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Question-7258




Question Number 7258 by Tawakalitu. last updated on 19/Aug/16
Commented by Yozzia last updated on 19/Aug/16
y′=(x/( (√(5−x^2 ))))⇒y=∫(x/( (√(5−x^2 ))))dx.  Let u=5−x^2 ⇒du=−2xdx⇒xdx=((−1)/2)du.  y=(1/2)∫((−du)/( (√u)))=((−1)/2)∫u^(−1/2) du=((−1)/2)×2u^(1/2) +c  y=c−(√(5−x^2 ))  The point (2,6) lies on the curve.  ∴ 6=c−(√(5−4))⇒c=7  ∴ y=7−(√(5−x^2 ))
$${y}'=\frac{{x}}{\:\sqrt{\mathrm{5}−{x}^{\mathrm{2}} }}\Rightarrow{y}=\int\frac{{x}}{\:\sqrt{\mathrm{5}−{x}^{\mathrm{2}} }}{dx}. \\ $$$${Let}\:{u}=\mathrm{5}−{x}^{\mathrm{2}} \Rightarrow{du}=−\mathrm{2}{xdx}\Rightarrow{xdx}=\frac{−\mathrm{1}}{\mathrm{2}}{du}. \\ $$$${y}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{−{du}}{\:\sqrt{{u}}}=\frac{−\mathrm{1}}{\mathrm{2}}\int{u}^{−\mathrm{1}/\mathrm{2}} {du}=\frac{−\mathrm{1}}{\mathrm{2}}×\mathrm{2}{u}^{\mathrm{1}/\mathrm{2}} +{c} \\ $$$${y}={c}−\sqrt{\mathrm{5}−{x}^{\mathrm{2}} } \\ $$$${The}\:{point}\:\left(\mathrm{2},\mathrm{6}\right)\:{lies}\:{on}\:{the}\:{curve}. \\ $$$$\therefore\:\mathrm{6}={c}−\sqrt{\mathrm{5}−\mathrm{4}}\Rightarrow{c}=\mathrm{7} \\ $$$$\therefore\:{y}=\mathrm{7}−\sqrt{\mathrm{5}−{x}^{\mathrm{2}} } \\ $$
Commented by Tawakalitu. last updated on 19/Aug/16
Thanks so much sir, i really appreciate.
$${Thanks}\:{so}\:{much}\:{sir},\:{i}\:{really}\:{appreciate}. \\ $$

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