Answer to Question #172104 in Calculus for Jasmine

Question

Find the area bounded by y=2; y=√x+2; and y=2-x.

Answer 1

$y=\ \sqrt x+2=>x=\left(y-2\right)^2=y^2-4y+4$

$y=2-x=>x=2-y$

$y^2-4y+4=2-y$

$y^2-3y+2=0$

$\left(y-1\right)\left(y-2\right)=0$

$y=1\ or\ y=2$

$\int_{1}^{2}{2-y-\left(y^2-4y+4\right)\ dy}$

$-2y+\frac{3}{2}y^2-\frac{y^3}{3}$

$-4+\frac{12}{2}-\frac{8}{3}-\left(-2+\frac{3}{2}-\frac{1}{3}\right)=\frac{1}{6}$