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Question Number 61229 by maxmathsup by imad last updated on 30/May/19

let f_n (a) =∫_0 ^a  x^n (√(a^2 −x^2 ))dx  with a>0  1) determine a explicit form of f(a)  2) let g_n (a) =f^′ (a)   give g_n (a) at form of integral and give its  value   3) find the value of  ∫_0 ^2  x^3 (√(4−x^2 ))dx  and ∫_0 ^(√3) x^4 (√(3−x^2 ))dx

$${let}\:{f}_{{n}} \left({a}\right)\:=\int_{\mathrm{0}} ^{{a}} \:{x}^{{n}} \sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }{dx}\:\:{with}\:{a}>\mathrm{0} \\ $$ $$\left.\mathrm{1}\right)\:{determine}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left({a}\right) \\ $$ $$\left.\mathrm{2}\right)\:{let}\:{g}_{{n}} \left({a}\right)\:={f}^{'} \left({a}\right)\:\:\:{give}\:{g}_{{n}} \left({a}\right)\:{at}\:{form}\:{of}\:{integral}\:{and}\:{give}\:{its} \\ $$ $${value}\: \\ $$ $$\left.\mathrm{3}\right)\:{find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\mathrm{2}} \:{x}^{\mathrm{3}} \sqrt{\mathrm{4}−{x}^{\mathrm{2}} }{dx}\:\:{and}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} {x}^{\mathrm{4}} \sqrt{\mathrm{3}−{x}^{\mathrm{2}} }{dx}\: \\ $$

Answered by perlman last updated on 30/May/19

1) put x=asin(t)  fn(a)=a^(n+1) ∫_0 ^(π/2) sin^n (t)(√((a^2 −a^2 sin^2 (t) )) cos(t)dt=  =a^(n+2) ∫_0 ^(π/2) sin^n (t)cos^2 (t)dt=a∫sin^n (t)dt−a^(n+2) ∫sin^(n+2) (t)dt  let I_n =∫_0 ^(π/2) sin^n (t)dt=∫sin(t)sin^(n−1) (t)dt=[−cos(t)sin^(n−1) (t)]+(n−1)∫cos^2 (t)sin^(n−2) (t)dt  (n−1)∫_0 ^(π/2) (1−sin^2 (t))sin^((n−2)) (t)dt=(n−1)I_(n−2) −(n−1)I_n =I_n   I_n =((n−1)/n)I_(n−2)   I_0 =(π/2)  I_1 =1  I_(2n) =((2n−1)/(2n))I_(2(n−1))   I_(2n) =((2n−1)/(2n)).((2(n−1)−1)/(2(n−1)))......((2−1)/2)I_0 =(((2n−1)(2n−3)....(1))/(2n.2(n−1)....2(1)))I_0   =((2n(2n−1)(2n−2).......1)/([2^n n!]^2 ))I_0 =(((2n)!)/(2^(2n) (n!)^2 ))(π/2)  I_(2n+1) =((2n)/(2n+1))I_(2n−1) =((2n)/(2n+1)).((2n−2)/(2n−1))......(2/3)I_1 =(((2^n n!)^2 .2)/((2n+1)!))=((2^(2n+1) (n!)^2 )/((2n+1)!))  fn(a)=a^(n+2) (I_n −I_(n+2) )  gn(a)=(d/da)∫_0 ^a x^n (√((a^2 −x^2 )))dx=(d/da)∫_0 ^1 a^n t^n a^2 (√((1−t^2 )))dt=∫_0 ^1 (d/da)(a^(n+2) t^n (√((1−t^2 )) dt)=  (n+2)∫_0 ^1 a^(n+1) t^n (√((1−t^2 )) dt  ∫_0 ^2 x^3 (√((4−x^2 )))dx  n=3 a=2   =2^5 (I1−I3)=2^5 (1−(2/3))=((32)/3)

$$\left.\mathrm{1}\right)\:{put}\:{x}={asin}\left({t}\right) \\ $$ $${fn}\left({a}\right)={a}^{{n}+\mathrm{1}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{{n}} \left({t}\right)\sqrt{\left({a}^{\mathrm{2}} −{a}^{\mathrm{2}} {sin}^{\mathrm{2}} \left({t}\right)\:\right.}\:{cos}\left({t}\right){dt}= \\ $$ $$={a}^{{n}+\mathrm{2}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{{n}} \left({t}\right){cos}^{\mathrm{2}} \left({t}\right){dt}={a}\int{sin}^{{n}} \left({t}\right){dt}−{a}^{{n}+\mathrm{2}} \int{sin}^{{n}+\mathrm{2}} \left({t}\right){dt} \\ $$ $${let}\:{I}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{{n}} \left({t}\right){dt}=\int{sin}\left({t}\right){sin}^{{n}−\mathrm{1}} \left({t}\right){dt}=\left[−{cos}\left({t}\right){sin}^{{n}−\mathrm{1}} \left({t}\right)\right]+\left({n}−\mathrm{1}\right)\int{cos}^{\mathrm{2}} \left({t}\right){sin}^{{n}−\mathrm{2}} \left({t}\right){dt} \\ $$ $$\left({n}−\mathrm{1}\right)\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{1}−{sin}^{\mathrm{2}} \left({t}\right)\right){sin}^{\left({n}−\mathrm{2}\right)} \left({t}\right){dt}=\left({n}−\mathrm{1}\right){I}_{{n}−\mathrm{2}} −\left({n}−\mathrm{1}\right){I}_{{n}} ={I}_{{n}} \\ $$ $${I}_{{n}} =\frac{{n}−\mathrm{1}}{{n}}{I}_{{n}−\mathrm{2}} \\ $$ $${I}_{\mathrm{0}} =\frac{\pi}{\mathrm{2}} \\ $$ $${I}_{\mathrm{1}} =\mathrm{1} \\ $$ $${I}_{\mathrm{2}{n}} =\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}{n}}{I}_{\mathrm{2}\left({n}−\mathrm{1}\right)} \\ $$ $${I}_{\mathrm{2}{n}} =\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}{n}}.\frac{\mathrm{2}\left({n}−\mathrm{1}\right)−\mathrm{1}}{\mathrm{2}\left({n}−\mathrm{1}\right)}......\frac{\mathrm{2}−\mathrm{1}}{\mathrm{2}}{I}_{\mathrm{0}} =\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{3}\right)....\left(\mathrm{1}\right)}{\mathrm{2}{n}.\mathrm{2}\left({n}−\mathrm{1}\right)....\mathrm{2}\left(\mathrm{1}\right)}{I}_{\mathrm{0}} \\ $$ $$=\frac{\mathrm{2}{n}\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{2}\right).......\mathrm{1}}{\left[\mathrm{2}^{{n}} {n}!\right]^{\mathrm{2}} }{I}_{\mathrm{0}} =\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }\frac{\pi}{\mathrm{2}} \\ $$ $${I}_{\mathrm{2}{n}+\mathrm{1}} =\frac{\mathrm{2}{n}}{\mathrm{2}{n}+\mathrm{1}}{I}_{\mathrm{2}{n}−\mathrm{1}} =\frac{\mathrm{2}{n}}{\mathrm{2}{n}+\mathrm{1}}.\frac{\mathrm{2}{n}−\mathrm{2}}{\mathrm{2}{n}−\mathrm{1}}......\frac{\mathrm{2}}{\mathrm{3}}{I}_{\mathrm{1}} =\frac{\left(\mathrm{2}^{{n}} {n}!\right)^{\mathrm{2}} .\mathrm{2}}{\left(\mathrm{2}{n}+\mathrm{1}\right)!}=\frac{\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} \left({n}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!} \\ $$ $${fn}\left({a}\right)={a}^{{n}+\mathrm{2}} \left({I}_{{n}} −{I}_{{n}+\mathrm{2}} \right) \\ $$ $${gn}\left({a}\right)=\frac{{d}}{{da}}\int_{\mathrm{0}} ^{{a}} {x}^{{n}} \sqrt{\left({a}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)}{dx}=\frac{{d}}{{da}}\int_{\mathrm{0}} ^{\mathrm{1}} {a}^{{n}} {t}^{{n}} {a}^{\mathrm{2}} \sqrt{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{dt}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{d}}{{da}}\left({a}^{{n}+\mathrm{2}} {t}^{{n}} \sqrt{\left(\mathrm{1}−{t}^{\mathrm{2}} \right.}\:{dt}\right)= \\ $$ $$\left({n}+\mathrm{2}\right)\int_{\mathrm{0}} ^{\mathrm{1}} {a}^{{n}+\mathrm{1}} {t}^{{n}} \sqrt{\left(\mathrm{1}−{t}^{\mathrm{2}} \right.}\:{dt} \\ $$ $$\int_{\mathrm{0}} ^{\mathrm{2}} {x}^{\mathrm{3}} \sqrt{\left(\mathrm{4}−{x}^{\mathrm{2}} \right)}{dx} \\ $$ $${n}=\mathrm{3}\:{a}=\mathrm{2}\: \\ $$ $$=\mathrm{2}^{\mathrm{5}} \left({I}\mathrm{1}−{I}\mathrm{3}\right)=\mathrm{2}^{\mathrm{5}} \left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}\right)=\frac{\mathrm{32}}{\mathrm{3}} \\ $$

Commented bymaxmathsup by imad last updated on 31/May/19

thanks sir.

$${thanks}\:{sir}. \\ $$

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