Question

calcular as areas do plano X 0 Y delimitada pelos pares de curvas y^2=4x e 2x-4=y
alguem me ajude por favor a resposta esta dando 19/3 u.a

Answer 1

$\Bmatrix {y^2=4x \\y=2x-4 \end$

encontrando o valor do intervalo:

$y=2x-4\\\\ \boxed{\boxed{x= \frac{y+4}{2} }}\\\\:::::::::::::::::::::\\\\ y^2=4x\\\\y^2=4* \frac{y+4}{2} \\\\y^2=2y+8\\\\y^2-2y-8=0\to \boxed{\boxed{y=-2 \;, y=4}}$

nesse intervalo y²=4x limita pela esquerda e  y=2x-4 limita pela direita

$\int\limits^{4}_{-2} { \frac{y+4}{2}-\frac{y^2}{4} } \, dx \\\\ \int\limits^{4}_{-2} { \frac{y}{2}+ 2 -\frac{y^2}{4} } \, dx = \left[ \frac{y^2}{4}+2y- \frac{y^3}{12} \right]^4_2 = \\\\ = \left( \frac{4^2}{4} + 2*4-\frac{4^3}{12} \right)-\left( \frac{(-2)^2}{4} + 2*(-2)-\frac{(-2)^3}{12} \right)=9 \; u.a$