Question

calcular longitud de curva​

Answer 1

Respuesta:

Explicación paso a paso:

$y=\frac{x^4}{8}+\frac{1}{4x^2}\\\\y'=\frac{x^3}{2}-\frac{1}{2x^3}$

$y'=\frac{x^6-1}{2x^3}$

$L_c=\int\limits^a_b {\sqrt{1+[f'(x)]^2}} \ dx\\\\L_c=\int\limits^D_A {\sqrt{1+(y')^2} \ dx$

$L_c=\int\limits^D_A {\sqrt{\frac{4x^6}{4x^6}+(\frac{x^6-1}{2x^3})^2} } \ dx\\\\L_c=\int\limits^D_A {\sqrt{\frac{4x^6}{4x^6}+\frac{x^{12}-2x^6+1}{4x^6}} } \ dx\\\\L_c=\int\limits^D_A {\sqrt{\frac{x^{12}+2x^6+1}{4x^6}} } \ dx\\\\L_c=\int\limits^D_A {\sqrt{\frac{(x^6+1)^2}{4x^6}} } \ dx\\\\L_c=\int\limits^D_A {\frac{x^6+1}{2x^3}} \ dx\\\\L_c=\frac{x^6-2}{8x^2}|_A^D$

$L_c=\frac{D^6-2}{8D^2}-\frac{A^2-2}{8A^2}$