Question Number 102169 by bemath last updated on 07/Jul/20

∫_0 ^1 x^(3/2)  (√(1−x)) dx ?

Answered by bemath last updated on 07/Jul/20

Answered by john santu last updated on 07/Jul/20

I=∫_0 ^1 x^(3/2)  (√(1−x)) dx   replace 1−x by x   I=∫_0 ^1  (1−x)^(3/2)  (√x) dx   ⇒2I = ∫_0 ^1 ( x^(3/2) (√(1−x)) +(1−x)^(3/2) (√(x )) )dx  2I= ∫_0 ^1 (√x) (√(1−x )) (x+1−x) dx   I = (1/2)∫_0 ^1  (√(x−x^2 )) dx   I= (1/2)∫_0 ^1  (√((1/4)−((1/4)−x+x^2 ))) dx  I=(1/2)∫_0 ^1  (√((1/4)−(x−(1/2))^2 )) dx   set x−(1/2) = (1/2)t ⇒I=(1/2)∫_(−1) ^1 (√((1/4)−(1/4)t^2 )) ((1/2)dt)  I = (1/8)∫_(−1) ^1 (√(1−t^2 )) dt = (π/(16)) ⊛  [ ∫_(−1) ^1 (√(1−x^2 )) dx = (1/2)π(1)^2  ]   the area of a half circle with radius 1 ]

Answered by 1549442205 last updated on 08/Jul/20

Putting (1/x)−1=t^2 ⇒−(1/x^2 )dx=2tdt  ⇒x=(1/(1+t^2 )),     (√(1−x))=(t/(√(1+t^2 ))),dx =−((2tdt)/((t^2 +1)^2 ))   F=−2∫((1/(1+t^2 )))^(3/2) .(t^2 /((1+t^2 )^2 (√(1+t^2 ))))dt  =−2∫((t^2 dt)/((1+t^2 )^4 ))=−2∫(((1+t^2 )−1)/((1+t^2 )^4 ))dt=−2∫(dt/((1+t^2 )^3 ))+2∫(dt/((1+t^2 )^4 ))  Applying the current formular:  I_n =∫(dt/((1+t^2 )^n ))=(1/2).(t/((1+t^2 )^(n−1) ))+(1/2).((2n−3)/(2n−2)).I_(n−1)   we get  I_2 =(1/(2(2−1))).(t/((1+t^2 )^(2−1) ))+((2.2−3)/(2.3−2))I_(2−1) =(1/2).(t/(t^2 +1))+(1/2)I_1   =(t/(2(t^2 +1)))+(1/2)∫(dt/(t^2 +1))=(t/(2(t^2 +1)))+(1/2)arctan t+C_1   I_3 =(1/(2(3−1))).(t/((t^2 +1)^(3−1) ))+((2.3−3)/(2.3−2))I_(3−1) =  (1/4)(t/((t^2 +1)^2 ))+(3/4)I_2 =(t/(4(t^2 +1)^2 ))+((3t)/(8(t^2 +1)))+(3/8)arctan t  I_4 =(1/(2(4−1))). (t/((t^2 +1)^(4−1) ))+((2.4−3)/(2.4−2)).I_3 =  =(1/6).(t/((t^2 +1)^3 ))+(5/6).I_3 =(t/(6(t^2 +1)^3 ))+((5t)/(24(t^2 +1)^2 ))+((5t)/(16(t^2 +1)))+(5/(16))arctan t  We get F=2(I_4 −I_3 )=(t/(8(t^2 +1)^3 ))−(t/(12(t^2 +1)^2 ))−(t/(8(t^2 +1)))+((arctan t)/8)  Hence,∫_0 ^1 x^(3/2) (√(1−x))dx=F(∞)−F(0)=(𝛑/(16))−0=(𝛑/(16))

Commented bymathmax by abdo last updated on 07/Jul/20

the functin x→x^(3/2) (√(1−x))is ≥0 on [0,1] ⇒∫_0 ^1 x^(3/2) (√(1−x))dx ≥0 ...!

Commented by1549442205 last updated on 08/Jul/20

Thank you sir.I shall check again later.  Now I corrected

Answered by mathmax by abdo last updated on 07/Jul/20

I =∫_0 ^1  x^(3/2) (√(1−x))dx  changement x =sin^2 θ give  I =∫_0 ^(π/2) sin^3 θ cosθ (2sinθcosθ)dθ  =2 ∫_0 ^(π/2)  sin^4 θ cos^2 θ dθ  we have B(p,q) =2 ∫_0 ^(π/2)  cos^(2p−1) θ sin^(2q−1) θ dθ  2p−1 =4 ⇒p =(5/2)  and 2q−1 =2 ⇒q =(3/2)   ⇒  I =B((5/2),(3/2)) =((Γ((5/2)).Γ((3/2)))/(Γ((5/2)+(3/2)))) =((Γ((5/2))Γ((3/2)))/(Γ(4)))  we have Γ(x+1) =xΓ(x) ⇒Γ((5/2)) =Γ(1+(3/2)) =(3/2)Γ((3/2)) =(3/2)Γ((1/2)+1)  =(3/4)Γ((1/2)) ⇒ I =(((3/4)Γ((1/2))×(1/2)Γ((1/2)))/(3!)) =(3/(8×3!))(Γ((1/2)))^2   Γ(x) =∫_0 ^∞  t^(x−1)  e^(−t)  dt ⇒Γ((1/2)) =∫_0 ^∞  t^(−(1/2))  e^(−t)  dt =∫_0 ^∞  (e^(−t) /(√t))dt =_((√t)=u)   =∫_0 ^∞  (e^(−u^2 ) /u)(2u)du =2 ∫_0 ^∞  e^(−u^2 ) du =2×((√π)/2) ⇒ I =(π/(16))

Answered by Dwaipayan Shikari last updated on 07/Jul/20

∫_0 ^1 (1−x)^(3/2) (√x)dx=∫_0 ^1 (((x−x^2 )^(3/2) )/x)=∫_0 ^1 (1−x)(√(x−x^2 ))=(1/2)∫_0 ^1 (1+1−2x)(√(x−x^2 ))  (1/2)∫_0 ^1 (√(x−x^2 ))+(1/2)∫_0 ^1 (1−2x)(√(x−x^2 ))dx=(1/2)∫_0 ^1 (√((1/4)−(x−(1/2))^2 ))+(1/3)(0)  (1/2)∫_0 ^1 (√(((1/2))^2 −(x−(1/2))^2 ))dx=[((x−(1/2))/4)(√(x−x^2 ))]_0 ^1 −(1/(16))sin^(−1) [((x−(1/2))/(1/2))]_0 ^1   =0−(π/(32))+((3π)/(32))=(π/(16))           {Note: ∫_0 ^1 (1−2x)(√(x−x^2 ))=0