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I-n-0-1-1-u-ud-u-Demonstrate-that-n-N-I-n-1-I-n-1-u-n-u-3-2-d-u-and-deduce-the-meaning-of-variations-of-I-n-N-

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Question Number 171090 by Kodjo last updated on 07/Jun/22
I_n =∫_0 ^1 (1−u)(√(ud(u))) Demonstrate that ∀n∈N, I_(n+1) −I_n =(1−u)^n u^(3/2) d(u) and deduce the meaning of variations of (I_n )∈N
$${I}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{u}\right)\sqrt{{ud}\left({u}\right)} \\ $$$${Demonstrate}\:{that}\:\forall{n}\in{N},\:{I}_{{n}+\mathrm{1}} −{I}_{{n}} =\left(\mathrm{1}−{u}\right)^{{n}} {u}^{\frac{\mathrm{3}}{\mathrm{2}}} {d}\left({u}\right)\:\:{and}\:{deduce}\:{the}\:{meaning}\:{of}\:{variations}\:{of}\:\left({I}_{{n}} \right)\in{N} \\ $$
Commented by Kodjo last updated on 07/Jun/22
Excuse me it's a mistake here is the real subject
Commented by Kodjo last updated on 07/Jun/22
Commented by Kodjo last updated on 07/Jun/22
Answered by aleks041103 last updated on 09/Jun/22
I_(n+1) =∫_0 ^1 (1−u)^(n+1) (√u) du= =∫_0 ^1 (1−u)(1−u)^n (√u) du= =∫_0 ^1 (1−u)^n (√u) du−∫_0 ^1 (1−u)^n u(√u) du= =I_n +∫_1 ^0 (1−u)^n u^(3/2) du ⇒I_(n+1) −I_n =∫_1 ^0 (1−u)^n u^(3/2) du
$${I}_{{n}+\mathrm{1}} =\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{u}\right)^{{n}+\mathrm{1}} \sqrt{{u}}\:{du}= \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{u}\right)\left(\mathrm{1}−{u}\right)^{{n}} \sqrt{{u}}\:{du}= \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{u}\right)^{{n}} \sqrt{{u}}\:{du}−\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{u}\right)^{{n}} {u}\sqrt{{u}}\:{du}= \\ $$$$={I}_{{n}} +\int_{\mathrm{1}} ^{\mathrm{0}} \left(\mathrm{1}−{u}\right)^{{n}} {u}^{\mathrm{3}/\mathrm{2}} {du} \\ $$$$\Rightarrow{I}_{{n}+\mathrm{1}} −{I}_{{n}} =\int_{\mathrm{1}} ^{\mathrm{0}} \left(\mathrm{1}−{u}\right)^{{n}} {u}^{\mathrm{3}/\mathrm{2}} {du} \\ $$$$ \\ $$
Commented by aleks041103 last updated on 09/Jun/22
I_(n+1) =∫_0 ^1 (1−u)^(n+1) u^(1/2) du= =∫_0 ^1 (1−u)^(n+1) d((2/3)u^(3/2) )= ={(2/3)u^(3/2) (1−u)^(n+1) }_0 ^1 −((2(n+1))/3)∫_1 ^0 (1−u)^n u^(3/2) du= =−((2(n+1))/3)(I_(n+1) −I_n ) ⇒((2n+5)/3)I_(n+1) =((2n+2)/3)I_n ⇒I_(n+1) =((2n+2)/(2n+5))I_n ⇒I_n =((2n)/(2n+3)) ((2n−2)/(2n+1)) ... ((2.0+2)/(2.0+5))I_0 I_0 =∫_0 ^1 (√u)du=(2/3)u^(3/2) ]_0 ^1 =(2/3) ⇒I_n =((2.2.4.6. ... . 2n)/(3.5.7. ... .(2n+3)))=((2.2^n (1.2. ... .n))/((2n+3)!!)) I_n =((2^(n+1) n!)/((2n+3)!!)) but (2n+1)!!=1.3. ... .(2n−1)(2n+1)= =((1.2.3. ... .(2n−1)(2n)(2n+1))/(2.4. ... .(2n−2)(2n)))= =(((2n+1)!)/(2^n n!)) ⇒(2n+3)!!=(2(n+1)+1)!!=(((2n+3)!)/(2^(n+1) (n+1)!)) ⇒I_n =((4^(n+1) (n+1)!(n!))/((2n+3)!))
$${I}_{{n}+\mathrm{1}} =\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{u}\right)^{{n}+\mathrm{1}} {u}^{\mathrm{1}/\mathrm{2}} {du}= \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{u}\right)^{{n}+\mathrm{1}} {d}\left(\frac{\mathrm{2}}{\mathrm{3}}{u}^{\mathrm{3}/\mathrm{2}} \right)= \\ $$$$=\left\{\frac{\mathrm{2}}{\mathrm{3}}{u}^{\mathrm{3}/\mathrm{2}} \left(\mathrm{1}−{u}\right)^{{n}+\mathrm{1}} \right\}_{\mathrm{0}} ^{\mathrm{1}} −\frac{\mathrm{2}\left({n}+\mathrm{1}\right)}{\mathrm{3}}\int_{\mathrm{1}} ^{\mathrm{0}} \left(\mathrm{1}−{u}\right)^{{n}} {u}^{\mathrm{3}/\mathrm{2}} {du}= \\ $$$$=−\frac{\mathrm{2}\left({n}+\mathrm{1}\right)}{\mathrm{3}}\left({I}_{{n}+\mathrm{1}} −{I}_{{n}} \right) \\ $$$$\Rightarrow\frac{\mathrm{2}{n}+\mathrm{5}}{\mathrm{3}}{I}_{{n}+\mathrm{1}} =\frac{\mathrm{2}{n}+\mathrm{2}}{\mathrm{3}}{I}_{{n}} \\ $$$$\Rightarrow{I}_{{n}+\mathrm{1}} =\frac{\mathrm{2}{n}+\mathrm{2}}{\mathrm{2}{n}+\mathrm{5}}{I}_{{n}} \\ $$$$\Rightarrow{I}_{{n}} =\frac{\mathrm{2}{n}}{\mathrm{2}{n}+\mathrm{3}}\:\frac{\mathrm{2}{n}−\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}\:…\:\frac{\mathrm{2}.\mathrm{0}+\mathrm{2}}{\mathrm{2}.\mathrm{0}+\mathrm{5}}{I}_{\mathrm{0}} \\ $$$$\left.{I}_{\mathrm{0}} =\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{{u}}{du}=\frac{\mathrm{2}}{\mathrm{3}}{u}^{\mathrm{3}/\mathrm{2}} \right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\Rightarrow{I}_{{n}} =\frac{\mathrm{2}.\mathrm{2}.\mathrm{4}.\mathrm{6}.\:…\:.\:\mathrm{2}{n}}{\mathrm{3}.\mathrm{5}.\mathrm{7}.\:…\:.\left(\mathrm{2}{n}+\mathrm{3}\right)}=\frac{\mathrm{2}.\mathrm{2}^{{n}} \left(\mathrm{1}.\mathrm{2}.\:…\:.{n}\right)}{\left(\mathrm{2}{n}+\mathrm{3}\right)!!} \\ $$$${I}_{{n}} =\frac{\mathrm{2}^{{n}+\mathrm{1}} {n}!}{\left(\mathrm{2}{n}+\mathrm{3}\right)!!} \\ $$$${but} \\ $$$$\left(\mathrm{2}{n}+\mathrm{1}\right)!!=\mathrm{1}.\mathrm{3}.\:…\:.\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)= \\ $$$$=\frac{\mathrm{1}.\mathrm{2}.\mathrm{3}.\:…\:.\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{2}.\mathrm{4}.\:…\:.\left(\mathrm{2}{n}−\mathrm{2}\right)\left(\mathrm{2}{n}\right)}= \\ $$$$=\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{\mathrm{2}^{{n}} {n}!} \\ $$$$\Rightarrow\left(\mathrm{2}{n}+\mathrm{3}\right)!!=\left(\mathrm{2}\left({n}+\mathrm{1}\right)+\mathrm{1}\right)!!=\frac{\left(\mathrm{2}{n}+\mathrm{3}\right)!}{\mathrm{2}^{{n}+\mathrm{1}} \left({n}+\mathrm{1}\right)!} \\ $$$$\Rightarrow{I}_{{n}} =\frac{\mathrm{4}^{{n}+\mathrm{1}} \left({n}+\mathrm{1}\right)!\left({n}!\right)}{\left(\mathrm{2}{n}+\mathrm{3}\right)!} \\ $$
Commented by Kodjo last updated on 09/Jun/22
Thanks
$${Thanks} \\ $$
Commented by ilhamQ last updated on 11/Jun/22
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$$\left.\right]\left.\right]\left.\right]\left.\right]\left.\right]\left.\right]\left.\right] \\ $$$$ \\ $$

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