Question

calcule a area do conjunto (x,y) limitado pelas curvas y=x2 e y=x-x2

Answer 1

Boa tarde!

Estabelecendo os limites 'inferior' e 'superior' da integral:
$x^2=x-x^2\\2x^2-x=0\\x(2x-1)=0\\x=0\\2x-1=0\\x=\dfrac{1}{2}$

Portanto, integral para obter a área de 0 a 1/2.

Calculando:
$\displaystyle{\int_{0}^{1/2}\left(x-x^2-x^2\right)\,\mathrm{d}x} \\ \\=\displaystyle{\int_{0}^{1/2}\left(x-2x^2\right)\,dx} \\ \\=\displaystyle{\left.\dfrac{x^2}{2}-\dfrac{2x^3}{3}\right\rvert_{0}^{1/2}} \\ \\=\displaystyle{\dfrac{\left(\dfrac{1}{2}\right)^2}{2}-\dfrac{2\cdot\left(\dfrac{1}{2}\right)^3}{3}} \\ \\=\displaystyle{\dfrac{1}{8}-\dfrac{1}{12}}\\ \\=\boxed{\boxed{\displaystyle{\dfrac{1}{24}}}}$

Espero ter ajudado!