Question Number 105026 by mohammad17 last updated on 25/Jul/20

Answered by john santu last updated on 25/Jul/20

Q−3  (dx/dy) = (y^2 /2)−(1/(2y^2 )) = ((y^4 −1)/(2y^2 ))  length of curve   = ∫_a ^(1.5a) (√(1+((dx/dy))^2 )) dy   = ∫_a ^(1.5a) (√(1+(((y^4 −1)/(2y^2 )))^2 )) dy   = ∫_a ^(1.5a) (√((y^8 +4y^4 −2y^4 +1)/(4y^4 ))) dy  = ∫_a ^(1.5a) (1/(2y^2 )) (√((y^4 +1)^2 )) dy   =∫_a ^(1.5a) ((y^4 +1)/(2y^2 )) dy = [(1/6)y^3 −(1/(2y))]_a ^(1.5a)   (JS ♠⧫)

Commented bymohammad17 last updated on 25/Jul/20

thank you sir

Answered by Dwaipayan Shikari last updated on 25/Jul/20

2)∫_0 ^a (√(4−9x^2 ))dx  =[((3x)/2)(√(4−9x^2 ))]_0 ^a −[(((2)^2 )/2)sin^(−1) ((3x)/2)]_0 ^a   =((3a)/2)(√(4−9a^2 ))−2sin^(−1) ((3a)/2)

Commented bymohammad17 last updated on 25/Jul/20

thank you sir

Answered by john santu last updated on 25/Jul/20

Q−4.(1)  ∫(ax+3a)^(a+3)  dx = (1/a)∫(ax+3a)^(a+3) d(ax+3a)  [ b = ax+3a ]   (1/a)∫ b^(a+3)  db = (b^(a+4) /(a(a+4))) + c   = (((ax+3a)^(a+4) )/(a^2 +4a)) + c

Answered by mathmax by abdo last updated on 25/Jul/20

4) ∫ (ax+3a)^(a+3) dx =∫a^(a+3) (x+3a)^(a+3) dx =a^(a+3)  ∫ (x+3a)^(a+3) dx  =(a^(a+3) /(a+4)) (x+3a)^(a+4)  +c

Commented bybemath last updated on 26/Jul/20

typo sir. (a^(a+3) /(a+4)) (x+3)^(a+4)  + c

Commented bymathmax by abdo last updated on 26/Jul/20

thank you sir

Answered by mathmax by abdo last updated on 25/Jul/20

2) I =∫_0 ^a (√(4−9x^2 ))dx  =2 ∫_0 ^a (√(1−(9/4)x^2 ))dx  we do the changement  (3/2)x =sint ⇒ I =2 ∫_0 ^(arcsin(((3a)/2))) (√(1−sin^2 t))×(2/3) cost dt  =(4/3) ∫_0 ^(arcsin(((3a)/2))) cos^2 t dt =(2/3)∫_0 ^(arcsin(((3a)/2)))  (1+cos(2t))dt  =(2/3) arcsin(((3a)/2)) +(1/3)[sin(2t)]_o ^(arcsin(((3a)/2)))   =(2/3) arcsin(((3a)/2)) +(2/3)[sint (√(1−sin^2 t))]_0 ^(arcsin(((3a)/2)))   =(2/3) arcsin(((3a)/2))+(2/3){((3a)/2)(√(1−((9a^2 )/4)))}  =(2/3) arcsin(((3a)/2)) +a(√(1−((9a^2 )/4)))