Question

calcule o comprimento da curva y=2/3x^3/2 e 0<=x<=1

Answer 1

Fórmula do comprimento da curva $y=f(x)$ no intervalo $[a,\,b]:$

$L=\displaystyle\int_a^b\sqrt{1+\left(\frac{dy}{dx} \right )^{\!\!2}}\,dx$

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Para esta questão, temos

$y=\dfrac{2}{3}\,x^{3/2}$


Derivando,

$\dfrac{dy}{dx}=\dfrac{2}{3}\cdot \dfrac{3}{2}\,x^{(3/2)-1}\\\\\\ \dfrac{dy}{dx}=x^{1/2}$


Elevando ao quadrado,

$\left(\dfrac{dy}{dx} \right )^{\!\!2}=\big(x^{1/2}\big)^{2}\\\\\\ \left(\dfrac{dy}{dx} \right )^{\!\!2}=x$


Substituindo na expressão que fornece o comprimento da curva, temos

$L=\displaystyle\int_0^1\sqrt{1+x}\,dx\\\\\\ =\int_0^1(1+x)^{1/2}\,dx~~~~~~\mathbf{(i)}$


Substituição:

$1+x=u~~\Rightarrow~~dx=du$


Mudando os extremos de integração:

$\text{Quando }x=0~\Rightarrow~u=1\text{Quando }x=1~\Rightarrow~u=2$


Substituindo em $\mathbf{(i)},$ obtemos

$=\displaystyle\int_1^2 u^{1/2}\,du\\\\\\ =\left.\left(\dfrac{u^{(1/2)+1}}{\frac{1}{2}+1} \right )\right|_1^2\\\\\\ =\left.\left(\dfrac{u^{3/2}}{\frac{3}{2}} \right )\right|_1^2\\\\\\ =\left.\left(\dfrac{2}{3}\,u^{3/2} \right)\right|_1^2\\\\\\ =\dfrac{2}{3}\cdot 2^{3/2}-\dfrac{2}{3}\cdot 1^{3/2}\\\\\\ =\dfrac{2}{3}\cdot 2\sqrt{2}-\dfrac{2}{3}\cdot 1\\\\\\ =\dfrac{2}{3}\big(2\sqrt{2}-1\big)\text{ unidades de comprimento.}$


Bons estudos! :-)