2}}+\frac{\mathrm{4}\alpha−\mathrm{1}}{\mathrm{8}}{ln}\mid\frac{\mathrm{2}\sqrt{\alpha}+\mathrm{1}}{\mathrm{2}\sqrt{\alpha}−\mathrm{1}}\mid\right] \\ $$$$=\frac{\sqrt{\alpha}}{\mathrm{4}}+\frac{\mathrm{4}\alpha−\mathrm{1}}{\mathrm{16}}{ln}\mid\frac{\mathrm{4}\alpha+\mathrm{4}\sqrt{\alpha}+\mathrm{1}}{\mathrm{4}\alpha−\mathrm{1}}\mid \\ $$$$\left.\mathrm{2}\right)\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{xdx}}{\:\sqrt{{x}^{\mathrm{2}} −{x}+\alpha}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\left(\mathrm{2}{x}−\mathrm{1}\right){dx}}{\:\sqrt{{x}^{\mathrm{2}} −{x}+\alpha}}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} −{x}+\alpha}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\:\sqrt{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\sqrt{\frac{\mathrm{4}\alpha−\mathrm{1}}{\mathrm{4}}}\right)^{\mathrm{2}} }} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\mid\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\sqrt{{x}^{\mathrm{2}} −{x}+\alpha}\mid\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\mid\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\alpha}\mid−{ln}\mid\sqrt{\alpha}−\frac{\mathrm{1}}{\mathrm{2}}\mid\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\mid\frac{\sqrt{\alpha}+\frac{\mathrm{1}}{\mathrm{2}}}{\:\sqrt{\alpha}−\frac{\mathrm{1}}{\mathrm{2}}}\mid\right]=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\frac{\mathrm{2}\sqrt{\alpha}+\mathrm{1}}{\mathrm{2}\sqrt{\alpha}−\mathrm{1}}\mid=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\frac{\mathrm{4}\alpha+\mathrm{4}\sqrt{\alpha}+\mathrm{1}}{\mathrm{4}\alpha−\mathrm{1}}\mid \\ $$
Commented by turbo msup by abdo last updated on 04/May/20
$${thankx} \\ $$