Question Number 7433 by Tawakalitu. last updated on 28/Aug/16
Commented by Yozzia last updated on 28/Aug/16
$$ \\ $$$${u}=\frac{\underset{{i}=\mathrm{1}} {\overset{\mathrm{99}} {\sum}}\sqrt{\mathrm{10}+\sqrt{{i}}}}{\underset{{i}=\mathrm{1}} {\overset{\mathrm{99}} {\sum}}\sqrt{\mathrm{10}−\sqrt{{i}}}} \\ $$$$\Rightarrow{u}\underset{{i}=\mathrm{1}} {\overset{\mathrm{99}} {\sum}}\sqrt{\mathrm{10}−\sqrt{{i}}}−\underset{{i}=\mathrm{1}} {\overset{\mathrm{99}} {\sum}}\sqrt{\mathrm{10}+\sqrt{{i}}}=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by sou1618 last updated on 29/Aug/16
$${u}\fallingdotseq\frac{\int_{\mathrm{1}} ^{\mathrm{100}} \sqrt{\mathrm{10}+\sqrt{{x}}}{dx}}{\int_{\mathrm{0}} ^{\mathrm{99}} \sqrt{\mathrm{10}−\sqrt{{x}}}{dx}}=\frac{\mathrm{403}.\mathrm{903}...}{\mathrm{168}.\mathrm{506}...}=\mathrm{2}.\mathrm{397} \\ $$$$=\mathrm{1}+\mathrm{1}.\mathrm{397}=\mathrm{1}+\sqrt{\mathrm{2}}\:\:?? \\ $$$${I}\:{use}\:{calculator}.{sorry}... \\ $$$${somebody}\:{solve}\:{it},{please}. \\ $$
Commented by sou1618 last updated on 29/Aug/16
![u≒((∫_0 ^(100) (√(10+(√x)))dx)/(∫_0 ^(100) (√(10−(√x)))dx)) (0≤x≤100) y_1 =(√(10+(√x_1 )))((√(10))≤y≤(√(20))) y^2 =10+(√x) x=(y^2 −10)^2 =y^4 −20y^2 +100 x_1 =y_1 ^4 −20y_1 ^2 +100 y_2 =(√(10−(√x_2 )))(0≤y≤(√(10))) x_2 =(10−y^2 )^2 x_2 =y_2 ^4 −20y_2 ^2 +100 ∫_0 ^(100) y_1 dx_1 =100×(√(20))−∫_(√(10)) ^(√(20)) y^4 −20y^2 +100dy =100(√(20))−[(1/5)y^5 −((20)/3)y^3 +100y]_(√(10)) ^(√(20)) =100(√(20))−{20(4(√(20))−(√(10)))−((200)/3)(2(√(20))−(√(10)))+100((√(20))−(√(10)))} =(√(20))(100−80+((400)/3)−100)+(√(10))(20−((200)/3)+100) =((160)/3)(√(20))+((160)/3)(√(10)) ∫_0 ^(100) y_2 dx_2 =∫_0 ^(√(10)) y^4 −20y^2 +100dy =[(1/5)y^5 −((20)/3)y^3 +100y]_0 ^(√(10)) =20(√(10))−((200)/3)(√(10))+100(√(10)) =((160)/3)(√(10)) u=(((√(20))+(√(10)))/(√(10))) =(√2)+1 ⇊gragh is { ((y=(√(10+(√x))))),((y=(√(10−(√x))))) :}≡{x=y^4 −20y^2 +100(0≤y≤(√(20)))](Q7437.png)
$${u}\fallingdotseq\frac{\int_{\mathrm{0}} ^{\mathrm{100}} \sqrt{\mathrm{10}+\sqrt{{x}}}{dx}}{\int_{\mathrm{0}} ^{\mathrm{100}} \sqrt{\mathrm{10}−\sqrt{{x}}}{dx}} \\ $$$$ \\ $$$$\left(\mathrm{0}\leqslant{x}\leqslant\mathrm{100}\right) \\ $$$${y}_{\mathrm{1}} =\sqrt{\mathrm{10}+\sqrt{{x}_{\mathrm{1}} }}\left(\sqrt{\mathrm{10}}\leqslant{y}\leqslant\sqrt{\mathrm{20}}\right) \\ $$$${y}^{\mathrm{2}} =\mathrm{10}+\sqrt{{x}} \\ $$$${x}=\left({y}^{\mathrm{2}} −\mathrm{10}\right)^{\mathrm{2}} ={y}^{\mathrm{4}} −\mathrm{20}{y}^{\mathrm{2}} +\mathrm{100} \\ $$$${x}_{\mathrm{1}} ={y}_{\mathrm{1}} ^{\mathrm{4}} −\mathrm{20}{y}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{100} \\ $$$$ \\ $$$${y}_{\mathrm{2}} =\sqrt{\mathrm{10}−\sqrt{{x}_{\mathrm{2}} }}\left(\mathrm{0}\leqslant{y}\leqslant\sqrt{\mathrm{10}}\right) \\ $$$${x}_{\mathrm{2}} =\left(\mathrm{10}−{y}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$${x}_{\mathrm{2}} ={y}_{\mathrm{2}} ^{\mathrm{4}} −\mathrm{20}{y}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{100} \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\mathrm{100}} {y}_{\mathrm{1}} {dx}_{\mathrm{1}} =\mathrm{100}×\sqrt{\mathrm{20}}−\int_{\sqrt{\mathrm{10}}} ^{\sqrt{\mathrm{20}}} {y}^{\mathrm{4}} −\mathrm{20}{y}^{\mathrm{2}} +\mathrm{100}{dy} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{100}\sqrt{\mathrm{20}}−\left[\frac{\mathrm{1}}{\mathrm{5}}{y}^{\mathrm{5}} −\frac{\mathrm{20}}{\mathrm{3}}{y}^{\mathrm{3}} +\mathrm{100}{y}\right]_{\sqrt{\mathrm{10}}} ^{\sqrt{\mathrm{20}}} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{100}\sqrt{\mathrm{20}}−\left\{\mathrm{20}\left(\mathrm{4}\sqrt{\mathrm{20}}−\sqrt{\mathrm{10}}\right)−\frac{\mathrm{200}}{\mathrm{3}}\left(\mathrm{2}\sqrt{\mathrm{20}}−\sqrt{\mathrm{10}}\right)+\mathrm{100}\left(\sqrt{\mathrm{20}}−\sqrt{\mathrm{10}}\right)\right\} \\ $$$$\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{20}}\left(\mathrm{100}−\mathrm{80}+\frac{\mathrm{400}}{\mathrm{3}}−\mathrm{100}\right)+\sqrt{\mathrm{10}}\left(\mathrm{20}−\frac{\mathrm{200}}{\mathrm{3}}+\mathrm{100}\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{160}}{\mathrm{3}}\sqrt{\mathrm{20}}+\frac{\mathrm{160}}{\mathrm{3}}\sqrt{\mathrm{10}} \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\mathrm{100}} {y}_{\mathrm{2}} {dx}_{\mathrm{2}} =\int_{\mathrm{0}} ^{\sqrt{\mathrm{10}}} {y}^{\mathrm{4}} −\mathrm{20}{y}^{\mathrm{2}} +\mathrm{100}{dy} \\ $$$$\:\:\:\:\:\:\:\:\:=\left[\frac{\mathrm{1}}{\mathrm{5}}{y}^{\mathrm{5}} −\frac{\mathrm{20}}{\mathrm{3}}{y}^{\mathrm{3}} +\mathrm{100}{y}\right]_{\mathrm{0}} ^{\sqrt{\mathrm{10}}} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{20}\sqrt{\mathrm{10}}−\frac{\mathrm{200}}{\mathrm{3}}\sqrt{\mathrm{10}}+\mathrm{100}\sqrt{\mathrm{10}} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{160}}{\mathrm{3}}\sqrt{\mathrm{10}} \\ $$$$ \\ $$$${u}=\frac{\sqrt{\mathrm{20}}+\sqrt{\mathrm{10}}}{\sqrt{\mathrm{10}}} \\ $$$$\:\:\:=\sqrt{\mathrm{2}}+\mathrm{1} \\ $$$$ \\ $$$$\downdownarrows{gragh}\:{is\begin{cases}{{y}=\sqrt{\mathrm{10}+\sqrt{{x}}}}\\{{y}=\sqrt{\mathrm{10}−\sqrt{{x}}}}\end{cases}}\equiv\left\{{x}={y}^{\mathrm{4}} −\mathrm{20}{y}^{\mathrm{2}} +\mathrm{100}\left(\mathrm{0}\leqslant{y}\leqslant\sqrt{\mathrm{20}}\right)\right. \\ $$
Commented by sou1618 last updated on 29/Aug/16
Commented by Tawakalitu. last updated on 29/Aug/16
$${Wow},\:{thanks}\:{so}\:{much}\:{sir}.\:{God}\:{bless}\:{you}. \\ $$