Question

⚠️Usando integral⚠️, calcule o comprimento do arco da curva dada por: y =x^(3/2)-4 que vai de A até B, conforme a figura a seguir:

Answer 1

       $\boxed{L=\int\limits^a_b {\sqrt{1+(\frac{dy}{dx} )^2} } \, dx }$     Usando a notação de Leibniz.

             $\boxed{y= x^{\frac{3}{2} } - 4}$

$\frac{dy}{dx} = \frac{d (x^{\frac{3}{2} }) }{dx} - \frac{d(0)}{dx}= \frac{d (x^{\frac{3}{2} }) }{dx} -0 = \frac{d (x^{\frac{3}{2} }) }{dx} = \frac{3}{2} .x^{\frac{1}{2} }$

substituindo esse valor na formula do arco.

$L=\int\limits^a_b {\sqrt{1+(\frac{dy}{dx} )^2} } \, dx = \int\limits^a_b {\sqrt{1+(\frac{3}{2} .x^{\frac{1}{2} } })^2 } \, dx = \int\limits^a_b {\sqrt{1+\frac{9}{4} .x} } \, dx$

                                    $\boxed{\int\limits^a_b {\sqrt{1+\frac{9}{4}x } } \, dx }$

substituindo:       u = $1+\frac{9}{4} x$   ,  então du =  $\frac{9}{4} dx$   logo dx =  $\frac{4}{9} .du$.

$\int\limits^a_b {\sqrt{u} } \, \frac{4}{9} du =\frac{4}{9} \int\limits^a_b {\sqrt{u} } \, du= \frac{4}{9} \int\limits^a_b {u^{\frac{1}{2} } } \,du = = \frac{4}{9} \int\limits^a_b {\frac{x^{\frac{1}{2} +1} }{\frac{1}{2} +1} } \, dx = \frac{4}{9} \int\limits^a_b {\frac{x^{\frac{3}{2} } }{\frac{3}{2} } } \, dx = \frac{4}{9} \int\limits^a_b {\frac{2}{3}.x^{\frac{3}{2} } } \, dx = \frac{4}{9} \frac{2}{3} .\int\limits^a_b {x^{\frac{3}{2} } } \, dx = \frac{8}{27} . x^{\frac{3}{2} } |^a_b$

Resposta:L= $\frac{8}{27} . x^{\frac{3}{2} } |^a_b$