Question Number 100065 by kungmikami last updated on 24/Jun/20

Answered by abdomathmax last updated on 24/Jun/20

B(m,n)=∫_0 ^1  x^(m−1) (1−x)^(n−1)  dx  changement 1−x=t give   B(m,n) =−∫_0 ^1 (1−t)^(m−1) t^(n−1) (−dt) =  ∫_0 ^1  t^(n−1) (1−t)^(m−1)  dt =B(n,m)

Answered by abdomathmax last updated on 24/Jun/20

I =∫_0 ^1  ∫_0 ^x^2   e^(y/x) dy dx ⇒I =∫_0 ^1 A(x)dx  A(x) =∫_0 ^x^2  e^(y/x)  dy =[xe^(y/x) ]_(y=0) ^x^2   =xe^x  ⇒  I = ∫_0 ^1  xe^x  dx =[xe^x ]_0 ^1 −∫_0 ^1  e^x  dx  =e−[e^x ]_0 ^1  =e−(e−1) =1

Commented bykungmikami last updated on 24/Jun/20

thanks man

Commented bymathmax by abdo last updated on 24/Jun/20

sorry A(x) =[xe^(y/x) ]_(y=0) ^x^2   =xe^x −x ⇒  I =∫_0 ^1 xe^x dx−∫_0 ^1  xdx =1−(1/2) ⇒ I =(1/2)

Answered by smridha last updated on 24/Jun/20

7.∫_0 ^𝛑 sin^2 x.(1+cosx)^4 dx  =2^6 ∫_0 ^𝛑 sin^2 ((x/2)).cos^(10) ((x/2))dx  let (x/2)=u we get  =2^7 .∫_0 ^(𝛑/2) sin^2 (u).cos^(10) (u)du  =2^7 .(1/2).((𝚪((3/2))𝚪(((11)/2)))/(𝚪(7)))  =2^6 .(1/(6!)).(1/2)(√𝛑).(9/2).(7/2).(5/2).(3/2).(1/2)(√𝛑)  =((9.7.5.3.1)/(6.5.4.3.2.1))𝛑=((63)/(48))𝛑

Answered by abdomathmax last updated on 24/Jun/20

J =∫_0 ^1  ∫_x ^(2−x)  (x/y)dy dx =∫_0 ^1 A(x)dx  A(x) =∫_x ^(2−x) (x/y)dy =x[lny]_x ^(2−x)  =x[ln(2−x)−lnx]  ⇒J =∫_0 ^1  xln(2−x)dx−∫_0 ^1  xlnx dx but  ∫_0 ^(1 ) xln(2−x)dx =[(x^2 /2)ln(2−x)]_0 ^1 +∫_0 ^1 (x^2 /2)(1/(2−x))dx  =(1/2)∫_0 ^1  (x^2 /(2−x))dx =−(1/2)∫_0 ^1  ((x^2 −4+4)/(x−2))dx  =−(1/2)∫_0 ^1 (x+2)−2∫_0 ^1  (dx/(x−2))  =−(1/2)[(x^2 /2)+2x]_0 ^1 −2[ln∣x−2∣]_0 ^1   =−(1/2)((5/2))−2(−ln2) =2ln2−(5/4)  ∫_0 ^1 xlnxdx =[(x^2 /2)lnx]_0 ^1 −∫_0 ^1  (x^2 /2)(dx/x)  =−(1/2)∫_0 ^1 xdx  =−(1/2)[(x^2 /2)]_0 ^1  =−(1/4) ⇒  J =2ln2−(5/4)+(1/4) ⇒J =2ln2−1

Answered by abdomathmax last updated on 24/Jun/20

we use 2 ∫_0 ^(π/2)  cos^(2m−1) x sin^(2n−1) xdx =B(m,n)=((Γ(m).Γ(n))/(Γ(m+n)))  ∫_0 ^π  sin^2 x(1+cosx)^4  dx  =∫_0 ^(π/(2 )) sin^2 x(1+cosx)^4  dx +∫_(π/2) ^π  sin^2 x(1+cosx)^4  dx  =A +B  A =∫_0 ^(π/2)  sin^2 x Σ_(k=0) ^4  C_4 ^k  cos^k  xdx  =Σ_(k=0) ^4  C_4 ^k  ∫_0 ^(π/2) cos^k x sin^2 x dx  but  2m−1 =k ⇒m=((k+1)/2) and 2n−1 =2 ⇒n =(3/2)  ⇒∫_0 ^(π/2)  cos^k x sin^2 x dx =∫_0 ^(π/2)  cos^(2(((k+1)/2))−1)  sin^(2((3/2))−1) 3dx  =(1/2)B(((k+1)/2),(3/2)) =((Γ(((k+1)/2)).Γ((3/2)))/(Γ(((k+1)/2)+(3/2)))) =((Γ(((k+1)/2))Γ((3/2)))/(Γ(((k+4)/2))))  ⇒A =(1/2)Σ_(k=0) ^4  C_4 ^k  B(((k+1)/2),(3/2))  B =∫_(π/2) ^π  sin^2 x(1+cosx)^4  dx =_(x =(π/2)+t)   =∫_0 ^(π/2)  cos^2 t(1−sint)^4  dt  =∫_0 ^(π/2)  cos^2 t Σ_(k=0) ^4  C_4 ^k  (−1)^(k ) sin^k t dt  =Σ_(k=0) ^4 (−1)^k  C_4 ^k   ∫_0 ^(π/2)  cos^2 t sin^k t dt=....

Answered by smridha last updated on 24/Jun/20

anothet way of 7  2^6 ∫_0 ^𝛑 sin^2 ((x/2))cos^(10) ((x/2))dx  let sin(x/2)=t so we get  =2^7 ∫_0 ^1 t^2 .(1−t^2 )^(5−(1/2)) dt  now let t^2 =u so that  =2^6 ∫_0 ^1 u^(1−(1/2)) (1−u)^(5−(1/2)) du  =2^6 ∫_0 ^1 u^((3/2)−1) (1−u)^(((11)/2)−1) du  =2^6 B((3/2),((11)/2)) or 2^6 ((𝚪((3/2)).Γ(((11)/2)))/(𝚪(7)))=((63)/(48))𝛑

Answered by smridha last updated on 24/Jun/20

8.𝚽=x^3 +y^3 +3xyz−3  ▽𝚽=(3x^2 +3yz)i^� +(3y^2 +3xz)j^� +3xyk^�   unit normal vector(1,2,−1)  n^� =((▽𝚽)/(∣▽𝚽∣))=((−3i^� +9j^� +6k^� )/(√((−3)^2 +9^2 +6^2 )))=(1/(√(14))) .[−i^� +3j^� +2k^� ]

Answered by smridha last updated on 25/Jun/20

9)dr of PQ^→  is (4,−2,1)  now 𝚽=x^2 −y^2 +2z^2   so the directional derivative is  =[▽𝚽]_((1,2,3)) .n^� [where n^�  is the unit vector along PQ^→ ]  =(2i^� −4j^� +12k^� ).(((4i^� −2j^� +k^� )/(√(4^2 +(−2)^2 +1^2 ))))  =(1/(√(21))).(8+8+12)=((28)/(√(21)))