Question

Exercício de integral

Answer 1

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

${( \frac{dy}{dx} )}^{2} = { (\frac{1}{3} {x}^{ - \frac{1}{3} } ) }^{2} = \frac{1}{9} {x}^{ - \frac{2}{3} }$

$s = \int\limits^b_a\sqrt{1 + {( \frac{dy}{dx} )}^{2} }dx$

$s = \int\limits^2_0 \sqrt{1 + \frac{1}{9} {x}^{ - \frac{2}{3} } } dx \\ s = \int\limits^2_0 \sqrt{1 + \frac{1}{9 {x}^{ \frac{2}{3} } } } dx$

$s = \int\limits^2_0 \sqrt{1 + \frac{1}{ { ({3x}^{ \frac{1}{3} }) }^{2} } }dx$

$z = 3 {x}^{ \frac{1}{3} } \\ {z}^{3} = 27x \\ x = \frac{ {z}^{3} }{27} \\ dx = \frac{3 {z}^{2} }{27}dz = \frac{ {z}^{2} }{9}$

$\int \sqrt{1 + \frac{1}{ {(3 {x}^{ \frac{1}{3} } )}^{2} } } dx \\ \int \sqrt{1 + \frac{1}{ {z}^{2} } } \: \frac{ {z}^{2} }{9}dz$

$\int \sqrt{ \frac{ {z}^{2} + 1 }{ {z}^{2} } } . \frac{ {z}^{2} }{9}dz \\ \int \frac{ \sqrt{ {z}^{2} + 1 } }{z}. \frac{ {z}^{2} }{9}dz$

$\frac{1}{9} \int z \sqrt{1 + {z}^{2} }dz = \frac{1}{27} {(1 + {z}^{2}) }^{ \frac{3}{2} } \\ \frac{1}{27} {(1 + 9 {x}^{ \frac{2}{3} } )}^{ \frac{3}{2} }$

Substituindo os limites de integração temos:

$\frac{1}{27} {(1 + 9. {0}^{ \frac{2}{3} } )}^{ \frac{3}{2} } = \frac{1}{27}.{1}^{ \frac{3}{2} } = \frac{1}{27} \\ = 0,03$

$\frac{1}{27} {(1 + {2}^{ \frac{2}{3} } )}^{ \frac{3}{2} } = 0,15$

Subtraindo temos

$0,15 - 0,03 = 0,12$

Portanto

$\int\limits^2_0 \sqrt{1 + \frac{1}{9 {x}^{ \frac{2}{3} } } } dx = 0,12$