Question

Calcule a área da figura gerada pelas funções y1 = -x2 + 6x e y2 = 3x2

Answer 1

$y_1=-x^2+6x;\ y_2=3x^2\\\\ y_1\cap y_2\ \to\ -x^2+6x=3x^2\ \to\ 4x^2-6x=0\ \to\ x=0\ ou\ x=\dfrac{3}{2}\\\\ S=|\int_0^{\frac{3}{2}}y_1-y_2\ dx\ |=\bigg|\bigg(\int-x^2+6x-3x^2\ dx\bigg)\bigg|_{x=0}^{\frac{3}{2}}=\\\\ =\bigg|\bigg(6\int x\ dx-4\int x^2\ dx\bigg)\bigg|_{x=0}^{\frac{3}{2}}=\bigg|\bigg(3x^2-\dfrac{4x^3}{3}\bigg)\bigg|_{x=0}^{\frac{3}{2}}=\\\\ =\bigg|0-\bigg(3\dfrac{9}{4}-\dfrac{4}{3}\dfrac{27}{8}\bigg)\bigg|=\bigg|\dfrac{18}{4}-\dfrac{27}{4}\bigg|=\bigg|\dfrac{-9}{4}\bigg|=\boxed{\dfrac{9}{4}\ u.a.}$