Question

encontrar a área da região de x+y=3 e y+x^2=3?

Answer 1

$\mathrm{x+y=3\ \to\ f(x)=3-x}\\ \mathrm{x^2+y=3\ \to\ g(x)=3-x^2}\\\\ \textrm{Pontos de intersec\c{c}\~ao:}\\\\ \mathrm{f(x)=g(x)\ \to\ 3-x=3-x^2}\\ \mathrm{3-x-3+x^2=0\ \to\ x^2-x=0\ \to\ x(x-1)=0}\\ \mathrm{x_1=0\ \| \ x_2-1=0\ \to\ x_2=1}\\ \mathbf{f(x)\cap g(x)=\{x_1,x_2\}=\{0,1\}}$

$\textrm{Calculando\ a\ \'area entre f(x) e g(x):}\\\\ \mathrm{\int\limits_0^1f(x)-g(x)\ dx=\int\limits_0^1(3-x)-(3-x^2)\ dx=}\\\\ \mathrm{=\int\limits_0^13-x-3+x^2\ dx=\int\limits_0^1x^2-x\ dx=}\\\\ \mathrm{=\int\limits_0^1x^2\ dx-\int\limits_0^1 x\ dx=\bigg(\dfrac{x^3}{3}-\dfrac{x^2}{2}\bigg)\mid_0^1\ =}\\\\ \mathrm{=\bigg(\dfrac{1^3}{3}-\dfrac{1^2}{2}\bigg)-\bigg(\dfrac{0^3}{3}-\dfrac{0^2}{2}\bigg)=\dfrac{1}{3}-\dfrac{1}{2}=-\dfrac{1}{6}}$

$\mathrm{A=\bigg|\int\limits_0^1x^2-x\ dx\bigg|=\bigg|-\dfrac{1}{6}\bigg|}\ \to\ \mathbf{A=\dfrac{1}{6}\ u.a.}$