Question 1022073
This is the integral between 0 and 2 of x*sqrt(4-x^2)
let u=4-x^2
du=-2xdx
(-1/2) du=xdx
integral of -(1/2)u^(1/2)
(-1/2)(2/3)u^(3/2)
(-1/3)*(4-x^2)^(3/2) evaluated at 0 and 2.
at 2, (-1/3)(4-x^2)^(3/2)=0.
at 0, -(-1/3)(4-x^2)^(3/2)=8, because 4^(3/2)=-(-8/3)=8/3
Area is 8/3
{{{graph(300,200,-5,5,-5,10,x*sqrt(4-x^2))}}}