Question

Aplicando integrais, calcule a área da região limitada pelas curvas: y=x^2 ; y=x^2/2 e a recta y=x

Answer 1

Resposta:

$\frac{3}{6}$ u.a

Explicação passo-a-passo:

$\int\limits^1_0 {x^{2} -\frac{x^{2} }{2} } \, dx \\\\\int {x^{2} -\frac{x^{2} }{2} dx } \\\\\int {\frac{2x^{2} -x^{2} }{2} dx }\\\\\int {\frac{x^{2} }{2} dx }\\\\ \frac{1}{2} \int {x^{2} dx }\\\\\frac{1}{2}\times \frac{x^{3} }{3} } \int dx }\\\\\frac{x^{3} }{6} \\\\\\substituindo ~os~ limites:\\\\\frac{1^{3} }{6}-\frac{0^{3} }{6}\\\\\frac{1}{6} \\\\\\Agora ~a~ pr\'oxima ~parte ~da~ figura:\\\\\int\limits^2_1 {x-\frac{x^{2} }{2} } \, dx \\\\\int {x}- \int{\frac{x^{2} }{2} } \ dx$

$\frac{x^{2} }{2} -\frac{x^{3} }{3} \\\\ (\frac{2^{2} }{2} -\frac{2^{3} }{3})-(\frac{x^{2} }{2} -\frac{x^{3} }{3}) \\\\\frac{1}{3}$

$\frac{1}{6}+\frac{1}{3}=\frac{1+2}{6} \\\\\frac{3}{6}$