Найти длину дуги кривой y = (1/3X^3/2) + 1, 0 ≤ X ≤ 1

Y = (1/3) x³/²+1 0≤x≤1

L=∫√ (1 + (y') ²dx

y' = ((1/3) x³/²) ' = (1/3) * (3/2) * x¹/²=√x/2 ⇒

L=∫₀¹ (√ (1 + (√x/2) ²) dx=∫₀¹ (√ (1+x/4) dx=∫₀¹ (√ ((x+4) / 4) dx = (1/2) * ∫₀¹√ (x+4) dx.

x+4=v dx=dv ⇒

(1/4) * ∫₀¹ (√v) dv = (1/4) * (2/3) v³/² |₀¹ = (1/3) * (x+4) ³/² |₀¹ = (1/3) * (5√5-8) ≈1,06

Ответ: ≈1,06 ед.