Question Number 7903 by tawakalitu last updated on 24/Sep/16
Commented by sou1618 last updated on 24/Sep/16
![S = (( Σ_(k=1) ^(99) (√(10+(√k))) )/( Σ_(k=1) ^(99) (√(10−(√k))) )) S≒S ′=(( ∫_0 ^(100) (√(10+(√x) )) dx )/( ∫_0 ^(100) (√(10−(√y) )) dy )) t=10+(√x) x=(t−10)^2 dx=2t−20 dt x(0→100)⇔t(10→20) u=10−(√y) y=(10−u)^2 dy=2t−20 du y(0→100)⇔u(10→0) S ′=((∫_(10) ^(20) (√(10+(√((t−10)^2 )))) (2t−20)dt)/(∫_(10) ^0 (√(10−(√((10−u)^2 )))) (2u+20)du)) =((∫_(10) ^(20) (√(10+∣t−10∣)) (2t−20)dt)/(∫_(10) ^0 (√(10−∣10−u∣)) (2u−20)du)) =((∫_(10) ^(20) 2(√t) (t−10)dt)/(∫_(10) ^0 2(√u) (u−10)du)) =((∫_(10) ^(20) (√t^3 )−10(√t) dt)/(∫_(10) ^0 (√u^3 )−10(√u) du)) =(( [(2/5)(√t^5 )−10(2/3)(√t^3 )]_(10) ^(20) )/( [(2/5)(√u^5 )−10(2/3)(√u^3 )]_(10) ^0 )) =(( (2/5)((√(20^5 ))−(√(10^5 )))−10(2/3)((√(20^3 ))−(√(10^3 ))) )/( 0−((2/5)(√(10^5 ))−10(2/3)(√(10^3 ))) )) ×((3(√(10)))/(3(√(10)))) =(((6/5)(4000(√2)−1000)−20(200(√2)−100))/(−(6/5)(1000)+20(100))) =((4800(√2)−1200−4000(√2)+2000)/(−1200+2000)) =((800(√2)+800)/(800)) =1+(√2) so S≒1+(√2)](https://www.tinkutara.com/question/Q7909.png)
$${S}\:=\:\frac{\:\:\underset{{k}=\mathrm{1}} {\overset{\mathrm{99}} {\sum}}\sqrt{\mathrm{10}+\sqrt{{k}}}\:}{\:\:\underset{{k}=\mathrm{1}} {\overset{\mathrm{99}} {\sum}}\sqrt{\mathrm{10}−\sqrt{{k}}}\:} \\ $$$${S}\fallingdotseq{S}\:'=\frac{\:\:\int_{\mathrm{0}} ^{\mathrm{100}} \sqrt{\mathrm{10}+\sqrt{{x}}\:}\:{dx}\:}{\:\:\int_{\mathrm{0}} ^{\mathrm{100}} \sqrt{\mathrm{10}−\sqrt{{y}}\:}\:{dy}\:} \\ $$$${t}=\mathrm{10}+\sqrt{{x}} \\ $$$${x}=\left({t}−\mathrm{10}\right)^{\mathrm{2}} \\ $$$$\:\:{dx}=\mathrm{2}{t}−\mathrm{20}\:{dt} \\ $$$$\:\:{x}\left(\mathrm{0}\rightarrow\mathrm{100}\right)\Leftrightarrow{t}\left(\mathrm{10}\rightarrow\mathrm{20}\right) \\ $$$$ \\ $$$${u}=\mathrm{10}−\sqrt{{y}} \\ $$$${y}=\left(\mathrm{10}−{u}\right)^{\mathrm{2}} \\ $$$$\:\:{dy}=\mathrm{2}{t}−\mathrm{20}\:{du} \\ $$$$\:\:{y}\left(\mathrm{0}\rightarrow\mathrm{100}\right)\Leftrightarrow{u}\left(\mathrm{10}\rightarrow\mathrm{0}\right) \\ $$$$ \\ $$$${S}\:'=\frac{\int_{\mathrm{10}} ^{\mathrm{20}} \sqrt{\mathrm{10}+\sqrt{\left({t}−\mathrm{10}\right)^{\mathrm{2}} }}\:\left(\mathrm{2}{t}−\mathrm{20}\right){dt}}{\int_{\mathrm{10}} ^{\mathrm{0}} \sqrt{\mathrm{10}−\sqrt{\left(\mathrm{10}−{u}\right)^{\mathrm{2}} }}\:\left(\mathrm{2}{u}+\mathrm{20}\right){du}} \\ $$$$=\frac{\int_{\mathrm{10}} ^{\mathrm{20}} \sqrt{\mathrm{10}+\mid{t}−\mathrm{10}\mid}\:\left(\mathrm{2}{t}−\mathrm{20}\right){dt}}{\int_{\mathrm{10}} ^{\mathrm{0}} \sqrt{\mathrm{10}−\mid\mathrm{10}−{u}\mid}\:\left(\mathrm{2}{u}−\mathrm{20}\right){du}} \\ $$$$=\frac{\int_{\mathrm{10}} ^{\mathrm{20}} \mathrm{2}\sqrt{{t}}\:\left({t}−\mathrm{10}\right){dt}}{\int_{\mathrm{10}} ^{\mathrm{0}} \mathrm{2}\sqrt{{u}}\:\left({u}−\mathrm{10}\right){du}} \\ $$$$=\frac{\int_{\mathrm{10}} ^{\mathrm{20}} \sqrt{{t}^{\mathrm{3}} }−\mathrm{10}\sqrt{{t}}\:{dt}}{\int_{\mathrm{10}} ^{\mathrm{0}} \sqrt{{u}^{\mathrm{3}} }−\mathrm{10}\sqrt{{u}}\:{du}} \\ $$$$=\frac{\:\:\left[\frac{\mathrm{2}}{\mathrm{5}}\sqrt{{t}^{\mathrm{5}} }−\mathrm{10}\frac{\mathrm{2}}{\mathrm{3}}\sqrt{{t}^{\mathrm{3}} }\right]_{\mathrm{10}} ^{\mathrm{20}} \:\:}{\:\:\left[\frac{\mathrm{2}}{\mathrm{5}}\sqrt{{u}^{\mathrm{5}} }−\mathrm{10}\frac{\mathrm{2}}{\mathrm{3}}\sqrt{{u}^{\mathrm{3}} }\right]_{\mathrm{10}} ^{\mathrm{0}} \:\:} \\ $$$$=\frac{\:\:\frac{\mathrm{2}}{\mathrm{5}}\left(\sqrt{\mathrm{20}^{\mathrm{5}} }−\sqrt{\mathrm{10}^{\mathrm{5}} }\right)−\mathrm{10}\frac{\mathrm{2}}{\mathrm{3}}\left(\sqrt{\mathrm{20}^{\mathrm{3}} }−\sqrt{\mathrm{10}^{\mathrm{3}} }\right)\:\:}{\:\:\mathrm{0}−\left(\frac{\mathrm{2}}{\mathrm{5}}\sqrt{\mathrm{10}^{\mathrm{5}} }−\mathrm{10}\frac{\mathrm{2}}{\mathrm{3}}\sqrt{\mathrm{10}^{\mathrm{3}} }\right)\:\:}\:×\frac{\mathrm{3}\sqrt{\mathrm{10}}}{\mathrm{3}\sqrt{\mathrm{10}}} \\ $$$$=\frac{\frac{\mathrm{6}}{\mathrm{5}}\left(\mathrm{4000}\sqrt{\mathrm{2}}−\mathrm{1000}\right)−\mathrm{20}\left(\mathrm{200}\sqrt{\mathrm{2}}−\mathrm{100}\right)}{−\frac{\mathrm{6}}{\mathrm{5}}\left(\mathrm{1000}\right)+\mathrm{20}\left(\mathrm{100}\right)} \\ $$$$=\frac{\mathrm{4800}\sqrt{\mathrm{2}}−\mathrm{1200}−\mathrm{4000}\sqrt{\mathrm{2}}+\mathrm{2000}}{−\mathrm{1200}+\mathrm{2000}} \\ $$$$=\frac{\mathrm{800}\sqrt{\mathrm{2}}+\mathrm{800}}{\mathrm{800}} \\ $$$$=\mathrm{1}+\sqrt{\mathrm{2}} \\ $$$${so} \\ $$$${S}\fallingdotseq\mathrm{1}+\sqrt{\mathrm{2}} \\ $$
Commented by tawakalitu last updated on 24/Sep/16
$${Wow}\:{thanks}\:{so}\:{much}.\:{i}\:{really}\:{appreciate}. \\ $$
Commented by Yozzia last updated on 24/Sep/16
$${How}\:{come}\:{S}\:\fallingdotseq\:{S}\:'\:? \\ $$
Commented by sou1618 last updated on 24/Sep/16
$$\sqrt{\mathrm{10}\pm\sqrt{{k}}}\fallingdotseq\int_{{k}−\mathrm{1}} ^{{k}} \sqrt{\mathrm{10}\pm\sqrt{{x}}}\:{dx} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{100}} {\sum}}\sqrt{\mathrm{10}\pm\sqrt{{k}}}\fallingdotseq\underset{{k}=\mathrm{1}} {\overset{\mathrm{100}} {\sum}}\int_{{k}−\mathrm{1}} ^{\:{k}} \sqrt{\mathrm{10}\pm\sqrt{{x}}}\:{dx} \\ $$$$\left(\underset{{k}=\mathrm{1}} {\overset{\mathrm{99}} {\sum}}\sqrt{\mathrm{10}\pm\sqrt{{k}}}\right)+\sqrt{\mathrm{10}\pm\mathrm{10}}\fallingdotseq\int_{\mathrm{0}} ^{\mathrm{100}} \sqrt{\mathrm{10}\pm\sqrt{{x}}}\:{dx} \\ $$$${now} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{99}} {\sum}}\sqrt{\mathrm{10}\pm\sqrt{{k}}}>>\sqrt{\mathrm{10}\pm\mathrm{10}} \\ $$$${I}\:{think}\:\sqrt{\mathrm{10}+\mathrm{10}}\:{is}\:{less}\:{than}\:\sqrt{\mathrm{2}}\%\:{of}\:\:\Sigma \\ $$$$\:{because}…\:{when}\:\mathrm{0}<{k}<\mathrm{100} \\ $$$$\:\:\:\sqrt{\mathrm{2}}\sqrt{\mathrm{10}+\sqrt{{k}}}>\sqrt{\mathrm{10}+\mathrm{10}} \\ $$$$ \\ $$$$ \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{99}} {\sum}}\sqrt{\mathrm{10}\pm\sqrt{{k}}}\fallingdotseq\int_{\mathrm{0}} ^{\mathrm{100}} \sqrt{\mathrm{10}\pm\sqrt{{x}}}\:{dx} \\ $$$${so} \\ $$$${S}\fallingdotseq{S}\:' \\ $$
Answered by prakash jain last updated on 02/Oct/16
$$\mathrm{answer}\:\mathrm{n}\:\mathrm{comments} \\ $$