Find the exact length of the curve.?

Assuming that you mean x = y^4/8 + 1/(4y^2):

dx/dy = y^3/2 - 1/(2y^3)

So, √(1 + (y^3/2 - 1/(2y^3))^2)

= √(1 + (y^6/4 - 1/2 + 1/(4y^6)))

= √(y^6/4 + 1/2 + 1/(4y^6))

= √(y^3/2 + 1/(2y^3))^2)

= y^3/2 + 1/(2y^3), since y > 0 (with y in [1, 2])

= (1/2)(y^3 + y^(-3)).

Thus, the arc length equals

∫(y = 1 to 2) (1/2)(y^3 + y^(-3)) dy

= (1/2)(y^4/4 - y^(-2)/2) {for y = 1 to 2}

= (1/8)(y^4 - 2/y^2) {for y = 1 to 2}

= 33/16.

I hope this helps!

x = (y⁴/8)+(1/4y²) = (1/8)[(y⁶+2)/y²],

Recall the formula : d (u/v) = v du - u dv)/v²;

dx = (1/8)[y² d(y⁶+2) - (y⁶+2) d(y²)]/y⁴

= {(1/8)[y² (6y⁵+0) - (y⁶+2) (2y)]/y⁴} dy.

= { (1/8)[6y³ - 2y³ - 4y‾³] } dy

= { (1/2)[y³ - y‾³] } dy.

ds = √[dx²+dy²].

Let y = ₑᵐ,

m = ℓn y,

dm = y‾¹ dy,

dy = y dm = ₑᵐ dm .. ... so that

dx = { (1/2)[ₑ³ᵐ - ₑ‾³ᵐ] } ₑᵐ dm

= Sinh(3m) ₑᵐ dm

Arc length, ds² = [dx² +dy²]

= [Sinh²(3m)+1]ₑ²ᵐ dm²

= Cosh²(3m) ₑ²ᵐ dm²;

ds = √[Cosh²(3m) ₑ²ᵐ dm²] = Cosh(3m) ₑᵐ dm.

Length (in 1<y<2) | (1<y<2) = (0 <m <ℓn2) are the limits & ₑˡᶯ² = 2.

= ∫ ds

= ∫Cosh(3m) ₑᵐ dm = ∫⅓ d[Sinh(3m)] ₑᵐ dm = {⅓ [ₑᵐ Sinh(3m)]ˡᶯ²ₒ} - ⅓ ∫ₑᵐ Sinh(3m) dm

= ⅓ {ₑˡᶯ²(ₑ³ˡᶯ²-ₑ‾³ˡᶯ²)/2 - ₑº(ₑº-ₑ‾º)/2} - ⅓ ∫⅓d(-Cosh(3m)) ₑᵐdm ... ∵ ₑ³ˡᶯ² = ₑˡᶯ⁸ = 8 & since 3ℓn2 = ℓn2³ = ℓn8.

= ⅓ {2(8 - 1/8) -]} + ⅓² Cosh(3m)) ₑᵐdm]ˡᶯ²ₒ +⅓²∫ₑᵐ Cosh(3m) dmᶥ͢ same as RHS;

∴ [⅓+⅓²]∫Cosh(3m) ₑᵐ dm = [⅓+⅓²] {16} - [⅓-⅓²] (1/4)] = (4/9)[16- 1/8],

(4/9) ∫Cosh(3m) ₑᵐ dm = (4/9)[16- 1/8] = 127/18,

∴∫Cosh(3m) ₑᵐ dm = 127/18.