Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 56280 by rahul 19 last updated on 13/Mar/19

Evaluate : 1) ((∫_0 ^( 1_ ) (1−(1−x^2 )^(100) )^(201) .xdx)/(∫_0 ^( 1) (1−(1−x^2 )^(100) )^(202) .xdx)) = ? 2) ((∫_0 ^( 1) (1−x^(200) )^(201) dx)/(∫_0 ^( 1) (1−x^(200) )^(202) dx)) = ?

$${Evaluate}\:: \\ $$$$\left.\mathrm{1}\right)\:\frac{\int_{\mathrm{0}} ^{\:\mathrm{1}_{} } \left(\mathrm{1}−\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{100}} \right)^{\mathrm{201}} \:.{xdx}}{\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\mathrm{1}−\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{100}} \right)^{\mathrm{202}} .{xdx}}\:=\:? \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\:\frac{\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\mathrm{1}−{x}^{\mathrm{200}} \right)^{\mathrm{201}} {dx}}{\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\mathrm{1}−{x}^{\mathrm{200}} \right)^{\mathrm{202}} {dx}}\:=\:? \\ $$

Commented by rahul 19 last updated on 13/Mar/19

Ans: 1) ((20201)/(20200 )) 2)((40401)/(40400)) .

$${Ans}: \\ $$$$\left.\mathrm{1}\left.\right)\:\frac{\mathrm{20201}}{\mathrm{20200}\:}\:\:\:\:\:\:\:\:\:\mathrm{2}\right)\frac{\mathrm{40401}}{\mathrm{40400}}\:. \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 13/Mar/19

rahul pls check the answer

$${rahul}\:{pls}\:{check}\:{the}\:{answer} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 13/Mar/19

2)t=x^(200) dt=200x^(199) dx dt=200×(t^(1/(200)) )^(199) dx dx=(dt/(200×t^((199)/(200)) )) ((∫_0 ^1 (1−x^(200) )^(201) dx)/(∫_0 ^1 (1−x^(200) )^(202) dx)) ((∫_0 ^1 (1−t)^(201) ×(dt/(200×t^((199)/(200)) )))/(∫_0 ^1 (1−t)^(202) ×(dt/(200×t^((199)/(200)) )))) ((∫_0 ^1 (1−t)^(201) ×t^((1/(200))−1) dt)/(∫_0 ^1 (1−t)^(202) ×t^((1/(200))−1) dt)) now look formula β function β(m,n)=∫_0 ^1 x^(m−1) (1−x)^(n−1) dx =((⌈(m)⌈(n))/(⌈(m+n))) =(((⌈(202)⌈((1/(200))))/(⌈(202+(1/(200)))))/((⌈(203)⌈((1/(200))))/(⌈(203+(1/(200)))))) now ⌈(n+1)=n! =(((201!)/((201+(1/(200)))!))/((202!)/((202+(1/(200)))!))) =((201!×(202+(1/(200)))!)/(202!×(201+(1/(200)))!)) =(1/(202))×(202+(1/(200)))×(201+(1/(200)))!×(1/((201+(1/(200)))!)) =1+(1/(200×202)) =l

$$\left.\mathrm{2}\right){t}={x}^{\mathrm{200}} \:\:{dt}=\mathrm{200}{x}^{\mathrm{199}} {dx} \\ $$$${dt}=\mathrm{200}×\left({t}^{\frac{\mathrm{1}}{\mathrm{200}}} \right)^{\mathrm{199}} {dx} \\ $$$${dx}=\frac{{dt}}{\mathrm{200}×{t}^{\frac{\mathrm{199}}{\mathrm{200}}} } \\ $$$$\frac{\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}^{\mathrm{200}} \right)^{\mathrm{201}} {dx}}{\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}^{\mathrm{200}} \right)^{\mathrm{202}} {dx}} \\ $$$$\frac{\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{t}\right)^{\mathrm{201}} ×\frac{{dt}}{\mathrm{200}×{t}^{\frac{\mathrm{199}}{\mathrm{200}}} }}{\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{t}\right)^{\mathrm{202}} ×\frac{{dt}}{\mathrm{200}×{t}^{\frac{\mathrm{199}}{\mathrm{200}}} }} \\ $$$$\frac{\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{t}\right)^{\mathrm{201}} ×{t}^{\frac{\mathrm{1}}{\mathrm{200}}−\mathrm{1}} {dt}}{\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{t}\right)^{\mathrm{202}} ×{t}^{\frac{\mathrm{1}}{\mathrm{200}}−\mathrm{1}} {dt}} \\ $$$${now}\:{look}\:{formula}\:\beta\:{function} \\ $$$$\beta\left({m},{n}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{m}−\mathrm{1}} \left(\mathrm{1}−{x}\right)^{{n}−\mathrm{1}} {dx} \\ $$$$=\frac{\lceil\left({m}\right)\lceil\left({n}\right)}{\lceil\left({m}+{n}\right)} \\ $$$$=\frac{\frac{\lceil\left(\mathrm{202}\right)\lceil\left(\frac{\mathrm{1}}{\mathrm{200}}\right)}{\lceil\left(\mathrm{202}+\frac{\mathrm{1}}{\mathrm{200}}\right)}}{\frac{\lceil\left(\mathrm{203}\right)\lceil\left(\frac{\mathrm{1}}{\mathrm{200}}\right)}{\lceil\left(\mathrm{203}+\frac{\mathrm{1}}{\mathrm{200}}\right)}} \\ $$$${now}\:\lceil\left({n}+\mathrm{1}\right)={n}! \\ $$$$=\frac{\frac{\mathrm{201}!}{\left(\mathrm{201}+\frac{\mathrm{1}}{\mathrm{200}}\right)!}}{\frac{\mathrm{202}!}{\left(\mathrm{202}+\frac{\mathrm{1}}{\mathrm{200}}\right)!}} \\ $$$$=\frac{\mathrm{201}!×\left(\mathrm{202}+\frac{\mathrm{1}}{\mathrm{200}}\right)!}{\mathrm{202}!×\left(\mathrm{201}+\frac{\mathrm{1}}{\mathrm{200}}\right)!} \\ $$$$=\frac{\mathrm{1}}{\mathrm{202}}×\left(\mathrm{202}+\frac{\mathrm{1}}{\mathrm{200}}\right)×\left(\mathrm{201}+\frac{\mathrm{1}}{\mathrm{200}}\right)!×\frac{\mathrm{1}}{\left(\mathrm{201}+\frac{\mathrm{1}}{\mathrm{200}}\right)!} \\ $$$$=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{200}×\mathrm{202}} \\ $$$$={l} \\ $$$$ \\ $$$$ \\ $$

Commented by rahul 19 last updated on 13/Mar/19

It will be = (1/(202))×(202+(1/(200)))= 1+(1/(202×200)) = ((40401)/(40400)) Ans.! Thank you Sir!

$${It}\:{will}\:{be}\: \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{202}}×\left(\mathrm{202}+\frac{\mathrm{1}}{\mathrm{200}}\right)=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{202}×\mathrm{200}} \\ $$$$=\:\frac{\mathrm{40401}}{\mathrm{40400}}\:{Ans}.! \\ $$$${Thank}\:{you}\:{Sir}! \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 13/Mar/19

thank you rahul...sometimes i loss consistsncy...

$${thank}\:{you}\:{rahul}...{sometimes}\:{i}\:{loss}\:{consistsncy}... \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 13/Mar/19

1)((∫_0 ^1 {1−(1−x^2 )^a }^(2a+1) .xdx)/(∫_0 ^1 {1−(1−x^2 )^a }^(2a+2) .xdx)) a=100 t=1−x^2 dt=−2xdx (((1/2)∫_1 ^0 (1−t^a )^(2a+1) ×−dt)/((1/2)∫_1 ^0 (1−t^a )^(2a+2) ×−dt)) ((∫_0 ^1 (1−t^a )^(2a+1) dt)/(∫_0 ^1 (1−t^a )^(2a+2) dt)) ((∫_0 ^1 (1−t^(100) )^(201) dt)/(∫_0 ^1 (1−t^(100) )^(202) dt)) look same type question of question no 2 but solving other way let t^(100) =sin^2 α →100t^(99) dt=2sinαcosαdα dt=((2sinαcosαdα)/(100(sin^2 α)^(1/(100)) ))=(1/(50))×(sinα)^(1−(1/(50))) ×cosαdα ((∫_0 ^(π/2) (1−sin^2 α)^(201) ×(1/(50))×(sinα)^(1−(1/(50))) ×cosαdα)/(∫_0 ^(π/2) (1−sin^2 α)^(202) ×(1/(50))×(sinα)^(1−(1/(50))) ×cosαdα))(/) ((∫^(π/2) cos^(402) α)(sinα)^(1−(1/(50))) ×cosαdα)/(∫_0 ^(π/2) cos^(404) ×sin^(1−(1/(50))) ×cosαdα)) ((∫_0 ^(π/2) (sinα)^(1−(1/(50))) ×(cosα)^(403) dα)/(∫_0 ^(π/2) (sinα)^(1−(1/(50))) ×(cosα)^(405) dα)) now formulA 2∫_0 ^(π/2) sin^(2p−1) αcos^(2q−1) αdα=((⌈(p)⌈(q))/(⌈(p+q))) 2p−1=1−(1/(50)) →2p=2−(1/(50)) →p=1−(1/(100)) 2q−1=403→2q=404→q=202 2q_1 −1=405→2q_1 =406→q_1 =203 q_1 =q+1→⌈(q_1 )=⌈(q+1)→q⌈(q) p+q_1 =p+q+1→⌈(p+q_1 )=⌈(p+q+1)→(p+q)⌈(p+q) ((2∫^(π/2) sin^(2p−1) αcos^(2q−1) αdα)/(2∫_0 ^(π/2) sin^(2p−1) αcos^(2q_1 −1) αdα)) =(((⌈(p)⌈(q))/(⌈(p+q)))/((⌈(p)⌈(q_1 ))/(⌈(p+q_1 )))) =((⌈(p)⌈(q)⌈(p+q_1 ))/(⌈(p)⌈(q_1 )⌈(p+q))) =((⌈(q)×(p+q)⌈(p+q))/(q⌈(q)×⌈(p+q))) =((p+q)/q)=((1−(1/(100))+202)/(202))=((100−1+20200)/(20200))=((20300−1)/(20200)) =((20299)/(20200)) Rahul pls check...

$$\left.\mathrm{1}\right)\frac{\int_{\mathrm{0}} ^{\mathrm{1}} \left\{\mathrm{1}−\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{{a}} \right\}^{\mathrm{2}{a}+\mathrm{1}} .{xdx}}{\int_{\mathrm{0}} ^{\mathrm{1}} \left\{\mathrm{1}−\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{{a}} \right\}^{\mathrm{2}{a}+\mathrm{2}} .{xdx}}\:\:\:{a}=\mathrm{100} \\ $$$${t}=\mathrm{1}−{x}^{\mathrm{2}} \:\:\:{dt}=−\mathrm{2}{xdx} \\ $$$$\frac{\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\mathrm{0}} \left(\mathrm{1}−{t}^{{a}} \right)^{\mathrm{2}{a}+\mathrm{1}} ×−{dt}}{\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\mathrm{0}} \left(\mathrm{1}−{t}^{{a}} \right)^{\mathrm{2}{a}+\mathrm{2}} ×−{dt}} \\ $$$$\frac{\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{t}^{{a}} \right)^{\mathrm{2}{a}+\mathrm{1}} {dt}}{\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{t}^{{a}} \right)^{\mathrm{2}{a}+\mathrm{2}} {dt}} \\ $$$$\frac{\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{t}^{\mathrm{100}} \right)^{\mathrm{201}} {dt}}{\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{t}^{\mathrm{100}} \right)^{\mathrm{202}} {dt}} \\ $$$${look}\:{same}\:{type}\:\:{question}\:\:{of}\:{question}\:{no}\:\mathrm{2} \\ $$$${but}\:{solving}\:{other}\:{way} \\ $$$${let}\:{t}^{\mathrm{100}} ={sin}^{\mathrm{2}} \alpha\:\:\:\rightarrow\mathrm{100}{t}^{\mathrm{99}} {dt}=\mathrm{2}{sin}\alpha{cos}\alpha{d}\alpha \\ $$$${dt}=\frac{\mathrm{2}{sin}\alpha{cos}\alpha{d}\alpha}{\mathrm{100}\left({sin}^{\mathrm{2}} \alpha\right)^{\frac{\mathrm{1}}{\mathrm{100}}} }=\frac{\mathrm{1}}{\mathrm{50}}×\left({sin}\alpha\right)^{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{50}}} ×{cos}\alpha{d}\alpha \\ $$$$\frac{\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{1}−{sin}^{\mathrm{2}} \alpha\right)^{\mathrm{201}} ×\frac{\mathrm{1}}{\mathrm{50}}×\left({sin}\alpha\right)^{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{50}}} ×{cos}\alpha{d}\alpha}{\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{1}−{sin}^{\mathrm{2}} \alpha\right)^{\mathrm{202}} ×\frac{\mathrm{1}}{\mathrm{50}}×\left({sin}\alpha\right)^{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{50}}} ×{cos}\alpha{d}\alpha}\frac{}{} \\ $$$$\frac{\left.\int^{\frac{\pi}{\mathrm{2}}} {cos}^{\mathrm{402}} \alpha\right)\left({sin}\alpha\right)^{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{50}}} \:×{cos}\alpha{d}\alpha}{\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}^{\mathrm{404}} ×{sin}^{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{50}}} ×{cos}\alpha{d}\alpha} \\ $$$$\frac{\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({sin}\alpha\right)^{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{50}}} ×\left({cos}\alpha\right)^{\mathrm{403}} {d}\alpha}{\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({sin}\alpha\right)^{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{50}}} ×\left({cos}\alpha\right)^{\mathrm{405}} {d}\alpha} \\ $$$$ \\ $$$${now}\:{formulA} \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{p}−\mathrm{1}} \alpha{cos}^{\mathrm{2}{q}−\mathrm{1}} \alpha{d}\alpha=\frac{\lceil\left({p}\right)\lceil\left({q}\right)}{\lceil\left({p}+{q}\right)} \\ $$$$ \\ $$$$\mathrm{2}{p}−\mathrm{1}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{50}}\:\rightarrow\mathrm{2}{p}=\mathrm{2}−\frac{\mathrm{1}}{\mathrm{50}}\:\rightarrow{p}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{100}} \\ $$$$\mathrm{2}{q}−\mathrm{1}=\mathrm{403}\rightarrow\mathrm{2}{q}=\mathrm{404}\rightarrow{q}=\mathrm{202} \\ $$$$\mathrm{2}{q}_{\mathrm{1}} −\mathrm{1}=\mathrm{405}\rightarrow\mathrm{2}{q}_{\mathrm{1}} =\mathrm{406}\rightarrow{q}_{\mathrm{1}} =\mathrm{203} \\ $$$${q}_{\mathrm{1}} ={q}+\mathrm{1}\rightarrow\lceil\left({q}_{\mathrm{1}} \right)=\lceil\left({q}+\mathrm{1}\right)\rightarrow{q}\lceil\left({q}\right) \\ $$$${p}+{q}_{\mathrm{1}} ={p}+{q}+\mathrm{1}\rightarrow\lceil\left({p}+{q}_{\mathrm{1}} \right)=\lceil\left({p}+{q}+\mathrm{1}\right)\rightarrow\left({p}+{q}\right)\lceil\left({p}+{q}\right) \\ $$$$\frac{\mathrm{2}\int^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{p}−\mathrm{1}} \alpha{cos}^{\mathrm{2}{q}−\mathrm{1}} \alpha{d}\alpha}{\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{p}−\mathrm{1}} \alpha{cos}^{\mathrm{2}{q}_{\mathrm{1}} −\mathrm{1}} \alpha{d}\alpha}\:=\frac{\frac{\lceil\left({p}\right)\lceil\left({q}\right)}{\lceil\left({p}+{q}\right)}}{\frac{\lceil\left({p}\right)\lceil\left({q}_{\mathrm{1}} \right)}{\lceil\left({p}+{q}_{\mathrm{1}} \right)}} \\ $$$$=\frac{\lceil\left({p}\right)\lceil\left({q}\right)\lceil\left({p}+{q}_{\mathrm{1}} \right)}{\lceil\left({p}\right)\lceil\left({q}_{\mathrm{1}} \right)\lceil\left({p}+{q}\right)} \\ $$$$=\frac{\lceil\left({q}\right)×\left({p}+{q}\right)\lceil\left({p}+{q}\right)}{{q}\lceil\left({q}\right)×\lceil\left({p}+{q}\right)} \\ $$$$=\frac{{p}+{q}}{{q}}=\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{100}}+\mathrm{202}}{\mathrm{202}}=\frac{\mathrm{100}−\mathrm{1}+\mathrm{20200}}{\mathrm{20200}}=\frac{\mathrm{20300}−\mathrm{1}}{\mathrm{20200}} \\ $$$$=\frac{\mathrm{20299}}{\mathrm{20200}}\:{Rahul}\:{pls}\:{check}... \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com