Question

ache a area da regiao compreendida pelas curvas x = y^2 e y = x-2

Answer 1

Ola Geovane

x = y² 
x = y + 2

y² = y + 2

y² - y - 2 = 0

delta
d² = 1 + 8 = 9
d = 3

y1 = (1 + 3)/2 = 4/2 = 2,  x1 = y1 + 2 = 4
y2 = (1 - 3)/2 = -2/2 = -1, x2 = y2 + 2 = 1

F(x) = ∫  y² dy = y³/3 + C

F(4) = 64/3 , F(1) = 1/3

área

A = F(4) - F(1) =  64/3 - 1/3 = 63/3 = 21 

.

Answer 2

Pontos de intersecção:

$\mathsf{{y}^{2}-2=y}\\\mathsf{{y}^{2}-y-2=0}\\\mathsf{\Delta=1+8=9}\\\mathsf{y=\dfrac{1\pm3}{2}}\\\mathsf{y_{1}=2~~y_{2}=-1}$

$\mathsf{x={y}^{2}~~x={(-1)}^{2}=1}\\\mathsf{x={y}^{2}~~x={2}^{2}=4}$

$\displaystyle\mathsf{A=-\int\limits_{-1}^{2}({y} ^{2}-(y+2)dy}$

$\displaystyle\mathsf{A=-\int\limits_{-1}^{2}({y} ^{2}-y-2)dy}$

$</p><p>\displaystyle\mathsf{A=-\left[\dfrac{1}{3}.{y}^{3}-\dfrac{1}{2}{y}^{2}-2y\right]_{-1}^{2}}$

$\mathsf{A=-(\dfrac{1}{3}.{2}^{3}-\dfrac{1}{2}.{2}^{2}-2.2-(\dfrac{1}{3}.{(-1)}^{3}-\dfrac{1}{2}.{(-1)}^{2}-2.(-1)))}$

$\mathsf{A=-(\dfrac{8}{3}-2-4+\dfrac{1}{3}+\dfrac{1}{2}-2)}$

$\mathsf{A=-(\dfrac{16-12-24+2+3-12}{6})}$

$\huge\boxed{\boxed{\mathsf{A=\dfrac{9}{2}~u.a}}}$