Ache o comprimento de arco do gráfico da equação de A a B.
5 y=\frac{2}{3} x^{2 / 3}; & A\left(1, \frac{2}{3}\right), & B\left(8,8, \frac{8}{3}\right)

Passo 1

O enunciado pede o comprimento de arco do gráfico de
y = \tfrac{2}{3}x^{2/3}
no intervalo de x=1 (ponto A) até x=8 (ponto B).
Para isso vamos usar a fórmula do comprimento de arco de uma função y(x)
S = \int_{x_1}^{x_2} \sqrt{1 + \bigl(y'(x)\bigr)^2}\,\mathrm{d}x.

Passo 2

Calculamos a derivada
y(x) = \tfrac{2}{3}x^{2/3} \quad\Longrightarrow\quad y'(x) = \tfrac{2}{3}\cdot\tfrac{2}{3}x^{-1/3} = \tfrac{4}{9}\,x^{-1/3}.

Passo 3

Montamos o integrando:
1 + \bigl(y'(x)\bigr)^2 = 1 + \Bigl(\tfrac{4}{9}x^{-1/3}\Bigr)^2 = 1 + \tfrac{16}{81}\,x^{-2/3} = \frac{81x^{2/3} + 16}{81x^{2/3}}.
Logo
\sqrt{1 + (y')^2} = \frac{\sqrt{81x^{2/3}+16}}{9\,x^{1/3}},
e
S = \int_{1}^{8} \frac{\sqrt{81x^{2/3}+16}}{9\,x^{1/3}}\,\mathrm{d}x.

Passo 4

Fazemos a troca de variável t = x^{1/3}, então x = t^3 e \mathrm{d}x = 3t^2\,\mathrm{d}t.
Quando x=1, t=1; quando x=8, t=2. Substituindo:
S = \int_{1}^{2} \frac{\sqrt{81t^2+16}}{9\,t}\;\bigl(3t^2\bigr)\,\mathrm{d}t = \frac{1}{3}\int_{1}^{2} t\sqrt{81t^2+16}\,\mathrm{d}t.

Passo 5

Para integrar \displaystyle\int t\sqrt{81t^2+16}\,\mathrm{d}t, usamos u=81t^2+16, \mathrm{d}u=162t\,\mathrm{d}t. Então
t\,\mathrm{d}t = \frac{\mathrm{d}u}{162},
\int t\sqrt{81t^2+16}\,\mathrm{d}t = \int u^{1/2}\,\frac{\mathrm{d}u}{162} = \frac{1}{162}\cdot \frac{u^{3/2}}{\tfrac{3}{2}} = \frac{2}{486}\,u^{3/2} = \frac{1}{243}\,u^{3/2}.
Voltando a t e incluindo o fator \tfrac{1}{3},
S = \frac{1}{3}\cdot \frac{1}{243} \Bigl[(81t^2+16)^{3/2}\Bigr]_{1}^{2} = \frac{1}{729}\Bigl[(81t^2+16)^{3/2}\Bigr]_{1}^{2}.

Passo 6

Avalia em t=2 e t=1:
81\cdot 2^2+16 = 324+16 = 340, \qquad 81\cdot1^2+16 = 97.
Portanto
S = \frac{1}{729}\Bigl(340^{3/2} - 97^{3/2}\Bigr) = \frac{340\sqrt{340} \;-\;97\sqrt{97}}{729}.

Resposta

S = \frac{340^{3/2} - 97^{3/2}}{729} \quad\text{ou}\quad S = \frac{340\sqrt{340} - 97\sqrt{97}}{729}.